Haritha

Chapter 19 Coordinate Geometry Internal And external Division Of A Straight Line Segment Exercise 19

Important Formulae :

1. Distance of (x, y) from the origin = \(\sqrt{x^2+y^2}\) units.

2. Distance between 2 points P(x₁,y₁) and Q(x₂,y₂)

= \(\overline{\mathrm{PQ}}=\sqrt{\left(\mathrm{x}_1-\mathrm{x}_2\right)^2+\left(\mathrm{y}_1-\mathrm{y}_2\right)^2} \text { units. }\)

3. If two points A(x₁,y₁) and B(x₂,y₂) are internally divided by point P(x, y) in ratio mn, then the co-ordinates of P are

P = \(\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}\right)\)

4. If two points A(x₁,y₁) and B(x₂,y₂) are externally divided by point P(x, y) in ratio mn, then the co-ordinates of P are

P = \(\left(\frac{m x_2-n x_1}{m-n}, \frac{m y_2-n y_1}{m-n}\right)\)

Question 1. Find the co-ordinate of the point which divides the line segment joining two points in the given ratio for the following:

Read and Learn More WBBSE Solutions For Class 9 Maths

1. (6, -14) and (-8, 10) in the ratio 3 4 internally

Solution: The coordinates of the point

\(\begin{aligned}
& =\left\{\frac{3 \times(-8)+4 \times 6}{3+4}, \frac{3 \times 10+4(-14)}{3+4}\right\} \\
& =\left(\frac{-24+24}{7}, \frac{30-56}{7}\right) \\
& =\left(0, \frac{-26}{7}\right)
\end{aligned}\)

Wbbse Class 9 Maths Chapter 19 Coordinate Geometry Solutions

∴ \(\left(0, \frac{-26}{7}\right)\)

2. (5, 3) and (-7, -2) in the ratio 2 3 internally

Solution: The coordinates of the point

\(\begin{aligned}
& =\left\{\frac{2 \times(-7)+3 \times 5}{2+3}, \frac{2(-2)+3 \times 3}{2+3}\right\} \\
& =\left(\frac{-14+15}{5}, \frac{-4+9}{5}\right) \\
& =\left(\frac{1}{5}, 1\right)
\end{aligned}\)

 

∴ \(\left(\frac{1}{5}, 1\right)\)

3. (-1, 2) and (4, -5) in the ratio 3: 2 externally

Solution: The coordinates of the point

\(\begin{aligned}
& =\left\{\frac{3 \times 4-2 \times(-1)}{3-2}, \frac{3(-5)-2 \times 2}{3-2}\right\} \\
& =\left(\frac{12+2}{1}, \frac{-15-4}{1}\right) \\
& =(14,-19)
\end{aligned}\)

= (14,-19)

∴ (14,-19)

Wbbse Class 9 Maths Coordinate Geometry Section Formula And Examples

4. (3, 2) and (6, 5) in the ratio externally

Solution: The coordinates of the point

=\(\left\{\frac{2 \times 6-1 \times 3}{2-1}, \frac{2 \times 5-1 \times 2}{2-1}\right\}\)

= \(\left(\frac{12-3}{1}, \frac{10-2}{1}\right)\)

= (9,8)

∴ (9,8)

Question 2. Find the co-ordinates of the mid-point of the line segment joining two points for the following:

1. (5, 4) and (3,-4)

Solution: The co-ordinates of the mid-point of the join of (5, 4) and (3,-4)

\(\begin{aligned}
& =\left(\frac{5+3}{2}, \frac{4-4}{2}\right) \\
& =\left(\frac{8}{2}, 0\right)
\end{aligned}\)

= (4, 0)

∴ (4, 0)

2. (6, 0) and (0, 7)

Solution: The co-ordinates of the midpoint of the join of (6,0) and (0,7)

\(\begin{aligned}
& =\left(\frac{6+0}{2}, \frac{0+7}{2}\right) \\
& =\left(3, \frac{7}{2}\right)
\end{aligned}\)

Wbbse Class 9 Coordinate Geometry Internal And External Division Solutions

Question 3. Let us calculate the ratio in which point (1, 3) divides the line segment joining points (4, 6) and (3, 5).

Solution: Let the required ratio = m:n.

∴ \(1=\frac{m \times 3+n \times 4}{m+n}\)

or, m+n=3m + 4n.
or, m-3m = 4n-n
or, -2m=3n

or, \(\frac{m}{n}=\frac{-3}{2}\)

or, m: n= (-3):2

Question 4. Let us calculate in what ratio is the line segment joining the points (7, 3) and (-9, 6) divided by the y-axis.

Solution: Here the abscissa of the point P = x

∴ Point P point abscissa of (x co-ordinate) = \(\frac{m \times(-9)+n \times 7}{m+n}\)

∴ Point P point is situated on y-axis.  ∴x=0.

∴ \(\frac{-9 m+7 n}{m+n}=0\)

or, -9m+7n = 0
or, -9m=-7n
or, 9m = 7n

or, \(\frac{m}{n}=\frac{7}{9}\)

∴ m:n = 7:9

Class 9 Wbbse Coordinate Geometry Chapter 19 Solved Exercises

Question 5. Prove that when the points A(7, 3), B(9, 6), C(10, 12) and D(8, 9) are joined in the order mentioned then they will form a parallelogram.

Solution: Mid-point of the diagonal AC is E.

\(\begin{aligned}
& =\left(\frac{7+10}{2}, \frac{3+12}{2}\right) \\
& =\left(\frac{17}{2}, \frac{15}{2}\right)
\end{aligned}\)

& the mid-point of the diagonal BD is F = \(\left(\frac{9+8}{2}, \frac{6+9}{2}\right)\)

= \(=\left(\frac{17}{2}, \frac{15}{2}\right)\)

Diagonals of the quadrilateral, AC & BD intersect each other at

= \(=\left(\frac{17}{2}, \frac{15}{2}\right)\)

∴ ABCD is a parallelogram. Proved

Class 9 Wbbse Maths Coordinate Geometry Straight Line Segment Solved Problems

Question 6. If the points (3, 2), (6, 3), (x, y) and (6, 5) when joined in the order mentioned and form a parallelogram, then let us calculate the point (x, y).

Solution:

Given

If the points (3, 2), (6, 3), (x, y) and (6, 5) when joined in the order mentioned and form a parallelogram,

Let A(3,2), B(6,3), C(x,y) & D(6,5) be the four points of the parallelogram ABCD. E and F are the mid-points of the diagonals AC & BD.

∴ E is the mid-point of AC.

∴ Co-ordinate of E= \(\left(\frac{3+x}{2}, \frac{2+y}{2}\right)\)

∵ F is the mid-point of BD.

∴ Co-ordinates of F = \(\left(\frac{6+6}{2}, \frac{3+5}{2}\right)\) = (6,4)

∴ ABCD is a parallelogram.

∴ AC and BD will bisect each other, i.e., E and F will be the same point.

∴ \(\frac{3+x}{2}=6\)

or, 3+ x = 12
or, x = 12-3=9

\(\frac{2+y}{2}=4\)

or, 2+ y = 8
or, y=8-2=6

∴ The required point is (9,6).

Question 7. If ((x₁,y₁), (x₂,y₂), (x₃,y₃)  and (x₄,y₄) points are joined in order to form a parallelogram, then prove that x₁ +x₃ = x₂ +x₄ and y₁ + y₃ = y₂+y₄

Solution:

Given

If ((x₁,y₁), (x₂,y₂), (x₃,y₃)  and (x₄,y₄) points are joined in order to form a parallelogram

Let A((x₁,y₁), B(x₂,y₂), C(x₃,y₃) & D(x₄,y₄) are the four points of the parallelogram ABCD.

∴ E is the mid-point of AC.

Co-ordinates of E = \(\left(\frac{x_1+x_3}{2}, \frac{y_1+y_3}{2}\right)\)

F is the mid-point of BD.

Co-ordinates of F = \(\left(\frac{\mathrm{x}_2+\mathrm{x}_4}{2}, \frac{\mathrm{y}_2+\mathrm{y}_4}{2}\right)\)

∵ ABCD is a parallelogram.
Diagonals AC & BD intersect at a point if E & F are the same points.

∴ \(\frac{x_1+x_3}{2}=\frac{x_2+x_4}{2} \text { or, } x_1+x_3=x_2+x_4\)

& \(\frac{y_1+y_3}{2}=\frac{y_2+y_4}{2} \text { or, } y_1+y_3=y_2+y_4\)

Wbbse Class 9 Maths Chapter 19 Coordinate Geometry Notes

Question 8. The coordinates of vertices A, B, and C of a triangle ABC are (-1, 3), (1, -1) and (5, 1) respectively, let us calculate the length of the median AD.

Solution:

 

Given

The coordinates of vertices A, B, and C of a triangle ABC are (-1, 3), (1, -1) and (5, 1) respectively

Let D is the mid-point of BC.

Co-ordinates of D = \(\left(\frac{1+5}{2}=\frac{-1+1}{2}\right)\) = (3,0)

∴ Length of median AD = \(\sqrt{(-1-3)^2+(3-0)^2} \text { unit }\)

\(\begin{aligned}
& =\sqrt{(-4)^2+(3)^2} \text { unit } \\
& =\sqrt{16+9} \text { unit } \\
& =\sqrt{25} \text { unit } \\
& =5 \text { unit }
\end{aligned}\)

 

Question 9. The coordinates of the vertices of the triangle are (2, 4), (6, -2) and (-4, 2) respectively. Let us find the length of three medians of a triangle.

Solution:

 

Given

The coordinates of the vertices of the triangle are (2, 4), (6, -2) and (-4, 2) respectively.

Let D, E, & F be the mid-points of BC, CA & AB respectively.

∴ Now co-ordinates of

D= \(\begin{aligned}
& D=\left(\frac{6-4}{2}=\frac{-2+2}{2}\right) \\
& =(1,0)
\end{aligned}\)

Co-ordinates of E = \(\left(\frac{2-4}{2}=\frac{-4+2}{2}\right)\) = (-1,-1)

Co-ordinates of F = \(\left(\frac{2+6}{2}=\frac{-4-2}{2}\right)\) = (4,-3)

∴ Length of median BE = \(\sqrt{(2-1)^2+(-4-0)^2} \text { unit }\)

\(\begin{aligned}
& =\sqrt{(1)^2+(-4)^2} \\
& =\sqrt{1+16} \text { unit } \\
& =\sqrt{17} \text { unit }
\end{aligned}\)

Length of median BE = \(\sqrt{(6+1)^2+(-2+1)^2} \text { unit }\)

\(\begin{aligned}
& =\sqrt{(7)^2+(-1)^2} \\
& =\sqrt{50} \text { unit } \\
& =5 \sqrt{2} \text { unit }
\end{aligned}\)

Length of median CF = \(\sqrt{(-4-4)^2+(2+3)^2} \text { unit }\)

\(\begin{aligned}
& =\sqrt{(-8)^2+(5)^2} \text { unit } \\
& =\sqrt{64+25} \text { unit } \\
& =\sqrt{89} \text { unit }
\end{aligned}\)

West Bengal Board Class 9 Coordinate Geometry Internal External Division Solutions

Question 10. The coordinates of the midpoints of sides of a triangle are (4, 3), (-2, 7) and (0, 11). Let us calculate the coordinates of its vertices.

Solution:

Given

The coordinates of the midpoints of sides of a triangle are (4, 3), (-2, 7) and (0, 11).

Let A (x₁,y₁) B (x₂,y₂) & C(x₃,y₃) are three vertices of the ΔABC
& P(4, 3); Q (-2, 7) & R (0, 11) are the three mid-points of the AB, BC & CA respectively.

 

 

\(\frac{x_1+x_2}{2}=4, \text { or, } x_1+x_2=8\) ….(1)

\(\frac{y_1+y_2}{2}=3, \text { or, } y_1+y_2=6\) …(2)

\(\frac{x_2+x_3}{2}=-2, \text { or, } x_2+x_3=-4\) ….(3)

\(\frac{y_2+y_3}{2}=7, \text { or, } y_2+y_3=14\) ….(4)

\(\frac{x_3+x_1}{2}=0, \text { or, } x_3+x_1=0\) ….(5)

\(\frac{y_3+y_1}{2}=11, \text { or, } y_3+y_1=22\) …(6)

Adding (1), (3) & (5), we get 2(x₁+ x₂+x₃)= 4

or, x₁+ x₂+x₃ = 4/2

or, x₁+ x₂+x₃ = 2 ….(7)

(6) – (1) => x₃ =-6
(6) – (3) => x₁= 6
(6) – (5) => x = 2

Again, by adding (2), (4) & (5), we get
2(y₁+ y₂+y₃)= 42

∴ y₁+ y₂+y₃ = 21 …… (8)
(8) – (2) => y₃= 15
(8) – (4) => y₁ = 7
(8) – (4) => y₂ = -1

∴ The coordinates of the vertices of the triangle are (6, 7) (2, -1), and (-6, 15).

Question 11. Multiple choice questions

1. The mid-point of line segment joining two points (1, 2m), and (-1 + 2m, 21 – 2m) is

1. (1, m)
2. (1, -m)
3. (m, -1)
4. (m, 1)

Solution; Mid-point

= \(\left(\frac{1-1+2 m}{2}, \frac{2 m+21-2 m}{2}\right)\) = (m, 1)

∴ 4. (m + 1)

Wbbse Class 9 Coordinate Geometry Internal And External Division Important Questions

2. The abscissa at point P which divides the line segment joining two points A(1, 5) and B(-4, 7) internally in the ratio 2:3 is

1. -1
2. 11
3. 1
4. -11

Solution:

Co-ordinates of P = \(\left(\frac{2 \times(-4)+3 \times 1}{2+3}, \frac{2 \times 7+3 \times 5}{2+3}\right)\)

= \(\left(-1, \frac{29}{5}\right)\)

∴ The abscissa is -1.

∴ – 1

3. The coordinates of the end points of a diameter of a circle are (7, 9) and (-1,-3). The coordinates of the centre of the circle is

1. (3, 3)
2. (4, 6)
3. (3, -3)
4. (4, -6)

Solution: Co-ordinates of the centre
\(=\left(\frac{7-1}{2}, \frac{9-3}{2}\right)\)

= (3, 3)

∴ (3, 3)

4. A point which divides the line segment joining two points (2,-5) and (-3,-2) externally in the ratio 4: 3. The ordinate of the point is

1. -18
2. -7
3. 18
4. 7

Solution: The coordinates of the point

=\(\left(\frac{4 \times(-3)-3 \times 2}{4-3}, \frac{4 \times(-2)-3 \times(-5)}{4-3}\right)\)

= (-18, 7)
∴ The ordinate of the point is 7.

4. 7

5. If the points P(1, 2), Q(4, 6), R(5, 7) and S(x, y) are the vertices of a parallelogram PQRS, then

1. x = 2, y = 4
2. x = 3, y=4
3. x = 2, y = 3
4. x = 2, y = 5

Solution: The mid-point of the diagonal PR is

= \(\left(\frac{1+5}{2}, \frac{2+7}{2}\right)=\left(3, \frac{9}{2}\right)\)

= \(\left(3, \frac{9}{2}\right)\)

The mid-point of the diagonal QS = \(\left(\frac{4+x}{2}, \frac{6+y}{2}\right)\)

∴ As mid-points PR & QS are the same point.

∴ \(\frac{4+x}{2}=3\)

or, 4+ x = 6
or, x = 2

or, \(\frac{6+y}{2}=\frac{9}{2}\)

or, 12+2y= 18
or, 2y= 18-12

or, y = 6/2=3

∴ 3. x = 2, y = 3

Question 12. Short answer type questions:

1. C is the centre of a circle and AB is its diameter; the coordinates of A and C are (6, -7) and (5, -2). Let us calculate the coordinates of B.

Solution:

Given

C is the centre of a circle and AB is its diameter; the coordinates of A and C are (6, -7) and (5, -2).

Let the co-ordinate of B = (x, y)

∴ The mid-point of AB = \(\left(\frac{6+x}{2}, \frac{-7+y}{2}\right)\)

∴As centre is the mid-point of the diameter.

∴ \(\frac{6+x}{2}=5[\latex]

or, 6+x=10
or, x = 4

and [latex]\frac{-7+y}{2}=-2\)

or, – 7+ y = -4
or, y = -4+7
or, y = 3

∴ The coordinates of B = (4,3).

Wbbse Class 9 Maths Coordinate Geometry Chapter 19

2. The points P and Q lie on 1st and 3rd quadrants respectively. The distance of the two points from the x-axis and y-axis is 6 units and 4 units respectively. Let us write the coordinates of the mid-point of line segment PQ.

Solution:

Given

The points P and Q lie on 1st and 3rd quadrants respectively. The distance of the two points from the x-axis and y-axis is 6 units and 4 units respectively.

As P lies in 1st quadrant
∴ The abscissa & the ordinate both are positive.
∴ The coordinates of P = (4, 6).

Again, Q lies in the 3rd quadrant.
∴ The abscissa & the ordinate both are negative.

∴ The coordinates of Q = (-4, -6).

∴ The mid point of PQ is = \(\left(\frac{4-4}{2}, \frac{6-6}{2}\right)=(0,0) .\)

= (0, 0).

3. points A and B lie in the 2nd and 4th quadrants respectively and the distance of each point from x-axis and y-axis are 8 units and 6 units respectively. Let us write the coordinate of the mid-point of line segment AB.

Solution:

Given

points A and B lie in the 2nd and 4th quadrants respectively and the distance of each point from x-axis and y-axis are 8 units and 6 units respectively.

As lies in the 2nd quadrant.
∴ The abscissa is negative, but the ordinate is positive.
∴ The coordinates of A = (-6, 8).

Again, B lies in the 4th quadrant.
∴ The abscissa is positive, but the ordinate is negative.

∴ The coordinates of B = (6, -8).

∴ The mid-point of A and B is \(\left(\frac{-6+6}{2}, \frac{8-8}{2}\right)=(0,0)\)

= (0, 0).

4. The point Plies on the segment AB and AP = PB; the coordinates of A and B are (3,-4) and (-5, 2) respectively. Let us write the coordinates of point P.

Solution:

Given

The point Plies on the segment AB and AP = PB; the coordinates of A and B are (3,-4) and (-5, 2) respectively.

As P lies on AB and AP = PB
∴ P is the mid-point of AB.

∴ The co-ordinates of P = \(\left(\frac{3-5}{2}, \frac{-4+2}{2}\right)\) =(-1,-1).

Wbbse 9th Class Maths Coordinate Geometry Internal And External Division Step By Step Solutions

5. The sides of rectangle ABCD are parallel to the coordinate axes. The coordinates of B and D are (7, 3) and (2, 6). Let us write the coordinates of A and C and the mid-point of diagonal AC.

Solution:

Given

The sides of rectangle ABCD are parallel to the coordinate axes. The coordinates of B and D are (7, 3) and (2, 6).

ABCD is a rectangle, AB II x-axis & AD II y-axis.
∴The coordinates of A = (2, 3).
∴ The coordinates of C

∴ The midpoint of AC = \(\left(\frac{2+7}{2}, \frac{3+6}{2}\right)=\left(\frac{9}{2}, \frac{9}{2}\right)\)

 

 

Chapter 20 Coordinate Geometry Area Of Triangular Region Exercise 20

Important Formulae :

1. Co-ordinates of the mid-point of A(x₁,y₁) and B(x₂,y₂) = \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)

2. In ΔABC, if the coordinates of the vertices are: A(x₁,y₁), B(x₂,y₂), and C(x₃,y₃) then the co-ordinates of its centroid are:

= \(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\)

3. In ΔABC if the coordinates of the vertices are: A(x₁,y₁), B(x₂,y₂), and C(x₃,y₃) then the area of the triangle ABC will be

= \(\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\)

4. Three points will be collinear if the area of the triangle formed by them is zero. (v) Area of the quadrilateral formed by the points (x₁,y₁), (x₂,y₂),(x₃,y₃), and (x₄,y₄)

\(\frac{1}{2}\left[\left(x_1 y_2+x_2 y_3+x_3 y_4+x_4 y_1\right)-\left(y_1 x_2+y_2 x_3+y_3 x_4+y_4 x_1\right)\right] \text { sq.units. }\)

Question 1. Find the area of a triangular region with the vertices given below:

Read and Learn More WBBSE Solutions For Class 9 Maths

1. (2,-2), (4, 2), and (-1, 3)

Solution: (2,-2), (4, 2), and (-1,3)

\(\begin{aligned}
& =\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\} \\
& =\frac{1}{2}\{2(2-3)+4(3+2)-1(-2-2)\} \text { sq. unit } \\
& =\frac{1}{2}(-2+20+4) \text { sq. unit } \\
& =\frac{1}{2} \times 22 \text { sq. unit } \\
& =11 \text { sq. unit }
\end{aligned}\)

2. (8, 9) (2, 6) and (9, 2)

Solution: (8, 9) (2, 6) and (9, 2)

\(\begin{aligned}
& =\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\} \\
& =\frac{1}{2}\{8(6-2)+2(2-9)+9(9-6)\} \text { sq. unit } \\
& =\frac{1}{2}(32-14+27) \text { sq. unit } \\
& =\frac{45}{2} \text { sq. unit } \\
& =22 \frac{1}{2} \text { sq. unit }
\end{aligned}\)

Wbbse Class 9 Maths Chapter 20 Coordinate Geometry Area Of Triangular Region Solutions

3. (1, 2), (3, 0) and origin (0,0)

Solution: (1, 2), (3, 0) and origin (0,0)

\(\begin{aligned}
& =\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\} \\
& =\frac{1}{2}\{1(0-0)+3(0-2)+0(2-0)\} \text { sq. unit } \\
& =\frac{1}{2}(-6) \text { sq. unit } \\
& =-3 \text { sq. unit }
\end{aligned}\)

Question 2. Prove that the points (3,-2), (-5, 4), and (-1, 1) are collinear.

Solution: Here the area of the triangle with vertices (3,-2), (-5, 4), and (-1, 1)

\(\begin{aligned}
& =\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\} \\
& =\frac{1}{2}\{3(4-1)-5(1+2)-1(-2-4)\} \text { sq. unit } \\
& =\frac{1}{2}(9-15+6) \text { sq. unit } \\
& =\frac{1}{2} \times 0 \text { sq. unit }=0
\end{aligned}\)

∴ The points are collinear.
∴ (3,-2), (-5, 4), and (-1, 1) are collinear points.

Question 3. Let us write by calculating for what value of K, the points (1, -1), (2, -1), and (K, -1) lie on the same straight line.

Solution: (1,-1), (2, -1), and (K, -1)

\(\begin{aligned}
& =\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\} \\
& =\frac{1}{2}\{1(-1+1)+2(-1+1)+K(-1+1)\} \text { sq. unit }=0 \text { sq. unit }
\end{aligned}\)

∴For any real value of K, the points are collinear.

Question 4. Let us prove that the line joining two points (1, 2) and (-2, -4) passes through the origin.

Solution: The coordinate of the origin is (0, 0).

∴ Area of the triangle formed by the points (1, 2), (-2, -4), and (0, 0) as vertices

\(\begin{aligned}
& =\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\} \\
& =\frac{1}{2}\{1(-4-0)-2(0-2)+0(2+4)\} \text { sq. unit } \\
& =\frac{1}{2}(-4+4+0) \text { sq. unit } \\
& =0
\end{aligned}\)

∴ The three points are collinear.

Wbbse Class 9 Coordinate Geometry Area Of Triangle Exercise Solutions

Question 5. Let us prove that the mid-point of the line segment joining two points (2, 1) and (6, 5) lies on the line joining two points (-4, -5) and (9, 8).

Solution: The mid-point of the line joining (2, 1) and (6, 5) is

= \(=\left(\frac{2+6}{2}, \frac{1+5}{2}\right)=(4,3)\)

∴ Now area of the triangle formed by (4, 3), (-4,-5), and (9,8)

\(\begin{aligned}
& =\frac{1}{2}\{4(-5-8)+(-4)(8-3)+9(3+5)\} \text { sq. unit } \\
& =\frac{1}{2}(-52-20+72) \text { sq. unit } \\
& =\frac{1}{2}(72-72) \text { sq. unit } \\
& =\frac{1}{2} \times 0=0
\end{aligned}\)

∴ The three points are collinear.
∴ The mid-point of the line joining (2, 1) & (6, 5) lies on the line joining two points (-4, -5) & (9,8).

Question 6. Let us find the area of the quadrilateral region formed by the line joining four given points each.

1. (1, 1), (3, 4), (5, -2), and (4, -7)

Solution:

Given

The four points are (1, 1), (3, 4), (5, -2), and (4, -7)

\(\begin{aligned}
& =\frac{1}{2}\left\{\left(x_1 y_2+x_2 y_3+x_3 y_4+x_4 y_1\right)-\left(y_1 x_2+y_2 x_3+y_3 x_4+y_4 x_1\right)\right\} \\
& =\frac{1}{2}[\{1.4+3(-2)+5(-7)+4.1\}-\{1.3+4.5+(-2) \cdot 4+(-7) .1\}] \\
& =\frac{1}{2}\{(4-6-35+4)-(3+20-8-7)\} \text { sq. unit } \\
& =\frac{1}{2}\{(8-41)-(23-15)\} \text { sq. unit } \\
& =\frac{1}{2}(-33-8) \text { sq. unit } \\
& =\frac{1}{2}(-41) \text { sq. unit } \\
& =\frac{41}{2} \text { sq. unit } \\
& =20 \frac{1}{2} \text { sq. unit }
\end{aligned}\)

2. (1, 4), (-2, 1), (2, 3), (3, 3)

Solution:

Given

The four points are = (1, 4), (-2, 1), (2, -3), and (3, 3)

=\(\begin{aligned}
& \left.=\frac{1}{2}\left\{x_1 y_2+x_2 y_3+x_3 y_4+x_4 y_i\right)-\left(y_1 x_2+y_2 x_3+y_3 x_4+y_4 x_1\right)\right\} \\
& =\frac{1}{2}\{(1+6+6+12)-(-8+2-9+3)\} \text { sq. unit } \\
& =\frac{1}{2}\{25-(-12)\} \text { sq. unit } \\
& =\frac{1}{2}(25+12) \text { sq. unit } \\
& =\frac{37}{2} \text { sq. unit } \\
& =18.5 \text { sq. unit }
\end{aligned}\)

Class 9 Wbbse Maths Coordinate Geometry Area Of Triangle Solved Problems

Question 7. The coordinates of three points A, B, and C are (3, 4) (-4, 3), and (8, -6) respectively. Let us find the area of the triangle and the perpendicular length drawn from point A on BC.

Solution:

Given

The coordinates of three points A, B, and C are (3, 4) (-4, 3), and (8, -6) respectively.

Area of the triangle ABC

\(=\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\} \text { sq. unit }\) \(\begin{aligned}
& =\frac{1}{2}\{3(3+6)-4(-6-4)+8(4-3)\} \text { sq. unit } \\
& =\frac{1}{2}(27+40+8) \text { sq. unit } \\
& =\frac{75}{2} \text { sq. unit } \\
& =37.5 \text { sq. unit }
\end{aligned}\)

Length of BC = \(\sqrt{(-4-8)^2+(3+6)^2} \text { unit }\)

\(\begin{aligned}
& =\sqrt{(-12)^2+(9)^2} \text { unit } \\
& =\sqrt{144+81} \text { unit } \\
& =\sqrt{225} \text { unit } \\
& =15 \text { unit }
\end{aligned}\)

∵ Area of ΔABC \(\frac{1}{2}\) x base x height

= \(\frac{1}{2}\) x BC x height

= \(\frac{1}{2}\) x 15x height

= \(\frac{75}{2}\)

Height of the triangle = \(\frac{75 \times 2}{2 \times 15}\) = 5 unit

Wbbse Class 9 Maths Chapter 20 Area Of Triangular Region Notes

Question 8. In triangle ABC, the coordinates of A are (2, 5) and the centroid of a triangle is (-2,1), let us find the coordinates of the mid-point of BC.

Solution:

 

Given

In triangle ABC, the coordinates of A are (2, 5) and the centroid of a triangle is (-2,1),

In ABC the coordinates of vertex A are (2, 5).
Co-ordinates of the centroid of AABC = G (-2, 1) & the co-ordinate of A = (2,5).

∵ We know the centroid intersects the median at G in the ratio 2: 1.

\(therefore \frac{2 \times x+1 \times 2}{2+1}=-2\)

or, \(\frac{2 x+2}{3}=-2\)

or, 2x+2 = -6
or, 2x = -6-2
or, 2x = -8

or, \(x=\frac{-8}{2}=-4\)

and \(\frac{2 \times y+1 \times 5}{2+1}=1\)

or, 2y+5=3
or, 2y=3-5
or, 2y=-2

or, \(y=\frac{-2}{2}=-1\)

∴ The coordinate of the mid-point of BC= (-4, -1).

West Bengal Board Class 9 Coordinate Geometry Area Of Triangle Chapter Solutions

Question 9. The coordinates of vertices of a triangle are (4, -3), (-5, 2), and (x, y); let us find the values of x and y if the centroid of a triangle is at the origin.

Solution:

Given

The coordinates of vertices of a triangle are (4, -3), (-5, 2), and (x, y);

We know the coordinates of the centroid

\(\begin{aligned}
& =\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) \\
& =\left(\frac{4-5+x}{3}, \frac{-3+2+y}{3}\right) \\
& =\left(\frac{-1+x}{3}, \frac{-1+y}{3}\right)
\end{aligned}\)

 

∵ The centroid of the triangle is the origin.
∴ The coordinate of the centroid is (0,0).

∴ \(\frac{-1+x}{3}=0\)

or, -1+x=0
or, x = 1

and \(\frac{-1+y}{3}=0\)

or, – 1+ y = 0
or, y = 1

∴ x=1,y=1

Wbbse Class 9 Area Of Triangular Region Important Questions

Question 10. The vertices at AABC are A(-1,5), B(3,1) and C(5,7). D, E, and F are the midpoints of BC, CA, and AB respectively. Let us find the area of the triangular region ΔDEF and prove that ΔABC=4ΔDEF.

Solution:

 

Given

The vertices at AABC are A(-1,5), B(3,1) and C(5,7). D, E, and F are the midpoints of BC, CA, and AB respectively.

As D, E & F are the mid-points of BC, CA & AB respectively.

∴ Co-ordinates of D = \(\left(\frac{3+5}{2}, \frac{1+7}{2}\right)=(4,4)\)

Co-ordinates of E = \(=\left(\frac{5-1}{2}, \frac{7+5}{2}\right)=(2,6)\)

Co-ordinates of F = \(\left(\frac{-1+3}{2}, \frac{5+1}{2}\right)=(1,3)\)

Area of ΔABC = \(\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\}\)

\(\begin{aligned}
& =\frac{1}{2}\{(-1)(1-7)+3(7-5)+5(5-1)\} \text { sq. unit } \\
& =\frac{1}{2}\{(-1) \times(-6)+3 \times 2+5 \times 4\} \text { sq. unit } \\
& ==\frac{1}{2}(6+6+20) \text { sq. unit } \\
& =\frac{1}{2} \times 32 \text { sq. unit } \\
& =16 \text { sq. unit }
\end{aligned}\)

 

Area of ΔDEF = \(\frac{1}{2}\{4(6-3)+2(3-4)+1(4-6)\} \text { sq. unit }\)

\(\begin{aligned}
& =\frac{1}{2}\{4 \times 3+2(-1)+1(-2)\} \text { sq. unit } \\
& =\frac{1}{2}(12-2-2) \text { sq. unit } \\
& =\frac{1}{2} \times 8 \text { sq. unit } \\
& =4 \text { sq. unit }
\end{aligned}\)

 

Area of ΔABC = 16 sq. unit
Area of 4ΔDEF = 4 x 4 sq. unit
= 16 sq. unit

∴ ΔABC = 4ΔDEF

Wbbse Class 9 Maths Coordinate Geometry Area Of Triangle Chapter 20

Question 11. Multiple choice question

1. The area of the triangular region formed by the three points (0, 4), (0, 0) and (-6, 0) is

1.  24 sq. unit
2. 12 sq. unit
3. 6 sq. unit
4. 8 sq. unit

Solution: (0.4), (0,0), and (-6, 0)

= \(\frac{1}{2}\) {(0(0-0)+0(0-4)+(-6)(-4-0)} sq. unit

= \(\frac{1}{2}\) (0+0+24) sq. unit

= 12 sq. unit

∴ 2. 12 sq. unit

2. The coordinates of the centroid of a triangle formed by the three points (7,-5), (-2, 5), and (4, 6) are

1. (3,-2)
2. (2, 3)
3. (3, 2)
4. (2,-3)

Solution: (7,-5), (-2, 5), and (4, 6)

\(\begin{aligned}
& =\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) \\
& =\left(\frac{7-2+4}{3}, \frac{-5+5+6}{3}\right)
\end{aligned}\)

 

= (3,2)

∴ 3. (3,2)

3. If ABC is a right-angled triangle whose ZABC 90°, co-ordinates of A and C are (0,4) and (3,0) respectively, then the area of triangle ABC is

1. 12 sq. unit
2. 6 sq. unit
3. 24 sq. unit
4. 8 sq. unit

 

 

Solution: In ABC ∠ABC = 90°

Co-ordinates of A = (0, 4)
Co-ordinates of B = (3,0)
Co-ordinates of C = (0, 0)

∴ \(\overline{\mathrm{AB}}\) = 4 unit

and \(\overline{\mathrm{BC}}\) = 3 unit

∴ Area of ΔABC = \(\frac{1}{2} \times \overline{\mathrm{BC}} \times \overline{\mathrm{AB}}\)

= \(\frac{1}{2}\) Χ 3 Χ 4 sq.unit = 6 sq.unit

Wbbse 9th Class Maths Coordinate Geometry Area Of Triangle Step By Step Solutions

4. If (0, 0), (4,-3), and (x, y) are collinear then

1. x = 8, y = -6
2. x = 8, y = 6
3. x = 4, y = -6
4. x = -8, y = -6

Solution: The area of the triangle formed by (0, 0), (4, -3), and (x, y) is zero.

∴ 0(-3- y) + 4 (y-0) + x (0 + 3) = 0
or, 3x + 4y = 0 ….(1)

Now, of the options given, putting in equation (i) only x = 8 and y = 6 satisfies the equation.
Because 3 x 8+4(-6)=24-24=0

∴ 1. x = 8, y = -6

5. If in triangle ABC, the co-ordinates of vertex A are (7, -4) and the centroid of the triangle (1,2), the the co-ordinates of mid-point of BC are

1. (-2,-5)
2. (-2,5)
3. (2,-5)
4. (5,-2)

 

 

Solution: D is the midpoint BC, let the coordinate of D be (x, y).
∴ G (1, 2) cuts median AD in the ratio 2:1

∴ \(\frac{2 \times x+1 \times 7}{2+1}=1\)

or, 2x+7=3
or, 2x = -4
or, 2x = -2

and \(\frac{2 \times y+1(-4)}{2+1}=2\)

or, 2y-4=6
or, 2y = 6+4

or, 2y= 10.
or, y = 5

∴ Co-ordinates of the midpoint of BC = (-2,5)

∴ 2. (-2, 5)

Question 12. Short answer type questions:

1. The coordinates of midpoints of the sides of a triangle ABC are (0,1), (1,1), and (1,0); let us find the coordinates of its centroid.

Solution:

Given

The coordinates of midpoints of the sides of a triangle ABC are (0,1), (1,1), and (1,0);

Let the co-ordinates of A= (x₁,y₁)
Co-ordinates of B = (x₂,y₂)
Co-ordinates of C = (x₃,y₃)

∴ The mid-point of AB = \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)

∴ \(\frac{x_1+x_2}{2}=0\) or, x₁+ x₂ = 0

and \(\frac{y_1+y_2}{2}\) or, y₁ + y₂ =0

The mid-point of BC = \(\left(\frac{\mathrm{x}_2+\mathrm{x}_3}{2}, \frac{\mathrm{y}_2+\mathrm{y}_3}{2}\right)\)

∴ \(\frac{x_2+x_3}{2}=1\) or, x₂+x₃=2

and \(\frac{y_2+y_3}{2}=1\) or, y₂+y₁ = 2

Again, the mid-point of CA = \(\left(\frac{x_3+\dot{x}_1}{2}, \frac{y_3+y_1}{2}\right)\)

∴ \(\frac{x_3+x_1}{2}=1\) or, x₃ +x₁ = 2

and \(\frac{y_3+y_1}{2}=0\) or, y₃+ y₁ =0

Adding (1), (3), and (5),

x₁ + x₂ + x₂ + x₃ + x₃ + x₁ = 0 + 2 + 2

or, 2x₁ + 2x₂ +2x₃ = 4

or, 2(x₁ + x₂+x₃)= 4

or, x₁ + x₂+x₃ \(\frac{4}{2}\) = 2

Again, adding (2), (4), and (6),

\(2 y_1+2 y_2+2 y_3=4\)

 

or, \(2\left(x_1+x_2+x_3\right)=4\)

or, \(y_1+y_2+y_3=\frac{4}{2}\)

or, \(y_1+y_2+y_3=2\)

Co-ordinates of the centroid of ΔABC =

\(=\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)=\left(\frac{2}{3}, \frac{2}{3}\right)\)

∵ \(\left[x_1+x_2+x_3=2 \text { and } y_1+y_2+y_3=2\right]\)

Wbbse Class 9 Maths Area Of Triangle Using Coordinate Geometry Formula And Examples

3. The coordinates of the centroid of a triangle are (6,9) and the two vertices are (15,0) and (0, 10); let us find the coordinates of the third vertex.

Solution:

Given

The coordinates of the centroid of a triangle are (6,9) and the two vertices are (15,0) and (0, 10);

Let the coordinates of the 3rd vertex of the triangle be (x, y).

Co-ordinates of the centroid

\(\begin{aligned}
& =\left(\frac{15+0+x}{3}, \frac{0+10+y}{3}\right) \\
& =\left(\frac{15+x}{3}, \frac{10+y}{3}\right)
\end{aligned}\)

 

By the problem,

\(\frac{15+x}{3}=6 \text { and } \frac{10+y}{3}=9\)

or, 15+ x = 18
or, x = 18-15
or, x = 3

or, 10+ y = 27
or, y = 27-10
or, y = 17

∴ The co-ordinate of the third vertex of the triangle is (3, 17)

3. If the three points (a, 0), (0, b), and (1, 1) are collinear then let us show that \(\frac{1}{a}+\frac{1}{b}\) = 1

Solution:

Given

If the three points (a, 0), (0, b), and (1, 1) are collinear

Area of the triangle formed by (a, 0), (0, b), and (1, 1) = 0

∴ \(\frac{1}{2}{a(b-1)+0(1-0)+1(0-b)}\)= 0

or, ab-a-b=0
or, – a – b = – ab
or, a + b = – ab

\(or, \frac{a}{a b}+\frac{b}{a b}=\frac{a b}{a b}
or, \frac{1}{b}+\frac{1}{a}=1\) \(\frac{1}{a}+\frac{1}{b}=1\)

Class 9 Wbbse Coordinate Geometry Area Of Triangular Region Chapter 20 Solved Exercises

4. Let us calculate the area of the triangular region formed by the three points (1, 4), (1, 2), and (-4, 1).

Solution: The area of the triangle formed by three points (1, 4), (-1, 2), and (-4, 1)

= \(\frac{1}{2}\) {1(2 − 1) + (− 1)(1 − 4) + (−4) (4 – 2)} sq. unit ((2-1)+(-1)(1-4)+(-4)(4-2)}

= \(\frac{1}{2}\) (1+3-8) sq. unit

= \(\frac{1}{2}\) (-4) sq. unit

= \(\frac{1}{2}\) Χ 4 sq. unit = 2 sq. unit

5. Let us write the coordinates of the centroid of a triangle formed by the three points (x-y. y-z), (-x, -y), and (y, z).

Solution: Co-ordinates of the centroid of the triangle formed by (x-y, y-z), (-x, -y), and (y, z)3

\(\begin{aligned}
& =\left(\frac{x-y-x+y}{3}, \frac{y-z-y+z}{3}\right) \\
& =\left(\frac{0}{3}, \frac{0}{3}\right) \\
& =(0,0) .
\end{aligned}\)

 

 

 

Chapter 21 Logarithm Exercise 21

Important Formulae:

1. If \(a^x=M\) (a and M are two natural nos. and a > 0, a ≠ 1 and M > 0) then x is a real number, called the base, and with respect to a, M is called logarithm and is written in form x = log_aM. It is read as ‘s is the logarithm of M to the base a’.

2. If  \(a^x=M\) then x = log_aM and conversely, if x = log_aM then \(a^x=M\)

3. log_aMN = log_aM+log_aN
4. log_aMNP = log_aM+log_aN + log_aP

5. \(\log _a \frac{M}{N}=\log _a M-\log _a N\)

6. \(\log _a M^C=C \log _a M\)

7. log_aM = log_aM x log_ab

8. \(\log _a^1=0\)

9. \(\log _a^a=0\)

10. \(a^{\log _a M}=M\)

11. \(\log _a^b \times \log _b a=1\)

12. \(\log _b a=\frac{1}{\log _a b}\)

13. \(\log _b M=\frac{\log _a M}{\log _a b}\)

14. \(\log _a\left(M_1 M_2 M_3 \ldots . . M_n\right)
=\log _a M_1+\log _a M_2+\log _a M_3 \ldots . \log _a M_n\)

[n = is a positive whole number]

15. If \(\log _a M=\log _a N \text { then } M=N\)

Read and Learn More WBBSE Solutions For Class 9 Maths

Question 1. Let us evaluate :

1. \(\log _{2 \sqrt{3}} 1728\)

Solution: Let, x = \(\log _{2 \sqrt{3}} 1728\)

∴ By definition, we get (2√3)^x = 1728

\(or, (2 \sqrt{3})^x=2^6 \times 3^3
or, (2 \sqrt{3})^x=2^6 \times(\sqrt{3})^6
or, (2 \sqrt{3})^x=(2 \sqrt{3})^6\)

 

∴ X = 6
∴ The value of \(\log _{2 \sqrt{3}} 1728\) is 6.

2. \(\log _{0.01} 0.000001\)

Solution: Let x = \(\log _{0.01} 0.000001\)

∴ By definition, we get (0.01)^x= 0.000001

or, (0.01)^x= (0.01)³
∴ X = 3

∴ The value of \(\log _{0.01} 0.000001\) is 3.

Wbbse Class 9 Maths Chapter 21 Logarithm Solutions

3. \(x=\log _{\sqrt{6}} 216\)

Solution: Let \(x=\log _{\sqrt{6}} 216\)

∴ By definition, we get (√6)^x =216

or, (√6)^x =(6)³
or, (√6)^x=  (√6)⁶

∴ X = 6
∴ The value of \(x=\log _{\sqrt{6}} 216\) is 6.

4. \(\log _4\left(\frac{1}{64}\right)\)

Solution: Let x = \(\log _4\left(\frac{1}{64}\right)\)

\(therefore(4)^x=\frac{1}{64}
or, (4)^x=\frac{1}{4^3}
or, (4)^x=(4)^{-3}\)

∴ X=-3
∴ The value of \(\log _4\left(\frac{1}{64}\right)\) is -3.

Wbbse Class 9 Logarithm Exercise Solutions

Question 2. Let us evaluate:

1. Let us write by calculating, and find its base when the logarithm of 625 is 4.

Solution: Let base be x.

\(\begin{aligned}
& therefore    \log _x 625=4 \\
& therefore    x^4=625
\end{aligned}
or, x^4=5^4
therefore     x=5
\)

 

∴ The required base is 5.

2. Let us write by calculating, and find its base if the logarithm of 5832 is 6. Solve: Let log, 5832 = 6

Solution: Let log_x 58326

∴ Required base is \( x=3 \sqrt{2}\)

Question 3. Let us evaluate:

1. If \(1+\log _{10} a=2 \log _{10} b\) then express a in terms of b.

Solution: \(1+\log _{10} a=2 \log _{10} b\)

Class 9 Wbbse Maths Logarithm Solved Problems

2. \(3+\log _{10} x=2 \log _{10} y\) then express x in terms of y.

Solution: \(3+\log _{10} x=2 \log _{10} y\)

Question 4. Let us evaluate:

1. \(\log _2\left[\log _2\left\{\log _3\left(\log _3 27^3\right)\right\}\right]\)

Solution: \(\log _2\left[\log _2\left\{\log _3\left(\log _3 27^3\right)\right\}\right]\)

 

2. \(\frac{\log \sqrt{27}+\log 8-\log \sqrt{1000}}{\log 1.2}\)

Solution: \(\frac{\log \sqrt{27}+\log 8-\log \sqrt{1000}}{\log 1.2}\)

Wbbse Class 9 Maths Chapter 21 Logarithm Notes

3. \(\log _3 4 \times \log _4 5 \times \log _5 6 \times \log _6 7 \times \log _7 3\)

Solution: \(\log _3 4 \times \log _4 5 \times \log _5 6 \times \log _6 7 \times \log _7 3\)

 

4. \(\log _{10} \frac{384}{5}+\log _{10} \frac{81}{32}+3 \log _{10} \frac{5}{3}+\log _{10} \frac{1}{9}\)

Solution: \(\log _{10} \frac{384}{5}+\log _{10} \frac{81}{32}+3 \log _{10} \frac{5}{3}+\log _{10} \frac{1}{9}\)

\(=\log _{10} 384-\log _{10} 5+\log _{10} 81-\log _{10} 32+3 \log _{10} 5-3 \log _{10} 3+\log _{10} 1-\log _{10} 9\)

 

\(=\log _{10}\left(3 \times 2^7\right)-\log _{10} 5+\log _{10} 3^4-\log _{10} 2^5+3 \log 5-3 \log 3+0-\log _{10} 3^2\)

 

\(=\log _{10} 3+\log _{10} 2^7-\log _{10} 5+4 \log _{10} 3-5 \log _{10} 2+3 \log 5-3 \log 3-2 \log _{10} 3\)

 

\(=5 \log _{10} 3-5 \log _{10} 3+7 \log _{10} 2-5 \log _{10} 2+2 \log _{10} 5\)

 

\(\begin{aligned}
& =2 \log _{10} 2+2 \log _{10} 5 \\
& =2 \log _{10}(2 \times 5) \\
& =2 \log _{10} 10 \\
& =2 \times 1 \\
& =2 \text { Ans. }
\end{aligned}\)

West Bengal Board Class 9 Logarithm Chapter Solutions

Question 5. Let us prove:

1. \(\log \frac{75}{16}-2 \log \frac{5}{9}+\log \frac{32}{243}=\log 2\)

Solution: L.H.S

= \(=\log \frac{75}{16}-2 \log \frac{5}{9}+\log \frac{32}{243}\)

= R.H.S proved.

2. \(\log _{10} 15\left(1+\log _{15} 30\right)+\frac{1}{2} \log _{10} 16\left(1+\log _4 7\right)-\log _{10} 6\left(\log _6 3+1+\log _6 7\right)=2\)

Solution:

 

3. \(\log _2 \log _2 \log _4 256+2 \log _{\sqrt{2}} 2=5\)

Solution:

\(\begin{aligned}
& \text { L.H.S. }=\log _2 \log _2 \log _4 256+2 \log _{\sqrt{2}} 2 \\
& =\log _2 \log _2 \log _4 4^4+2 \log _{\sqrt{2}}(\sqrt{2})^2
\end{aligned}\)

 

\(\log _2 \log _2 4 \log _4 4+2 \times 2 \log _{\sqrt{2}}(\sqrt{2})\)

 

\(\log _2 \log _2 4 \log _4 4+4\)

 

\(\begin{aligned}
& =\log _2 \log _2 4+4 \\
& =\log _2 \log _2 2^2+4 \\
& =\log _2 2 \log _2 2+4 \\
& =\log _2 2+4
\end{aligned}\)

 

= 1 + 4 = 5 R.H.s Proved

 

\(\begin{aligned}
&\begin{aligned}
& {\left[because \log _{\sqrt{2}} \sqrt{2}=1\right]} \\
& {\left[because \log _4 4=1\right]}
\end{aligned}\\
&\left[because \log _2 2=1\right]
\end{aligned}\)

Wbbse Class 9 Logarithm Important Questions

4. \(\log _{x^2} x \times \log _{y^2} y \times \log _{z^2} z=\frac{1}{8}\)

Solution:
L.H.S.= \(\log _{x^2} x \times \log _{y^2} y \times \log _{z^2} z\)

\(\frac{1}{8}\) = R.H.S. Proved

5. \(\log _{b^3} a \times \log _{c^3} b \times \log _{a^3} c=\frac{1}{27}\)

Solution:

L.H.S = \(\log _{b^3} a \times \log _{c^3} b \times \log _{a^3} c\)

= R.H.S (Proved)

6. \(\frac{1}{\log _{x y}(x y z)}+\frac{1}{\log _{y z}(x y z)}+\frac{1}{\log _{z x}(x y z)}=2\)

L.H.S.= \(\frac{1}{\log _{x y}(x y z)}+\frac{1}{\log _{y z}(x y z)}+\frac{1}{\log _{z x}(x y z)}\)

= R.H.S (Proved)

7. \(\log \frac{a^2}{b c}+\log \frac{b^2}{c a}+\log \frac{c^2}{a b}=0\)

Solution:

L.H.S = \(\log \frac{a^2}{b c}+\log \frac{b^2}{c a}+\log \frac{c^2}{a b}\)

R.H.S Proved

Wbbse Class 9 Maths Logarithm Chapter 21

8. \(x^{\log y-\log z} \times y^{\log z-\log x} \times z^{\log x-\log y}=1\)

Solution:

\(\text { Let } P=x^{\log y-\log z} \times y^{\log z-\log x} \times z^{\log x-\log y}=1\)

Taking log on both sides

 

Question 6. Let us prove

1. If \(\log \frac{x+y}{5}=\frac{1}{2}(\log x+\log y)\), then let us show that \(\frac{x}{y}+\frac{y}{x}=23\)

Solution:

Given

\(\log \frac{x+y}{5}=\frac{1}{2}(\log x+\log y)\)

 

⇒ \(\log \frac{x+y}{5}=\frac{1}{2}(\log x+\log y)\)

 

Wbbse 9th Class Maths Logarithm Step By Step Solutions

2. If \(a^4+b^4=14 a^2 b^2\) then let us show that xyz = 1.

Solution: 

Given

\(a^4+b^4=14 a^2 b^2\)

 

Question 7. If \(\frac{\log x}{y-z}=\frac{\log y}{z-x}=\frac{\log z}{x-y}\) then let us show that xyz = 1.

Solution: \(\frac{\log x}{y-z}=\frac{\log y}{z-x}=\frac{\log z}{x-y}\) = k

∴ \(\frac{\log x}{y-z}=k \Rightarrow \log x=k(y-z)\) …….(1)

∴ \(\frac{\log y}{z-x}=k \Rightarrow \log y=k(z-x)\) ….(2)

∴ \(\frac{\log z}{x-y}=k \Rightarrow \log z=k(x-y)\) …(3)

Adding equations (1), (2) and (3) we get

logx+logy + logz = k(y-z) + k(z-x) + k(x-y)
log(xyz) = k(y-Z+Z-x+x-y)
log(xyz) = k.0 = 0

∴ log(xyz) log1
∴ xyz= 1 Proved

Question 8. If, \(\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}\)  then let us show that:

1. \(x^{b+c} \cdot y^{c+a} \cdot z^{a+b}=1\)

2. \(x^{b^2+b c+c^2} \cdot y^{c^2+c a^2+a^2} \cdot z^{a^2+a b+b^2}=1\)

Solution:

\(\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}=k\)

 

logx = k(b-c)
logy = k (c-a)
logz= k ( a – b)

1. \([\log \left(x^{b+c} \cdot y^{c+a} \cdot z^{a+b}\right)=\log x^{b+c}+\log y^{c+a}+\log z^{a+b}\)

 

2. \(\log \left(x^{b^2+b c+c^2} \cdot y^{c^2+c a^2+a^2} \cdot z^{a^2+a b+b^2}\right)\)

= \(\log \dot{x}^{b^2+b c+c^2}+y^{c^2+c a^2+a^2}+z^{a^2+a b+b^2}\)

 

 

Question 9. If a^3-x.b^5x = a^5+x.b^3x, then let us show that \(\)

Solution:

Given

a^3-x.b^5x = a^5+x.b^3x

\(or, \frac{b^{5 x}}{b^{3 x}}=\frac{a^{5+x}}{a^{3-x}}
or, b^{5 x-3 x}=a^{(5+x)-(3-x)}
or, b^{5 x-3 x}=a^{5+x-3+x}
or, b^{2 x}=a^{2+2 x}
or, b^{2 x}=a^2 \cdot a^{2 x}\) \(\begin{aligned}
& \text { or, } \frac{b^{2 x}}{a^{2 x}}=a^2 \\
& \left(\frac{b}{a}\right)^{2 x}=a^2 \\
& \log \left(\frac{b}{a}\right)^{2 x}=\log a^2 \\
& 2 x \log \left(\frac{b}{a}\right)=2 \log a \\
& x \log \frac{b}{a}=\log a
\end{aligned}\)

Wbbse Class 9 Maths Logarithm Laws And Examples

Question 10. Let us evalute:

1. \(\log _8\left[\log _2\left\{\log _3\left(4^x+17\right)\right\}\right]=\frac{1}{3}\)

Solution:

Given

 

2. \(\log _8 x+\log _4 x+\log _2 x=11\)

Solution: \(\log _8 x+\log _4 x+\log _2 x=11\)

\(\text { or, } \frac{1}{\log _x 8}+\frac{1}{\log _x 4}+\frac{1}{\log _x 2}=11\)

 

\(\text { or, } \frac{1}{\log _x 2^3}+\frac{1}{\log _x 2^2}+\frac{1}{\log _x 2}=11\)

 

\(\text { or, } \frac{1}{3 \log _x 2}+\frac{1}{2 \log _x 2}+\frac{1}{\log _x 2}=11\)

 

\(\text { or, } \frac{1}{\log _x 2}\left(\frac{1}{3}+\frac{1}{2}+1\right)=11\)

 

\(\text { or, } \frac{1}{\log _x 2}\left(\frac{2+3+6}{6}\right)=11\)

 

\(\text { or, } \frac{1}{\log _x 2}\left(\frac{11}{6}\right)=11\)

 

\(\text { or, } \frac{1}{\log _x 2}=11 \times \frac{6}{11}\)

 

\(\text { or, } \frac{1}{\log _x 2}=6\)

 

\(or, \log _2 x=6
or, x=2^6
or, x=64 \)

 

Question 11. Let us show that the value of \(\log _{10} 2\) lies between \(\frac{1}{4} \text { and } \frac{1}{3}\)

Solution: Let \(\log _{10} 2=x\)

10^x = 2

Question 12. Multiple choice questions

1. If \(\log _{\sqrt{x}} 0.25=4\) then the value of x will be

1. 0.5
2. 0.25
3. 4
4. 16

Solution: \(\log _{\sqrt{x}} 0.25=4\)

\(therefore(\sqrt{x})^4=0.25
or, \left\{(\sqrt{x})^2\right\}^2=0.25
or, \mathrm{x}^2=0.25
or, x=\sqrt{0.25}
or, x=0.5\)

∴ 1. 0.5

2. If log₁₀(7x-5)= 2, then the value of x will be

1. 10
2. 12
3. 15
4. 18

Solution: log₁₀(7x-5)=2

\(or, 10^2=7 x-5
or, (7 x-5)=100
or, 7 x=105
or, x=15\)

3. If  log₂3 = a, then the value of log₈27 is

1. 3a
2. \(\frac{1}{a}\)
3. 2a
4. a

Solution: log₂3 = a

\(\begin{aligned}
& \log _8 27=\log _8 3^3 \\
& =3 \log _8 3 \\
& =3 \times \frac{1}{\log _3 8} \\
& =3 \times \frac{1}{\log _3 2^3} \\
& =3 \times \frac{1}{3 \log _3 2} \\
& =\log _2 3
\end{aligned}\)

 

log₈27 = a

∴ 4. a

Class 9 Wbbse Logarithm Chapter 21 Solved Exercises

4. If \(\log _{\sqrt{2}} x=a\), then the value of \(\log _{2 \sqrt{2}} x\) is

1. \(\frac{a}{3}\)
2. a
3. 2a
4. 3a

Solution: \(\log _{\sqrt{2}} x=a\)

∴ \(\begin{aligned}
therefore  & \log _{2 \sqrt{2}} x=\frac{1}{\log _x 2 \sqrt{2}} \\
& =\frac{1}{\log _x\left\{(\sqrt{2})^2 \cdot \sqrt{2}\right\}} \\
& =\frac{1}{\log _x(\sqrt{2})^3} \\
& =\frac{1}{3 \log _x \sqrt{2}} \\
& =\frac{1}{3} \log _{\sqrt{2}} x \\
& =\frac{a}{3} \\
therefore  & \log _{2 \sqrt{2}} x=\frac{a}{3}
\end{aligned}\)

∴ 1. \(\frac{a}{3}\)

5. If \(\log _x \frac{1}{3}=-\frac{1}{3}\) then the value of x is

1. 27
2. 9
3. 3
4. \(\frac{1}{27}\)

Solution: \(\log _x \frac{1}{3}=-\frac{1}{3}\)

\(or, \log _x 3^{-1}=-\frac{1}{3}\) \( or, -\log _x 3=-\frac{1}{3}\) \( or, \log _x 3=\frac{1}{3}\) \( or, x^{\frac{1}{3}=3}\) \( or, \left(x^{\frac{1}{3}}\right)^3=3^3\)

X = 27

∴ 1. 27

Question13. Short answer type questions:

1. Let us calculate the value of \(\log _4 \log _4 \log _4 256\)

Solution: \(\log _4 \log _4 \log _4 256\)

=\(\log _4 \log _4 \log _4 4^4\)

\(\begin{aligned}
& =\log _4 \log _4 4 \log _4 4 \\
& =\log _4 \log _4 4 \\
& =\log _4 1
\end{aligned}\)

 

∴\(\log _4 \log _4 \log _4 256\) = 0

2. Let us calculate the value of \(\log \frac{a^n}{b^n}+\log \frac{b^n}{c^n}+\log \frac{c^n}{a^n}\)

Solution: \(\log \frac{a^n}{b^n}+\log \frac{b^n}{c^n}+\log \frac{c^n}{a^n}\)

\(\begin{aligned}
& =\log \left(\frac{a}{b}\right)^n+\log \left(\frac{b}{c}\right)^n+\log \left(\frac{c}{a}\right)^n \\
& =n \log \frac{a}{b}+n \log \frac{b}{c}+n \log \frac{c}{a} \\
& =n(\log a-\log b)+n(\log b-\log c)+n(\log c-\log a) \\
& =n(\log a-\log b+\log b-\log c+\log c-\log a) \\
& =n .0 \\
& =0
\end{aligned}\)

 

3. Let us show that \(a^{\log _a x}=x .\)

Solution: Let  log_a = x

\(\begin{aligned}
& a^u=x \\
& a^{\log _a x}=x
\end{aligned}\)

 

\(\left[because \mathrm{u}=\log _{\mathrm{a}} \mathrm{x}\right]\)

 

4. If \(\log _e 2 \cdot \log _x 25=\log _{10} 16 \cdot \log _9 10 \) then let us calculate the value of x. 

Solution: \(\log _e 2 \cdot \log _x 25=\log _{10} 16 \cdot \log _9 10 \)

 

 

Class IX Maths Solutions WBBSE Chapter 5 Linear Simultaneous Equations Exercise 5.2

Question 1. Let us draw the graph of the following equations and write whether they are solvable or not and if they are solvable, let us write that particular solution or solutions if they have an infinite number of solutions:

1. 2x + 3y-7=0, 3x+2y-8=0

Solution:

Given

2x+3y-7=0 ….(1)
3x+2y-8=0….(2)

From equation ….(1)
2x + 3y-7=0

or, \(x=\frac{7-3 y}{2}\)

Read and Learn More WBBSE Solutions For Class 9 Maths

x	2	8	-1
y	1	-3	3

 

From equation ……(2)

3x + 3y-7=0

or, \(x=\frac{8-2 y}{3}\)

x	2	0	4
y	1	4	-2

 

Wbbse Class 9 Maths Chapter 5.2 Linear Simultaneous Equations Solutions

∴ The equations are solvable.
∴ x = 2;
y =1

 

Class IX Maths Solutions WBBSE

2. 4x-y=11, -8x+2y=-22

Solution:

Given

4x – y = 11 ….(1)

⇒ \(x=\frac{2 y+22}{8}\)

x	4	3	2
y	5	1	-3

 

– 8x + 2y = -22 ….(2)

⇒ \(x=\frac{2 y+22}{8}\)

x	3	5	1
y	1	9	-7

 

Here, \(\frac{4}{-8}=\frac{-1}{2}=\frac{11}{-22}\)

∴ The equations are solvable and have an infinite number of solutions, i.e., x = 2, y = -3; x = 3, y=1; and x = 4, y = 5.

 

Class IX Maths Solutions WBBSE

3. 7x + 3y = 42,

Solution: \(x=\frac{42-3 y}{7}\)

x	6	3	9
y	0	7	-7

 

21x+9y=42

\(x=\frac{42-9 y}{21}\)

 

x	2	-1	5
y	0	7	-7

 

Class 9 Math Chapter 5 WBBSE

Here = \(\frac{7}{21}=\frac{3}{9} \neq \frac{42}{42}\)

∴ There two equations are inconsistent, i.e., they are unsolvable.

 

Wbbse Class 9 Linear Simultaneous Equations Exercise 5.2 Solutions

4. 5x + y = 13, 5x +5y = 12

Solution:

Given

5x + y = 13 ….(1)

\(y=\frac{13-5 x}{1}\)

Class 9 Math Chapter 5 WBBSE

→AB

x	2	0	4
y	3	13	-7

 

5x+5y= 12 ……(2)

⇒ \(y=\frac{12-5 x}{5}\)

→ CD

x	4	0	\(\frac{53}{20}\)
y	-1.6	2.4	\(\frac{-1}{4}\)

 

The equations are solvable.

∴ x = \(\frac{53}{20}\)

y = \(\frac{-1}{4}\)

 

Class 9 Wbbse Maths Linear Equations Exercise 5.2 Solved Problems

Question 2. By comparing the coefficients of the same variables and constants of the following pairs of equations, let us write whether the pair of equations is solvable or not and check them by drawing the graphs of the equations.

1. x+5y=7, x+5y=20

Solution:

Given

x+5y=7, x+5y=20

x + 5y = 7 …(1)

∴ x=7-5y

Class 9 Math Chapter 5 WBBSE

→AB

x	2	-8	7
y	1	3	0

x+5y=20  ….(2)

→CD

Class 9 Wbbse Linear Equations Chapter 5.2 Solved Exercises

x	0	10	-10
y	4	2	6

Here, \(\frac{1}{1}=\frac{5}{5} \neq \frac{7}{20}\)

∴ There two equations are in consistent, i.e., they are unsolvable.

2. 2x + y = 8, y=8-2x

Solution:

Given

2x + y = 8, y=8-2x

2x + y = 8 ….(1)

→AB

x	2	0	-1	4
y	4	8	10	0

-3x+2y=-5 ……(2)

\(y=\frac{3 x-5}{2}\)

Class 9 Maths WB Board

→CD

x	1	3	-1
y	-1	2	-4

 

\(\frac{2}{-3} \neq \frac{1}{2} \neq \frac{8}{-5}\)

∴ The two equations are solvable. x=3, y=2

Wbbse Class 9 Maths Chapter 5.2 Linear Equations Notes

3. 5x + 8y = 14, 15x + 24y = 42

Solution: 5x+8y = 14  …(1)

⇒ \(y=\frac{14-5 x}{8}\)

x	-2	6	2
y	3	-2	5

15x + 24y = 42  ….(2)

⇒ \(y=\frac{42-15 x}{24}\)

→PQ

x	2	-2
y	5	3

 

\(\frac{5}{15}=\frac{8}{24}=\frac{14}{42}\)

Class 9 Maths WB Board

∴ The two equations are solvable and have an infinite number of solutions.

4. 3x+2y= 6, 12x + 8y = 24

Solution: 3x+2y=6…(1)

⇒ \(y=\frac{6-3 x}{2}\)

x	0	-2	6
y	3	6	-6

 

12x + 8y = 24  …(2)

⇒ \(y=\frac{24-12 x}{8}\)

x	0	-2	6
y	3	6	-6

 

\(\frac{3}{12}=\frac{2}{8}=\frac{6}{24}\)

Class 9 Maths WB Board

∴ The two equations are solvable and have infinite number of solutions.

Question 3. Let us determine the relations of ratios of the same variable and constants of the following pairs of equations and write whether the graphs of the equations will be parallel or intersecting or overlapping.

1. 5x + 3y = 11, 2x – 7y = – 12

Solution: 5x+3y= 11  ….(1)
2x-7y=-12 …(2)

\(\frac{5}{2} \neq \frac{3}{-7} \neq \frac{11}{-12}\)

∴ The graphs of the equations will be intersecting.

2. 6x – 8y = 2, 3x-4y = 1

Solution: 6x-8y = 2  …(1)
3x-4y = 1 ….(2)

Here, = \(\frac{6}{3}=\frac{-8}{-4}=\frac{2}{1}\)

∴The graphs of the equations are one straight lines, i.e., they are coincident.

West Bengal Board Class 9 Linear Simultaneous Equations Exercise 5.2 Solutions

3. 8x-7y = 0, 8x – 7y = 56

Solution: 8x-7y=0 ….(1)
8x – 7y = 56 …(2)

\(\frac{8}{8}=\frac{-7}{7} \neq \frac{0}{56}\)

∴ The graphs of the equations are parallel.

4. 4x-3y= 6, 4y – 5x = -7

Solution: 4x-3y=6 ….(1)
– 5x + 4y = -7 …(2)

Here \(\frac{4}{-5} \neq \frac{-3}{4} \neq \frac{6}{-7}\)

∴ The graphs of two equations will intersect each other.

Class 9 Mathematics West Bengal Board

Question 4. Let us solve the following pairs of equations by drawing graphs which are solvable and write 3 solutions which have infinite number of solutions.

1. 4x + 3y = 20, 8x +6y= 40

Solution: 4x + 3y = 20 …(1)
8x + 6y= 40 …(2)

From equation…….(1)
4x + 3y = 20

⇒ \(y=\frac{20-3 y}{4}\)

x	-1	2	5
y	8	4	0

From equation  …(2)
8x+6y= 40

⇒ \(x=\frac{40-6 y}{8}\)

There two equations are solvable, has infinite of solvitions, i.e., (1) x = 5, y = 0;
(2) x=-1, y = 8 (3) x = 2, y = 4.

 

Wbbse Class 9 Linear Equations Exercise 5.2 Important Questions

2. 4x + 3y = 20, 12x + 9y=20

Solution: 4x+3y=20 ….(1)

⇒ \(y=\frac{20-4 x}{3}\)

x	-1	2	5
y	8	4	0

 

12x + 9y=20 ….(2)

⇒ \(y=\frac{20-12 x}{9}\)

Here, \(\frac{4}{12}=\frac{3}{9} \neq \frac{20}{20}\)

∴ These two equations have no general solutions; they are parallel straight lines.

Class 9 Mathematics West Bengal Board

3. 4x + 3y = 20, \(\frac{3 x}{4}-\frac{y}{8}=1\)

Solution: 4x+3y=20   ….(1)

\(\frac{3 x}{4}-\frac{y}{8}=1\)  ….(2)

4x = 20-3y (1)

or, \(x=\frac{20-3 y}{4}\)

→CD

x	5	2	8
y	0	4	-4

 

\(\frac{3 x}{4}-\frac{y}{8}=1\)  ….(2)

or, \(x=\frac{4}{3}\left(1+\frac{y}{8}\right)\)

→ PQ

x	4	1.5	2
y	16	1	4

 

Here, \(\frac{4}{3 / 4}=\frac{3}{-1 / 8}=\frac{20}{1}\)

∴ The two equations are solvable.

Wbbse Class 9 Maths Linear Equations Exercise 5.2

4. p-q=3, \(\frac{p}{3}+\frac{q}{2}=6\)

Solution: p-q=3 …(1)

\(\frac{p}{3}+\frac{q}{2}=6\) ….(2)

 

p=3+q …(1)

→AB

p	3	0	-6
q	0	-3	3

 

Class 9 Mathematics West Bengal Board

\(\frac{p}{3}+\frac{q}{2}=6\)  …(2)

⇒ \(p=3\left(6-\frac{q}{2}\right)\)

→ CD

p	15	18	12
q	2	0	4

Here, \(\frac{1}{1 / 3} \neq \frac{-1}{1 / 2} \neq \frac{3}{6}\)

 

Wbbse 9th Class Maths Linear Simultaneous Equations Step By Step Solutions Exercise 5.2

5. p − q = 3, \(\frac{p}{5}-\frac{q}{5}=3\)

Solution: p-q=3
⇒p=3+q   …(1)

→PQ

p	4	3	1
q	1	0	-2

 

 

\(\frac{p}{5}+\frac{q}{5}=3\)

 

⇒ \(p=5\left(3+\frac{q}{5}\right)\) …(2)

Wbbse Class 9 Maths Graphical Method For Linear Simultaneous Equations Exercise 5.2

→RS

p	20	5	10
q	5	-10	-5

 

Here, \(\frac{1}{1 / 5}=\frac{-1}{-1 / 5} \neq \frac{3}{3}\)

∴ The two equations have no solutions, they are parallel straight lines.
99

6. p-q=3, 8p-8q = 5

Solution: p-q=3 ⇒ p=3+q …(1)

p	3	0	5
q	0	-3	2

 

Class 9 Mathematics West Bengal Board

8p-8q= 5

⇒ \(p=\frac{5+8 q}{8}\)

∴ The two equations have no solutions, they are parallel straight lines.

Question 5. Tathagata has written a linear equation in two variables x + y = 5. I write another linear equation in two variables, so that, the graphs of two equations will be
1. Parallel to each other,
2. Intersecting,
3. Overlapping.

Solution: x + y = 5
1. Equation of parallel straight line is 5x + 5y = 20
2. Equation of intersecting straight line is 2x+5y = 16
3. Equation of overlapping straight line is 2x + 2y = 10

 

Maths WBBSE Class 9 Solutions Chapter 5 Linear Simultaneous Equations Exercise 5.1

Let me frame the simultaneous equations in each of the following cases and see whether they can be solved or not.

Question 1. The sum of my elder sister’s present age and my father’s present age is 55 years. By calculating, I observed that after 16 years, my father’s age will be two times more than my sister’s.

1. Let me draw the graph after framing simultaneous equations.
2. With the help of a graph, let me find out whether the general solution of two equations can be determined.
3. Let me write the present age of my elder sister and father from the graph.

Read and Learn More WBBSE Solutions For Class 9 Maths

Solutions:

Given

The sum of my elder sister’s present age and my father’s present age is 55 years. By calculating, I observed that after 16 years, my father’s age will be two times more than my sister’s.

Let the elder sister’s present age x years and the father’s present age be y years.

According to the conditions,

x + y = 55……(1)

2(x+16) = y + 16 …………….(2)

From equation..(1)

→AB

x	25	35	40
y	30	20	15

 

Wbbse Class 9 Maths Chapter 5.1 Linear Simultaneous Equations Solutions

From equation..(2)

2x-y = -16

→CD

x	-5	0	-8
y	6	16	0

 

General solution x = 13, 4 = 42

∴ The present age of sister (x) = 13 years and the present age of father (y) = 42 years.

 

Wbbse Class 9 Linear Simultaneous Equations Exercise 5.1 Solutions

Question 2. Mita bought 3 pens and 4 pencils at Rs. 42 from the shop of Jadabkaku. I bought 9 pens and 1 dozen pencils at the same rate to give gifts to my friends at the cost of Rs. 126.

1.  Let me draw the graph after forming the simultaneous equations.
2. With the help of a graph, let me find whether the general solution of two equations can be determined.
3. Let me write the price of 1 pen and 1 pencil separately from the graph. Ans. Let the cost price of one pen = Rs. x and the cost price of one pencil = Rs. y.

Solution:

Given

Mita bought 3 pens and 4 pencils at Rs. 42 from the shop of Jadabkaku. I bought 9 pens and 1 dozen pencils at the same rate to give gifts to my friends at the cost of Rs. 126.

∴ Required simultaneous equations are
3x + 4y = 42…..(1)
9x+12y= 126…(2)

From equation ….(1)
3x + 4y = 42

or, \(x=\frac{42-4 y}{3}\)

→AB

x	14	10	6
y	0	3	6

 

Class 9 Wbbse Maths Linear Equations Exercise 5.1 Solved Problems

From equation…………. (2)
9x=126-12y

or, \(x=\frac{126-12 y}{9}\)

→CD

x	10	14	6
y	3	0	6

 

Wbbse Class 9 Maths Chapter 5.1 Linear Equations Notes

2. Infinite number of general solutions x = 10, y = 3 or, x = 6, y = 6

3. price of 1 pen = Rs. 10 & price of 1 pencil = Rs. 3
or, price of 1. pen = Rs. 6 & price of 1 pen = Rs. 6.

 

West Bengal Board Class 9 Linear Simultaneous Equations Exercise 5.1 Solutions

Question 3. Today we shall draw the pictures as we like in our school. For this, I bought 2 art paper and 5 sketch pens at Rs. 16. But Dola has bought 4 art paper and 10 sketch pens of the same rate and from the same shop.

1. Let me form simultaneous equations and draw the graph.
2. Let me see whether the general solution of the equations can be found out from the graph.
3. Let me write whether I can get the price of 1 art paper and 1 sketch pen.

Solution:

Given

Today we shall draw the pictures as we like in our school. For this, I bought 2 art paper and 5 sketch pens at Rs. 16. But Dola has bought 4 art paper and 10 sketch pens of the same rate and from the same shop.

Let the price of art paper = Rs. x & that of one sketch pen = Rs. y.

Class IX Maths Solutions WBBSE

∴ Required equations,
2x+5y= 16

\(x=\frac{16-5 y}{2}\)

From equation….(2)
4x + 10y = 28

or, x = \(x=\frac{28-10 y}{4}\)

2. will have no general solutions

3. No definite value of one art paper & one sketch pen.

Wbbse Class 9 Linear Equations Exercise 5.1 Important Questions

 

Class IX Maths Solutions WBBSE  Chapter 5 Linear Simultaneous Equations Exercise 5.6

Let us solve the following linear equations in two variables by applying the cross-multiplication method :

Question 1. 8x+5y= 11, 3x – 4y = 10

Solution:

Given

8x+5y = 11 …(1)

3x-4y = 10…..(2)

From equation (1) 8x+5y-11=0
From equation (2) 3x-4y-10=0

or,\(\frac{x}{5 \times(-10)-(-11) \times(-4)}=\frac{y}{(-11) \times 3-8(-10)}=\frac{1}{8(-4)-5 \times 3}\)

or,\(\frac{x}{-50-44}=\frac{y}{-33+80}=\frac{1}{-32-15}\)

or,\(\frac{x}{-94}=\frac{y}{47}=\frac{1}{-47}\)

∴ \(\frac{x}{-94}=\frac{1}{-47} \text { and } \frac{y}{47}=\frac{1}{-47}\)

or,\(x=\frac{-94}{-47} \text { and } y=\frac{47}{-47}\)

or, X = 2  and y = -1

∴ y=-1 , x = 2

Read and Learn More WBBSE Solutions For Class 9 Maths

Question 2. 3x4y = 1, 4x = 3y+6

Solution:

Given

3x-4y = 1 ….(1)

4x = 3y+ 6 …(2)

From equation (1) 3x-4y-1-0

From equation (2) 4x-3y-6=0

or, \(\frac{x}{(-4) \times(-6)-(-1)(-3)}=\frac{y}{(-1) \times 4-3(-6)}=\frac{1}{3(-3)-(-4) \times 4}\)

or, \( \quad \frac{x}{24-3}=\frac{y}{-4+18}=\frac{1}{-9+16}\)

or, \(\quad \frac{x}{21}=\frac{y}{14}=\frac{1}{7}\)

∴ \(\frac{x}{21}=\frac{1}{7} \text { and } \frac{y}{14}=\frac{1}{7}\)

or, \(x=\frac{21}{7} \text { and } y=\frac{14}{7}\)

or, x=3 and y = 2

∴ X = 3
y = 2

Class IX Maths Solutions WBBSE

Question 3. 5x + 3y = 11, 2x-7y = – 12

Solution:

Given

5x + 3y = 11….(1)

2x-7y=-12….(2)

From equation (1) 5x + 3y-11=0
From equation (2) 2x-7y+12=0.

∴ \(\frac{x}{3 \times 12-(-11) \times(-7)}=\frac{y}{-11 \times 2-5 \times 12}=\frac{1}{5 \times(-7)-3 \times 2}\)

or, \(\frac{x}{36-77}=\frac{y}{-22-60}=\frac{1}{-35-6}\)

or, \(\frac{x}{-41}=\frac{y}{-82}=\frac{1}{-41}\)

or, \(\frac{x}{41}=\frac{y}{82}=\frac{1}{41}\)

∴ \(\frac{x}{41}=\frac{1}{41} \text { and } \frac{y}{82}=\frac{1}{41}\)

or, \(x=\frac{41}{41} \text { and } y=\frac{82}{41}\)

∴ x= 1 and y = 2

Wbbse Class 9 Maths Chapter 5.6 Linear Simultaneous Equations Solutions

Question 4. 7x-3y-31=0, 9x-5y-41=0

Solution:

Given

7x-3y31= 0 ….(1)

9x-5y-41=0 …..(2)

or, \(\frac{x}{(-3) \times(-41)-(-31)(-5)}=\frac{y}{(-31) \times 9-7 \times(-41)}=\frac{1}{7 \times(-5)-(-3) \times 9}\)

or, \(\frac{x}{123-155}=\frac{y}{-279+287}=\frac{1}{-35+27}\)

or, \(\frac{x}{-32}=\frac{y}{8}=\frac{1}{-8}\)

∴ \(\frac{x}{-32}=\frac{1}{-8} \text { and } \frac{y}{8}=\frac{1}{-8}\)

or,\(x=\frac{-32}{-8} \text { and } y=\frac{8}{-8}\)

∴ X = 4 and y = -1

Class IX Maths Solutions WBBSE

Question 5. \(\frac{x}{6}-\frac{y}{3}=\frac{x}{12}-\frac{2 y}{3}=4\)

Solution:

Given

\(\frac{x}{6}-\frac{y}{3}=4\) …(1)

\(\frac{x}{12}-\frac{2 y}{3}=4\) …(2)

 

From equation (1) \(\frac{x}{6}-\frac{y}{3}=4\)

or, \(\frac{x-2 y}{6}=4\)

or, x-2y = 24…(3)

From equation (2) \(\frac{x}{12}-\frac{2 y}{3}=4\)

or, \(\frac{x-8 y}{12}=4\)

or,X-8y = 48 ….(3)

From equation (1) x-2y-24=0
From equation (4) x-8y-48=0

∴ \(\frac{x}{(-2) \times(-48)-(-24)(-8)}=\frac{y}{(-24) \times 1-1 \times(-48)}=\frac{1}{1 \times(-8)-(-2) \times 1}\)

or,\(\frac{x}{96-192}=\frac{y}{-24+48}=\frac{1}{-8+2}\)

or, \(\frac{x}{-96}=\frac{y}{24}=\frac{1}{-6}\)

∴ \(\frac{x}{-96}=\frac{1}{-6} \text { and } \frac{y}{24}=\frac{1}{-6}\)

or, \(x=\frac{-96}{-6} \text { and } y=\frac{24}{-6}\)

or, x =16, y= -4

∴ x= 6
y = -4

Wbbse Class 9 Linear Simultaneous Equations Exercise 5.6 Solutions

Question 6. \(\frac{x}{5}+\frac{y}{3}=\frac{x}{4}-\frac{y}{3}-\frac{3}{20}=0\)

Solution:

Given

 

\(\frac{x}{5}+\frac{y}{3}=0\) …(1)

\(\frac{x}{4}-\frac{y}{3}-\frac{3}{20}=0\)…(2)

Class IX Maths Solutions WBBSE

From equation (1) \(\frac{x}{5}+\frac{y}{3}=0\)

or, \(\frac{3 x+5 y}{15}=0\)

or, 3x+5y =0 ….(3)

From equation (2) \(\frac{x}{4}-\frac{y}{3}-\frac{3}{20}=0\)

or, \(\frac{15 x-20 y-9}{60}=0\)

or, 15x-20y -9 =0 …(4)

From equation (3) 3x+5y+0=0
From equation (4) 15x-20y-9=0

∴ \(\frac{x}{5 \times(-9)-0 \times(-20)}=\frac{y}{0 \times 15-3 \times(-9)}=\frac{1}{3 \times(-20)-5 \times 15}\)

or, \(\frac{x}{-45-0}=\frac{y}{0+27}=\frac{1}{-60-75}\)

or,\(\frac{x}{-45}=\frac{y}{27}=\frac{1}{-13}\)

∴\( \frac{x}{-45}=\frac{1}{-135} \text { and } \frac{y}{27}=\frac{1}{-135}\)

or,\(x=\frac{-45}{-135} \text { and } y=\frac{27}{-135}\)

or,\(x=\frac{1}{3} \text { and } y=-\frac{1}{5}\)

∴ \(x=\frac{1}{3} \text { and } y=-\frac{1}{5}\)

Class 9 Wbbse Maths Linear Equations Exercise 5.6 Solved Problems

Question 7. \(\frac{x+2}{7}+\frac{y-x}{4}=2 x-8, \quad \frac{2 y-3 x}{3}+2 y=3 x+4\)

Solution:

Given

\(\frac{x+2}{7}+\frac{y-x}{4}=2 x-8\) ….(1)

\(\frac{2 y-3 x}{3}+2 y=3 x+4\) ….(2)

Class IX Maths Solutions WBBSE

From equation(1) \(\frac{x+2}{7}+\frac{y-x}{4}=2 x-8\)

or, \(\frac{4 x+8+7 y-7 x}{28}=2 x-8\)

or, \(\frac{-3 x+7 y+8}{28}=2 x-8\)

or,56-224= -3x + 7y+8
or,56x+3x-7y = 8 +224

or,59x-7y = 232…(3)

From equation (2) \( \frac{2 y-3 x}{3}+2 y=3 x+4 \)

Or, \( \frac{2 y-3 x+6 y}{3}=3 x+4 \)

or, 9x+12=8y-3x

or, 9x + 3x – 8y = – 12
or,12x-8y=-12
or,3x-2y=-3 ….(4)

From equation (3) 59x-7y-232 = 0
From equation (4) 3x-2y+3-0

\( \frac{x}{(-7) \times 3-(-232) \times(-2)}=\frac{y}{(-232) \times 3-59 \times 3}=\frac{1}{59 \times(-2)-(-7) \times 3} \)

Class 9 Mathematics West Bengal Board

or, \( \frac{x}{-21-464}=\frac{y}{-696-177}=\frac{1}{-118+21} \)

or, \( \frac{x}{-485}=\frac{y}{-873}=\frac{1}{-97} \)

\( \frac{x}{-485}=\frac{1}{-97} \text { and } \frac{y}{-873}=\frac{1}{-97} \)

or, \( x=\frac{-485}{-97} \text { and } y=\frac{-873}{-97} \)

or, x = 5 and y = 9
x = 5,y=9

Wbbse Class 9 Maths Chapter 5.6 Linear Equations Notes

Question 8. x+5y = 36, \( \frac{x+y}{x-y}=\frac{5}{3} \)

Solution:

Given

x+5y = 36 ….(1)

\( \frac{x+y}{x-y}=\frac{5}{3} \) …(2)

 

or, 5x-5y=3x+3y
or, 5x-3x-5y-3y=0
or, 2x – 8y = 0
or,X-4y=0 …(4)

From equation (1) x+5y-36=0
From equation (3) x-4y+0=0

\( \frac{x}{5 \times 0-(-36) \times(-4)}=\frac{y}{(-36) \times 1-1 \times 0}=\frac{1}{1 \times(-4)-5 \times 1} \)

 

or, \( \frac{x}{0-144}=\frac{y}{-36-0}=\frac{1}{-4-5} \)

or, \( \frac{x}{-144}=\frac{y}{-36}=\frac{1}{-9} \)

or, \( \frac{x}{144}=\frac{y}{36}=\frac{1}{9} \)

\( \frac{x}{144}=\frac{1}{9} \text { and } \frac{y}{36}=\frac{1}{9} \)

Class 9 Mathematics West Bengal Board

or, \( x=\frac{144}{9} \text { and } y=\frac{36}{9} \)

or,  x=16 and y = 4

Question 9. 13x12y+15=0, 8x-7y = 0

Solution:

Given

13x-12y+15=0 …(1)

8x-7y+0=0…(2)

\( \frac{x}{-12 \times 0-15 \times(-7)}=\frac{y}{15 \times 8-13 \times 0}=\frac{1}{13 \times(-7)-(-12) \times 8} \)

 

Or, \( \frac{x}{0+105}=\frac{y}{120-0}=\frac{1}{-91+96} \)

Or, \( \frac{x}{105}=\frac{y}{120}=\frac{1}{5} \)

\( \frac{x}{105}=\frac{1}{5} \text { and } \frac{y}{120}=\frac{1}{5} \) \( x=\frac{105}{5} \text { and } y=\frac{120}{5} \)

Or, x=21 and y = 24
x = 21,
y = 24

West Bengal Board Class 9 Linear Simultaneous Equations Exercise 5.6 Solutions

Question 10. x + y = 2b, x – y = 2a

Solution:

Given

x + y = 2b …(1)

x – y = 2a …(2)

From equation (1) x+y-2b=0
From equation (2) xy- 2a =0

\( \frac{x}{1 \times(-2 a)-(2 b)(-1)}=\frac{y}{(-2 b) \times 1-1 \times(-2 a)}=\frac{1}{1 \times(-1)-1 \times 1} \)

Class 9 Mathematics West Bengal Board

Or, \( \frac{x}{-2 a-2 b}=\frac{y}{-2 b+2 a}=\frac{1}{-1-1} \)

Or, \( \frac{x}{-2(a+b)}=\frac{y}{-2(b-a)}=\frac{1}{-2} \)

Or, \( \frac{x}{2(a+b)}=\frac{y}{2(b-a)}=\frac{1}{2} \)

\( \frac{x}{2(a-b)}=\frac{1}{2} \text { and } \frac{y}{2(b-a)}=\frac{1}{2} \)

 

Or,\( x=\frac{2(a-b)}{2} \text { and } y=\frac{2(b-a}{2} \)

X= a-b, y= b-a

Question 11. x – y = 2a, ax + by = a² + b²

Solution:

Given

x – y = 2a …(1)
ax + by = a² + b² …(2)

From equation (1) x-y-2a=0
From equation (2) ax + by-(a²+b²)=0

\( \frac{x}{(-1)\left\{-\left(a^2+b^2\right)\right\}-(-2 a) \times b}=\frac{y}{-2 a \times a-1 \times\left\{-\left(a^2+b^2\right)\right\}}=\frac{1}{1 \times b-(-1) \times a} \)

Class 9 Maths WB Board

Or, \( \frac{x}{a^2+b^2+2 a b}=\frac{y}{-2 a^2+a^2+b^2}=\frac{1}{b+a} \)

Or, \( \frac{x}{(a+b)^2}=\frac{y}{b^2-a^2}=\frac{1}{b+a} \)

Or, \( \frac{x}{(a+b)(a+b)}=\frac{y}{(b+a)(b-a)}=\frac{1}{b+a} \)

\( \frac{x}{(a+b)(a+b)}=\frac{1}{b+a} \text { and } \frac{y}{(b+a)(b-a)}=\frac{1}{b+a} \)

Or, \( x=\frac{(a+b)(a+b)}{(a+b)} \text { and } y=\frac{(b+a)(b-a)}{(b+a)} \)

or, x = a + b and y = b-a
x = a + b
y = b-a

Wbbse Class 9 Linear Equations Exercise 5.6 Important Questions

Question 12. \( \frac{x}{a}+\frac{y}{b}=2 \), ax-by=a²-b²

Solution:

Given

\( \frac{x}{a}+\frac{y}{b}=2 \) ….(1)
ax-by=a²-b² …(2)

From equation (1) \( \frac{x}{a}+\frac{y}{b}=2 \)

Or, \( \frac{b x+a y}{a b}=2 \)

Or, bx+ay = 2ab ….(3)

From equation (3) bx+ay-2ab = 0
From equation (4) ax + by – (a² + b²) = 0

\( \frac{x}{a\left(b^2-a^2\right)-(-2 a b) \times x(-b)}=\frac{y}{(-2 a b) \times a-b\left(b^2-a^2\right)}=\frac{1}{b \times(-b)-a \times a} \)

 

Or, \( \frac{x}{a b^2-a^3-2 a b^2}=\frac{y}{-2 a^2 b-b^3+a^2 b}=\frac{1}{-b^2-a^2} \)

Or, \( \frac{x}{-a^3-a b^2}=\frac{y}{-a^2 b-b^3}=\frac{1}{-a^2-b^2} \)

Or, \( \frac{x}{-a\left(a^2+b^2\right)}=\frac{y}{-b\left(a^2+b^2\right)}=\frac{1}{-\left(a^2+b^2\right)} \)

Or,\( \frac{x}{a\left(a^2+b^2\right)}=\frac{y}{b\left(a^2+b^2\right)}=\frac{1}{\left(a^2+b^2\right)} \)

\( \frac{x}{a\left(a^2+b^2\right)}=\frac{1}{a^2+b^2} \text { and } \frac{y}{b\left(a^2+b^2\right)}=\frac{1}{a^2+b^2} \)

Class 9 Maths WB Board

Or, \( x=\frac{a\left(a^2+b^2\right)}{\left(a^2+b^2\right)} \text { and } y=\frac{b\left(a^2+b^2\right)}{\left(a^2+b^2\right)} \)

X=4

Y= b

Question 13. ax + by = 1, \( b x+a y=\frac{2 a b}{a^2+b^2} \)

Solution: ax + by = 1 ….(1)

\( b x+a y=\frac{2 a b}{a^2+b^2} \)…..(2)

 

From equation (1) ax + by – 1 = 0
From equation (2) b(a²+ b²)x + a(a²+ b²)y – 2ab = 0

\( \frac{x}{b(-2 a b)-(-1) a\left(a^2+b^2\right)}=\frac{y}{(-1) b\left(a^2+b^2\right)-(a)(-2 a b)}=\frac{1}{a \cdot a\left(a^2+b^2\right)-b \cdot b\left(a^2+b^2\right)} \)

Wbbse Class 9 Maths Linear Equations Exercise 5.6

Or,\( \frac{x}{-2 a b^2+a^3+a b^2}=\frac{y}{-a^2 b-b^3+2 a^2 b}=\frac{1}{a^4+a^2 b^2-a^2 b^2-b^4} \)

Or,\( \frac{x}{a\left(a^2-b^2\right)}=\frac{y}{b\left(a^2-b^2\right)}=\frac{1}{\left(a^2+b^2\right)\left(a^2-b^2\right)} \)

Or,\( \frac{x}{a\left(a^2-b^2\right)}=\frac{y}{b\left(a^2-b^2\right)}=\frac{1}{\left(a^2+b^2\right)\left(a^2-b^2\right)} \)

\( \frac{x}{a\left(a^2-b^2\right)}=\frac{1}{\left(a^2+b^2\right)\left(a^2-b^2\right)} \text { and } \frac{y}{b\left(a^2-b^2\right)}=\frac{1}{\left(a^2+b^2\right)\left(a^2-b^2\right)} \)

 

Or, \( x=\frac{a\left(a^2-b^2\right)}{\left(a^2+b^2\right)\left(a^2-b^2\right)} \text { and } y=\frac{b\left(a^2-b^2\right)}{\left(a^2+b^2\right)\left(a^2-b^2\right)} \)

Or,\( x=\frac{a}{a^2+b^2} \text { and } y=\frac{b}{a^2+b^2} \)

\( \begin{aligned}
& therefore x=\frac{a}{a^2+b^2} \\
& y=\frac{b}{a^2+b^2}
\end{aligned} \)

 

 

Class IX Maths Solutions WBBSE Chapter 5 Linear Simultaneous Equations Exercise 5.5

Question 1. Let us express the value of the variable x from the equation \(\frac{2}{x}+\frac{3}{y}=1\) in terms of the variable y.

Solution: \(\frac{2}{x}+\frac{3}{y}=1\)

\(\frac{2}{x}=1-\frac{3}{y}\) \(\frac{2}{x}=\frac{y-3}{y}\)

Read and Learn More WBBSE Solutions For Class 9 Maths

or, (y-3)x = 2y

or, \(x=\frac{2 y}{y-3}\)

Question 2. Let us write the value of x by putting \(\frac{7-4 x}{-5}\) -4x -5 instead of y in the equation 2x + 3y = 9.

Solution: 2x+3y=9

\(or, \quad 2 x+\frac{3(7-4 x)}{-5}=9
or, \quad 2 x-\frac{3(7-4 x)}{5}=9
or, \quad \frac{10 x-21+12 x}{5}=9
or, \quad 22 x-21=45\)

Class IX Maths Solutions WBBSE

or, 22x = 45 +21
or,22x=66

or, \(x=\frac{66}{22}\)

∴ x = 3.

Wbbse Class 9 Maths Chapter 5.5 Linear Simultaneous Equations Solutions

Question 3. Let us solve the following equations in two variables by substitution method and check them graphically.

1. 3x – y = 7, 2x + 4y = 0

Solutioin:

Given

3x – y = 7 …(1)

2x + 4y = 0…(2)

From equation (1)-y=-3x+7
or, y = 3x-7 …..(3)

Putting the value of y in equation (2),

2x + 4(3x-7)= 0

or, 2x+12x-28=0

or, 14x= 28

or, \(x=\frac{28}{14}\)

x = 2

Wbbse Class 9 Linear Simultaneous Equations Exercise 5.5 Solutions

Putting the value of y in equation (4),

y=3×2-7

or, y = 6-7

or,y = -1

∴ x = 2

y=-1

From equation (3)

y = 3x-7

x	2	4	-1
y	-1	5	-10

 

From equation (2) 4y = -2x

or, \(y=\frac{-x}{2}\)

x	2	6	-2
y	-1	-3	1

 

2. \(\frac{x}{2}+\frac{y}{3}=2=\frac{x}{4}+\frac{y}{2}\)

Solution:

\(\frac{x}{2}+\frac{y}{3}=2\) …(1)

\(\frac{x}{4}+\frac{y}{2}= \) …(2)

Class IX Maths Solutions WBBSE

From equation (1) \(\frac{x}{2}=2-\frac{y}{3}\)

or, \(\frac{x}{2}=\frac{6-y}{3}\)

or, \(x=\frac{12-2 y}{3}\) …(3)

Putting the value of y in equation (4),

\(\frac{\frac{12-2 y}{3}}{4}+\frac{y}{2}=2\)

 

or, \(\frac{2(6-y)}{3} \times \frac{1}{4}+\frac{y}{2}=2\)

or, \(\frac{6-y}{6}+\frac{y}{2}=2\)

or, \(\frac{6-y+3 y}{6}=2\)

or, 6+ 2y = 12
or, 2y=12-6

or, y = 6/2

∴ y = 3

Putting the value of y in equation (4),

\(x=\frac{12-2 \times 3}{3}\)

 

or, \(x=\frac{12-6}{3}\)

or, \(x=\frac{6}{3}\)

∴ x = 2, y=2

⇒ y = 3

From equation (3)

\(x=\frac{12-2 y}{3}\)

 

x	4	2	-6
y	0	3	-3

 

From equation (2)

\(\frac{x}{4}=2-\frac{y}{2}\)

 

or, x = 8- 2y

x	4	0	12	2
y	2	4	-2	3

 

Class IX Maths Solutions WBBSE

Question 4. Let us solve the following equations in two variables by substitution method and check whether the solutions satisfy the equations:

1. \(2 x+\frac{3}{y}=1,5 x-\frac{2}{y}=\frac{11}{12}\) 

Solution:

Given

\(2 x+\frac{3}{y}=1\) …(1)

\(5 x-\frac{2}{y}=\frac{11}{12}\) …..(2)

From equation (1) \(\frac{3}{y}=1-2 x\)

or, (1-2x)y = 3

or, \(y=\frac{3}{1-2 x}\)

Putting the value of y in equation (2),

\(5 x-\frac{2(1-2 x)}{3}=\frac{11}{12}\)

 

or, \(\frac{15 x-2+4 x}{3}=\frac{11}{12}\)

or, \(\quad 19 x-2=\frac{11}{4}\)

or, \(19 x=\frac{11}{4}+2\)

or, \(19 x=\frac{11+8}{4}\)

or, \(x=\frac{19}{19 \times 4}\)

or, \(x=\frac{1}{4}\)

Putting the value of x in equation (3)

\(y=\frac{3}{1-2 \times \frac{1}{4}}\)

 

or, \(y=\frac{3}{1-\frac{1}{2}}\)

or, y = 6

⇒ \(x=\frac{1}{4}\)

y = 6

From equation (1) L. H.S. = \(2 x+\frac{3}{y}\)

\(\begin{aligned}
& =2 \times \frac{1}{4}+\frac{3}{6} \\
& =\frac{1}{2}+\frac{1}{2} \\
& =\frac{1+1}{2} \\
& =\frac{2}{2}
\end{aligned}\)

Class IX Maths Solutions WBBSE

= 1
= R. H. S.

Again from equation (2) L. H. S. = \(5 x-\frac{2}{y}\)

\(\begin{aligned}
& =5 \times \frac{1}{4}-\frac{2}{6} \\
& =\frac{5}{4}-\frac{1}{3} \\
& =\frac{15-4}{12}
\end{aligned}\)

 

= \(\frac{11}{12}\)
= R. H. S.

∴ \(x=\frac{1}{4}\) & y = 6 satify the equations (1) & (2).

Class 9 Wbbse Maths Linear Equations Exercise 5.5 Solved Problems

2. \(\frac{2}{x}+\frac{3}{y}=2, \frac{5}{x}+\frac{10}{y}=5 \frac{5}{6}\)

Solution:

Given

\(\frac{2}{x}+\frac{3}{y}=2\) …(1)

 

\(\frac{5}{x}+\frac{10}{y}=5 \frac{5}{6}\)….(2)

From equation (1) \(\frac{3}{y}=2-\frac{2}{x}\)

or, \(\frac{3}{y}=\frac{2 x-2}{x}\)

or,(2x-2)y = 3 x

or, \(y=\frac{3 x}{2 x-2}\) ….(3)

Putting the value of x in equation (2),

\(\frac{5}{x}+\frac{10(2 x-2)}{3 x}=\frac{35}{6}\)

Class 9 Mathematics West Bengal Board

or, \(\frac{15+20 x-20}{3 x}=\frac{35}{6}\)

or, \(\frac{20 x-5}{x}=\frac{35}{2}\)

or, 40x-10=35x
or, 40x-35x=10
or, 5x = 10

or, \(x=\frac{10}{5}\)

or, x  = 2

Putting the value of x in equation (3),

or, \(y=\frac{3 \times 2}{2 \times 2-2}\)

or, \(y=\frac{6}{4-2}\)

or, \(y=\frac{6}{2}\)

∴ y = 3, x = 2

From equation (1) L. H.S. = \(\frac{2}{x}+\frac{3}{y}\)

= \(\frac{2}{2}+\frac{3}{3}\)

= 1 + 1 = 2
= R. H. S.

Again, from equation (2) L. H. S. \(\frac{5}{x}+\frac{10}{y}\)

\(\begin{aligned}
& =\frac{5}{2}+\frac{10}{3} \\
& =\frac{15+20}{6} \\
& =\frac{35}{6} \\
& =5 \frac{5}{6}
\end{aligned}\)

= R. H. S.

3. \(\frac{x+y}{x y}=3, \frac{x-y}{x y}=1\)

Solution:

Given

\(\frac{x+y}{x y}=3\) …(1)

\(\frac{x-y}{x y}=1\)   ….(2)

From equation (1) \(\frac{x}{x y}+\frac{y}{x y}=3\)

\(or, \frac{1}{y}+\frac{1}{x}=3
or, \frac{1}{y}=3-\frac{1}{x}
or, \quad \frac{1}{y}=\frac{3 x-1}{x}\)

Wbbse Class 9 Maths Chapter 5.5 Linear Equations Notes

Class 9 Mathematics West Bengal Board

or, (3x-1)y = x

or, \(y=\frac{x}{3 x-1}\) ….(3)

From equation (1) \(\frac{x}{x y}-\frac{y}{x y}=1\)

or, \(\frac{1}{y}-\frac{1}{x}=1\)  …..(4)

Putting the value \(y=\frac{x}{3 x-1}\) in equation (4)

or, \(\frac{1}{\frac{x}{3 x-1}}-\frac{1}{x}=1\)

or, \(\frac{3 x-1}{x}-\frac{1}{x}=1\)

or, \(\frac{3 x-1-1}{x}=1\)

or, 3x-2=x
or,3x-x=2
or,2x = 2

or, \(x=\frac{2}{2}\)

or, X = 1

Putting the value of x in equation (3),

or, \(y=\frac{1}{3 \times 1-1}\)

or, \(y=\frac{1}{3-1}\)

or, \(y=\frac{1}{2}\)

∴ x =1

\(y=\frac{1}{2}\)

From equation (1) L. H.S. = \(\frac{x+y}{x y}\)

= \(=\frac{1+\frac{1}{2}}{1 \times \frac{1}{2}}\)

\(\begin{aligned}
& \frac{2+1}{2} \\
= & \frac{1}{2} \\
= & \frac{3}{2} \times \frac{2}{1}
\end{aligned}\)

Class 9 Mathematics West Bengal Board

= 3
= R. H. S.

Again, from equation (2) L. H. S. = \(\frac{x-y}{x y}\)

\(\begin{aligned}
& =\frac{1-\frac{1}{2}}{1 \times \frac{1}{2}} \\
& =\frac{\frac{2-1}{2}}{\frac{1}{2}} \\
& =\frac{1}{2} \times \frac{2}{1}
\end{aligned}\)

= 1
= R. H. S.

4.  \(\frac{x+y}{x-y}=\frac{7}{3}, \quad x+y=\frac{7}{10}\)

Solution:

Given

\(\frac{x+y}{x-y}=\frac{7}{3}\)   ….(1)

\(x+y=\frac{7}{10}\) ……(2)

From equation (1) \(\frac{x+y}{x-y}=\frac{7}{3}\)

or, 7x-7y=3x+3y
or, 7x-3x=7y+ 3y
or, 4x = 10y

or, \(x=\frac{10 y}{4}\)

or, \(x=\frac{5 y}{2}\)  …..(3)

Putting the value of y in equation (2),

\(or,\frac{5 y}{2}+y=\frac{7}{10}
or,\quad\frac{5 y+2 y}{2}=\frac{7}{10}
or,\frac{7 y}{1}=\frac{7}{5}
or,\quad y=\frac{7}{5 \times 7}
or, \quad y=\frac{1}{5}\)

Class 9 Math Chapter 5 WBBSE

Putting the value of y in equation (2),

\(or, \quad x+\frac{1}{5}=\frac{7}{10}
or,\quad x=\frac{7}{10}-\frac{1}{5}
or,\quad x=\frac{7-2}{10}
or,\quad x=\frac{5}{10}
or,\quad x=\frac{1}{2}
therefore  x=\frac{1}{2}
y=\frac{1}{5}\)

 

From equation (1) L. H.S.  \(=\frac{x+y}{x-y}\)

= \(=\frac{\frac{1}{2}+\frac{1}{5}}{\frac{1}{2}-\frac{1}{5}}\)

= \(=\frac{\frac{7}{10}}{\frac{3}{10}}\)

= \(=\frac{7}{3}\)

= R. H. S.

Again, from equation (1) L. H. S. = x + y

\(\begin{aligned}
& =\frac{1}{2}+\frac{1}{5} \\
& =\frac{5+2}{10} \\
& =\frac{7}{10}
\end{aligned}\)

= R. H. S.

West Bengal Board Class 9 Linear Simultaneous Equations Exercise 5.5 Solutions

Question 5. Let us solve the following equations in two variables by substitution method:

1. 2(x-y) = 3, 5x + 8y = 14

Solution:

Given

2(x-y)=3 ….(1)

5x + 8y = 14  …..(2)

From equation (1) 2x-2y=3
or, 2x = 3 + 2y

or, \(x=\frac{3+2 y}{2}\)  ….(3)

Putting the value of x in equation (2),

\(\frac{5(3+2 y)}{2}+8 y=14\)

Class 9 Math Chapter 5 WBBSE

or, \(\frac{15+10 y+16 y}{2}=14\)

or, 15+26y=28
or, 26y=28-15
or, 26y = 13

\(or, \quad y=\frac{13}{26}
or, \quad y=\frac{1}{2}\)

Putting the value of y in equation (3),

\(x=\frac{3+2 \times \frac{1}{2}}{2}\)

 

or, \(x=\frac{3+1}{2}\)

or, \(x=\frac{4}{2}\)

or, x =2

∴ x =2 , y= 1/2

Wbbse Class 9 Linear Equations Exercise 5.5 Important Questions

2. \(2 x+\frac{3}{y}=5,5 x-\frac{2}{y}=3\)

Solution:

\(2 x+\frac{3}{y}=5 \) …..(1)

\(5 x-\frac{2}{y}=3\)   …..(2)

Class 9 Math Chapter 5 WBBSE

From equation (1) \(2 x=5-\frac{3}{y}\)

or, \(2 x=\frac{5 y-3}{y}\)

or, \(x=\frac{5 y-3}{2 y}\) ….(3)

Putting the value of x in equation (2),

\(\frac{5(5 y-3)}{2 y}-\frac{2}{y}=3\)

 

or, \(\frac{25 y-15-4}{2 y}=3\)

or, 25y-19 = 6y
or,25y-6y 19
or,19y= 19

or, y = \(\frac{19}{19}\)

or, y = 1

Putting the value of y in equation (3),

\(x=\frac{5-3}{2}\)

 

or, \(x=\frac{5-3}{2}\)

or, \(x=\frac{2}{2}\)

or, x = 1

∴ x = , y = 1

Wbbse Class 9 Maths Linear Equations Exercise 5.5

3. \(\frac{x}{2}+\frac{y}{3}=1, \frac{x}{3}+\frac{y}{2}=1\)

Solution: \(\frac{x}{2}+\frac{y}{3}=1\) ….(1)

\(\frac{x}{3}+\frac{y}{2}=1\) ….(2)

 

From equation (1) \(\frac{x}{2}=1-\frac{y}{3}\)

or,\(\frac{x}{2}=\frac{3-y}{3}\)

or, 3x = 6-2y

or, \(x=\frac{6-2 y}{3}\) …(3)

Putting the value of y in equation (3),

\(\frac{\frac{6-2 y}{3}}{3}+\frac{y}{2}=1\)

Class 9 Math Chapter 5 WBBSE

or, \(\frac{6-2 y}{3} \times \frac{1}{3}+\frac{y}{2}=1\)

or, \(\frac{12-4 y+9 y}{18}=1\)

or, 12 + 5y = 18
or, 5y = 18-12

or, \(y=\frac{6}{5}\)

Putting the value of y in equation (3),

\(x=\frac{6-2 \times \frac{6}{5}}{3}\)

 

\(or, x=\frac{\frac{30-12}{5}}{3}
or, \quad x=\frac{18}{5} \times \frac{1}{3}
or, \quad x=\frac{6}{5}\)

 

∴ \(x=\frac{6}{5}\)

∴ \(y=\frac{6}{5}\)

Wbbse 9th Class Maths Linear Simultaneous Equations Step By Step Solutions Exercise 5.5

4. \(\frac{x}{3}=\frac{y}{4}\), 7x-5y = 2

Solution:

Given

\(\frac{x}{3}=\frac{y}{4}\) ….(1)

7x-5y – 2 ….(2)

From equation(1) 4x = 3y

or, \(x=\frac{3 y}{4}\) …..(3)

Putting the value of x in equation. (2),

\(7 \times \frac{3 y}{4}-5 y=2\)

 

or, \(\frac{21 y-20 y}{4}=2\)

or, y = 8

Putting the value of y in equation (3),

\(x=\frac{3 \times 8}{4}\)

Class 9 Math Solution WBBSE In English

∴ x = 6, y = 8

5. \(\frac{2}{x}+\frac{5}{y}=1, \frac{3}{x}+\frac{2}{y}=\frac{19}{20}\)

Solution:

Given

\(\frac{2}{x}+\frac{5}{y}=1\) …(1)

\(\frac{3}{x}+\frac{2}{y}=\frac{19}{20}\)……..(2)

 

From equation (1) \(\frac{2}{x}=1-\frac{5}{y}\)

or, \(\frac{2}{x}=\frac{y-5}{y}\)

or, (y-5)x = 2y

or, \(x=\frac{2 y}{y-5}\) ….(3)

Putting the value of x in equation (2),

\(\frac{3}{\frac{2 y}{y-5}}+\frac{2}{y}=\frac{19}{20}\)

 

or, \(\frac{3(y-5)}{2 y}+\frac{2}{y}=\frac{19}{20}\)

or, \(\frac{3 y-15+4}{2 y}=\frac{19}{20}\)

or, \(\frac{3 y-11}{y}=\frac{19}{10}\)

or, 30y-110 = 19y
or,30y-19y=110
or,11y=110

or, \(y=\frac{110}{11}\)

or,y = 10

Putting the value of y in equation (3),

\(x=\frac{2 \times 10}{10-5}\)

Class 9 Math Solution WBBSE In English

or,\(x=\frac{20}{5}\)

or,X = 4

∴ X=4
y = 10

Wbbse Class 9 Maths Methods To Solve Linear Simultaneous Equations Exercise 5.5

6. \(\frac{1}{3}\)(x − y) = \(\frac{1}{3}\) (y-1), \(\frac{1}{3}\)(4x-5y) = x – 7

Solution:

Given

\(\frac{1}{3}\)(x − y) = \(\frac{1}{3}\) (y-1) ….(1)

\(\frac{1}{3}\)(4x-5y) = x – 7 …..(2)

From equation (1) \(\frac{x-y}{3}=\frac{y-1}{4}\)

or, 4x-4y=3y-3
or, 4x=3y – 3+ 4y
or, 4x = 7y-3

or, \(x=\frac{7 y-3}{4}\)  ….(3)

Putting the value of x in equation (2),

\(\frac{1}{7}\left\{4 \frac{(7 y-3)}{4}-5 y\right\}=\frac{7 y-3}{4}-7\)

Class 9 Math Solution WBBSE In English

or, \(\frac{1}{7}(7 y-3-5 y)=\frac{7 y-3-28}{4}\)

or, \(\frac{2 y-3}{7}=\frac{7 y-31}{4}\)

or, 49y-217 = 8y-12
or, 49y-8y= -12+217
or, 41y=205

or, \(y=\frac{205}{41}\)

or, y = 5

Putting the value of y in equation (3),

\(x=\frac{7 \times 5-3}{4}\)

 

\(or, x=\frac{35-3}{4} or, x=\frac{32}{4}\)

 

or, x =8

∴ x = 8, y =5

Class 9 Wbbse Linear Equations Chapter 5.5 Solved Exercises

7. \(\frac{x}{14}+\frac{y}{18}=1, \frac{x+y}{2}+\frac{3 x-5 y}{4}=2\)

Solution:

\(\frac{x}{14}+\frac{y}{18}=1\) ….(1)

\(\frac{x+y}{2}+\frac{3 x-5 y}{4}=2\) ….(2)

Class 9 Math Solution WBBSE In English

From equation, (1) \(\frac{x}{14}+\frac{y}{18}=1\)

or, \(\frac{9 x+7 y}{126}=1[/latex

or, 9x + 7y = 126
or, 9x=126-7y

or, [latex]x=\frac{126-7 y}{9}\)   ….(3)

From equation (1) \(\frac{x+y}{2}+\frac{3 x-5 y}{4}=2\)

or, \(\frac{2 x+2 y+3 x-5 y}{4}=2\)

or, 5x – 3y = 8 (4)

From equation (4) \(x=\frac{126-7 y}{9}\)

\(\frac{5(126-7 y)}{9}-3 y=8\)

 

or, \(\frac{630-35 y-27 y}{9}=8\)

or, – 62y + 630 = 72
or,-62y=72-630
or, -62y=-558
or, 62y=558

or, \(y=\frac{558}{62}\)

or, y=9

Putting the value of y in equation (3),

\(x=\frac{126-7 \times 9}{9}\)

Class 9 Math Solution WBBSE In English

\(or, \quad x=\frac{126-63}{9}
or, x=\frac{63}{9}
or, \quad x=7\)

∴ x = 7, y=9

8. p (x+y)= q(xy) =2pq

Solution:

Given

P(x+y)=2pq ….(1)

q (x – y) = 2pq…..(2)

From equation (1) p (x+y)=2pq

or, \(x+y=\frac{2 p q}{p}\)

or, x + y = 2q
or, x = 2q-y ….(3)

Putting the value of x in equation(2)
q(2q – y-y) = 2pq

or, \(2 q-2 y=\frac{2 p q}{q}\)

or,-2y=2p-2q

or, \(y=\frac{2(p-q)}{-2}\)

or,y=q-p

Putting the value of y in equation (3),
x = 2q – (q.-p)
or, x=2q-q+p
or, x=p+q
=x=p+q
y=q-p

 

Class 9 Math Chapter 5 WBBSE Chapter 5 Linear Simultaneous Equations Exercise 5.4

Question 1. Let us express the variable x of the equation \(\frac{x}{3}+\frac{y}{2}=8\) in term of variable y.

Solution:

Given

\(\frac{x}{3}+\frac{y}{2}=8\) \(or, \quad \frac{x}{3}=8-\frac{y}{2}
or, \frac{x}{3}=\frac{16-y}{2}
or, x=\frac{3}{2}(16-y)\)

 

Question 2. \(\frac{2}{x}+\frac{7}{y}=1\)=1 Express the value of y in terms of y.

Solution:

Given

\(\frac{2}{x}+\frac{7}{y}=1\) \(or,\quad \frac{7}{y}=1-\frac{2}{x}
or,\quad \frac{7}{y}=\frac{x-2}{x}
or,\quad(x-2) y=7 x
or,\quad y=\frac{7 x}{x-2}\)

Wbbse Class 9 Linear Simultaneous Equations Exercise 5.4 Solutions

Question 3. Let us solve the following equations by comparison method and check whether the solutions satisfy the equations.

1. 2(x − y) = 3, 5x + 8y = 14

Solution:

Given

2(x-y)=3 ….(1)
5x + 8y = 14 ….(2)

From equation (1) 2x-2y=3
or,2x = 3 + 2y

or, \(x=\frac{3+2 y}{2}\) ….(3)

From equation (2) 5x+8y = 14
or, 5x = 14-8y

or, \(x=\frac{3+2 y}{2}\) ….(4)

Comparing the value of x from equations (3) & (4),

\(\frac{3+2 y}{2}=\frac{14-8 y}{5}\)

Wbbse Class 9 Maths Chapter 5.4 Linear Simultaneous Equations Solutions

or, 15+10y=28-16y
or, 10y+ 16y=28-15
or, 26y=13

or, \(y=\frac{13}{26}\)

or, \(y=\frac{1}{2}\)

Putting the value of y in equation (3)

\(x=\frac{3+2 \times \frac{1}{2}}{2}\)

 

\(\begin{aligned}
& x=\frac{3+1}{2} \\
& x=\frac{4}{2} \\
& x=2 \\
& x=2
\end{aligned}\)

Class 9 Math Chapter 5 WBBSE

\(y=\frac{1}{2}\)

From equation (1) L. H.S. = 2(x-y)

\(\begin{aligned}
& =2\left(2-\frac{1}{2}\right) \\
& =2\left(\frac{4-1}{2}\right) \\
& =3
\end{aligned}\)

 

= R. H. S.

Again, from equation (1), L. H. S.= 5x + 8y

= \(5 \times 2+8 \times \frac{1}{2}\)

= 10 +4
= 14

= R. H. S.

Class 9 Wbbse Maths Linear Equations Exercise 5.4 Solved Problems

2.  \(2 x+\frac{3}{y}=5,5 x-\frac{2}{y}=3\)

Solution:

Given

\(2 x+\frac{3}{y}=5\) ….(1)

\(5 x-\frac{2}{y}=3\) …(2)

From equation (1) \(\frac{3}{y}=5-2 x\)

or, \(\frac{1}{y}=\frac{5-2 x}{3}\) …..(3)

From equation (2) \(\frac{-2}{y}=3-5 x\)

or, \(\frac{2}{y}=5 x-3\)

or, \(\frac{1}{y}=\frac{5 x-3}{2}\) ….(4)

Comparing the value of \(\frac{1}{y}\) from equations (3)& (4) we get,

\(\frac{5-2 x}{3}=\frac{5 x-3}{2}=\)

or, 15x –  9 = 10-4x
or, 15x+4x= 10 +9
or, 19x= 19

or, x = 19/19
or,  = 1

Putting the value of x in equation (3)

or, \(\frac{5-2 x}{3}=\frac{5 x-3}{2}\)

or, \(\frac{1}{y}=\frac{3}{3}\)

or, \(\frac{1}{y}=1\)

or, y = 1
X = 1

From equation (1)L. H.S. = \(2 x+\frac{3}{y}\)

= \(2 \times 1+\frac{3}{1}\)

=2+3
= 5
= R. H. S.

From equation (2) L. H. S.= \(5 x-\frac{2}{y}\)

= \(5 \times 1-\frac{2}{1}\)

=5-2
= 3
= R. H. S.

Wbbse Class 9 Maths Chapter 5.4 Linear Equations Notes

3. \(\frac{x}{2}+\frac{y}{3}=1, \frac{x}{3}+\frac{y}{2}=1\)

Solution:

Given

\(\frac{x}{2}+\frac{y}{3}=1\) …(1)

\(\frac{x}{3}+\frac{y}{2}=1\) …(2)

From equation (1) \(\frac{x}{2}=1-\frac{y}{3}\)

or, \(\frac{x}{2}=\frac{3-y}{3}\)

or, \(x=\frac{3(2-y)}{2}\) ….(3)

From equation(2) \(\frac{x}{3}=1-\frac{y}{2}\)

or, \(\frac{x}{3}=\frac{2-y}{2}\)

or, \(x=\frac{3(2-y)}{2}\)….(4)

Comparing the value of x in equations (3) & (4),

\(\frac{2(3-y)}{3}=\frac{3(2-y)}{2}\)

Class 9 Math Chapter 5 WBBSE

or, 4(3-y) 9 (2-y)
or, 12-4y=18-9y
or, 9y-4y=18-12
or, 5y = 6

or, \(y=\frac{6}{5}\)

Putting the value of y in equation (3) we get,

\(x=\frac{2\left(3-\frac{6}{5}\right)}{3}\)

or, [/latex]x=\frac{2\left(\frac{15-6}{5}\right)}{3}
or, \quad x=2 \times \frac{9}{5} \times \frac{1^{-}}{3}[/latex]

or, \(\begin{aligned}
& x=\frac{6}{5} \\
& =x=\frac{6}{5}
\end{aligned}
y=\frac{6}{5}
\)

Class IX Maths Solutions WBBSE

From equation (1 ) L. H.S. =\(\frac{x}{2}+\frac{y}{3}\)

\(\begin{aligned}
& =\frac{\frac{6}{5}}{2}+\frac{\frac{6}{5}}{3} \\
& =\frac{6}{5} \times \frac{1}{2}+\frac{6}{5} \times \frac{1}{3} \\
& =\frac{3}{5}+\frac{2}{5} \\
& =\frac{5}{5} \\
& =1
\end{aligned}\)

 

= R. H. S.

From equation (2) L. H. S. = \(\frac{x}{3}+\frac{y}{2}\)

\(\begin{aligned}
& =\frac{\frac{6}{5}}{3}+\frac{\frac{6}{5}}{2} \\
& =\frac{6}{5} \times \frac{1}{3}+\frac{6}{5} \times \frac{1}{2} \\
& =\frac{2}{5}+\frac{3}{5} \\
& =\frac{2+3}{5} \\
& =\frac{5}{5}
\end{aligned}\)

Class IX Maths Solutions WBBSE

= 1
= R. H. S.

4. 4x-3y = 18, 4y – 5x = -7

Solution:

Given

4x-3y = 18 …(1)

4y – 5x = -7 ….(2)

From equation (1) 4x = 18+ 3y

or, \(x=\frac{18+3 y}{4}\) ….(3)

From equation (2) -5x=-4y-7.
or, 5x = 4y+ 7

or, \(x=\frac{4 y+7}{5}\) ….(4)

Comparing the value of ‘x’ from equations (3) & (4),

or, \(\frac{18+3 y}{4}=\frac{4 y+7}{5}\)

or,16y+28 = 90+ 15y
or,16y 15y = 90-28
or, y = 62

Putting the value of y, we get

\(x=\frac{18+3 \times 62}{4}\)

 

\(or, \quad x=\frac{18+186}{4}
or, x=\frac{204}{4}
or, \quad x=51\)

x =51,y= 62

From equation (1) L. H.S. = 4x-3y
= 4 x 51-3 x 62
= 204
= 18
=R. H. S.

From equation (2) L. H. S.= 4y-5x
= 4 x 62-5 x 51
=248-255
=-7
= R. H. S.

West Bengal Board Class 9 Linear Simultaneous Equations Exercise 5.4 Solutions

Question 4. Let us solve the equations 2x + y = 8 and 2y-3x=-5 by comparison method and justify them by solving graphically.

Solution:

2x + y = 8 ….(1)
2y-3x=-5…(2)

Equation y = 8-2x …(3)

Equation 2y=3x-5

or, \(y=\frac{3 x-5}{2}\) …(4)

From equations (3) & (4),

\(8-2 x=\frac{3 x-5}{2}\)

or, 3x-5=16-4x
or, 3x+4x=16+5
or, 7x=21

or, \(x=\frac{21}{7}\)

or, x = 3

Putting this value of x in equation (3),
y=8-2×3

or,y = 2
x = 3

⇒ y = 8-2x

x	0	2	-1
y	8	4	10

 

Class IX Maths Solutions WBBSE

\(y=\frac{3 x-5}{2}\)

 

x	1	3	-7
y	-1	2	-13

 

Question 5. Let us solve the following equations in two variables by comparison method:

1. 3x-2y= 2, 7x + 3y = 43

Solution:

Given

3x-2y=2 …(1)
7x + 3y = 43 …(2)

From equation (1) 3x=2+2y

or, \(x=\frac{2+2 y}{3}\) …(3)

From equation (2) 7x=43-3y

or, \(x=\frac{43-3 y}{7}\) ….(4)

Comparing the value of x from equations (3) & (4)

\(\frac{2+2 y}{3}=\frac{43-3}{7}\)

 

or,14+ 14y = 129-9y
or, 14y+9y=129-14
or, 23y=115

or, \(y=\frac{115}{23}\)

or,y = 5

Putting the value of y in equation (3),

\(x=\frac{2+2 \times 5}{3}\)

 

\(x=\frac{12}{3}\)

X = 4

∴x=4&y=5

2. 2x-3y=8, \(\frac{x+y}{x-y}=\frac{7}{3}\)

Solution: 2x-3y=8 …(1)

\(\frac{x+y}{x-y}=\frac{7}{3}\)….(2)

Wbbse Class 9 Linear Equations Exercise 5.4 Important Questions

From equation (1) 2x=8+3y

or, \(x=\frac{8+3 y}{2}\) ….(3)

From equation (2) 7x-7y = 3x + 3y
or, 7x-3x=3y+ 7y.
or, 4x = 10y

or, \(x=\frac{10 y}{4}\)

or, \(x=\frac{5 y}{2}\) …(4)

Companing the value of x from equations (3) & (4),

\(\frac{8+3 y}{2}=\frac{5 y}{2}\)

 

or, 8+3y = 5y
or, 8+ 3y = 5y
or, 3y-5y=-8
or,-2y=-8

or, \(y=\frac{-8}{-2}\)

y = 4

Putting the value of y in equation (4),

\(x=\frac{5 \times 4}{2}\)

or, X = 10

∴ X = 10
y = 4

3. \(\frac{1}{3}(x-y)=\frac{1}{4}(y-1), \frac{1}{7}(4 x-5 y)=x-7\)

Solution:

\(\frac{1}{3}(x-y)=\frac{1}{4}(y-1)\) ….(1)

 

\(\frac{1}{7}(4 x-5 y)=x-7\) …(2)

Class 9 Mathematics West Bengal Board

From equation (1) \(\frac{x-y}{3}=\frac{y-1}{4}\)

or,4x-4y=3y-3
or,4x 3y+4y-3

or, \(x=\frac{7 y-3}{4}\) ….(3)

From equation (2) 7(x-7)=4x-5y
or, 7x-49=4x-5y
or, 7x-4x=49-5y
or, 3x=49-5y

or, \(x=\frac{49-5 y}{3}\) ……(4)

Comparing the value of x from equations (3) & (4),

\(\frac{7 y-3}{4}=\frac{49-5 y}{3}\)

 

or, 21y-9-196-20y
or, 21y+20y 196 +9
or, 41y=205

or, \(y=\frac{205}{4}\)

or, y = 5

Putting the value of y in equation (3),

\(x=\frac{7 \times 5-3}{4}\)

Class 9 Mathematics West Bengal Board

\(or, x=\frac{35-3}{4}
or, \quad x=\frac{32}{4}
or, x=8 \)

Required solution x = 8,y = 5

Wbbse Class 9 Maths Linear Equations Exercise 5.4

4. \(\frac{x+1}{y+1}=\frac{4}{5}, \frac{x-5}{y-5}=\frac{1}{2}\)

Solution:

\(\frac{x+1}{y+1}=\frac{4}{5}\) …….(1)

 

\(\frac{x-5}{y-5}=\frac{1}{2}\) ….(2)

From equation (1) 5x+5=4y+4
or, 5x = 4y+ 4-5
or, 5x= 4y-1

or, \(x=\frac{4 y-1}{5}\) ….(3)

From equation (2) 2x-10-y-5
or, 2x=y-5+10
or, 2x = y +5

or, \(x=\frac{y+5}{2}\) …(4)

Comparing the value of x from equations (2) & (4),

\(\frac{4 y-1}{5}=\frac{y+5}{2}\)

or, 8y-2=5y+25
or, 8y – 5y = 25+ 2
or, 3y=27

or, \(y=\frac{27}{3}\)

or,y=9

Putting the value of y in equation (4),

\(\begin{aligned}
&x=\frac{9+5}{2}\\
&x=\frac{14}{2}
\end{aligned}\)

Class 9 Mathematics West Bengal Board

x =7,y =9

5.  x + y = 11, y+2= \(\frac{1}{8}\) (10y+x)

Solution: x + y = 11……(1)

y+2= \(\frac{1}{8}\) (10y+x) …..(2)

From equation (1) x = 11-y ….(3)

From equation (2) \(\frac{(y+2)}{1}=\frac{(10 y+x)}{8}\)

or, 10y + x = 8y + 16
or, x = 8y + 16-10y
or, x = -2y+16 ….(4)

Comparing the value of x from equations (3) & (4),
11 – y = -2y+ 16
or, -y+2y= 16-11
or,y = 5

Putting the value of y in equation (3),
X=11-5
or,X = 6

∴ X = 6
y = 5

Wbbse 9th Class Maths Linear Simultaneous Equations Step By Step Solutions Exercise 5.4

6. \(\frac{x}{3}+\frac{y}{4}=1\), 2x + 4y = 11

Solution: \(\frac{x}{3}+\frac{y}{4}=1\)…..(1)

2x + 4y = 11 …(2)

From equation (1) \(\frac{x}{3}+\frac{y}{4}=1\),

or, \(\frac{4 x+3 y}{12}=1\)

or, 4x + 3y = 12
or, 4x=12-3y

or, \(x=\frac{12-3 y}{4}\)…(3)

From equation (2) 2x = 11-4y

or, \(x=\frac{11-4 y}{2}\) ….(4)

Comparing the value of x from equations (3) & (4),

or, \(\frac{12-3 y}{4}=\frac{11-4 y}{2}\)

or, 24 – 6y = 44 – 16y
or, 16y – 6y = 44-24
or,10y=20

or, \(y=\frac{20}{10}\)

∴ y = 2

\(x=\frac{12-3 \times 2}{4}\)

Class 9 Mathematics West Bengal Board

or, \(x=\frac{6}{4}\)

∴ \(x=\frac{3}{2}\), y = 2

7. \(x+\frac{2}{y}=7,2 x-\frac{6}{y}=9\)

Solution:

\(x+\frac{2}{y}=7\) …..(1)

\(2 x-\frac{6}{y}=9\) …(2)

From equation (1) \(x=7-\frac{2}{y}\)

or, \(x=\frac{7 y-2}{y}\)

From equation (2) \(2 x=9+\frac{6}{y}\)

or, \(2 x=\frac{9 y+6}{y}\)

or, \(x=\frac{9 y+6}{2 y}\) …..(4)

Comparing the value of x from equations (3) & (4),

\(\begin{aligned}
& \frac{7 y-2}{y}=\frac{9 y+6}{2 y} \\
& \frac{7 y-2}{1}=\frac{9 y+6}{2}
\end{aligned}\)

 

or, 14y- 4= 9y+6
or,14y-9y=6+4
or, 5y = 10

or, \(y=\frac{10}{5}\),

∴ y = 2

Putting the value of y in equation (3),

\(\begin{aligned}
& x=\frac{7 \times 2-2}{2} \\
& x=\frac{14-2}{2} \\
& x=\frac{12}{2}
\end{aligned}\)

Class 9 Maths WB Board

8. \(\frac{1}{x}+\frac{1}{y}=\frac{5}{6}, \frac{1}{x}-\frac{1}{y}=\frac{1}{6}\)

Solution: \(\frac{1}{x}+\frac{1}{y}=\frac{5}{6}\)…(1)

 

\(\frac{1}{x}-\frac{1}{y}=\frac{1}{6}\) ….(2)

 

From equation (1) \(\frac{1}{x}=\frac{5}{6}-\frac{1}{y}\) …(3)

From equation (2) \(\frac{1}{x}=\frac{1}{6}+\frac{1}{y}\) …(4)

Comparing the value of x from equations (3) & (4),

\(\begin{aligned}
& \frac{5}{6}-\frac{1}{y}=\frac{1}{6}+\frac{1}{y} \\
& \frac{5}{6}-\frac{1}{6}=\frac{1}{y}+\frac{1}{y} \\
& \frac{5-1}{6}=\frac{1+1}{y} \\
& \frac{4}{6}=\frac{2}{y} \\
& 4 y=12 \\
& y=\frac{12}{4} \\
& y=3
\end{aligned}\)

 

Putting the value of y in equation (3).

\(or, \frac{1}{x}=\frac{5-2}{6}
or, \frac{1}{x}=\frac{3}{6}
or, \quad 3 x=6
or, \quad x=\frac{6}{3}
or, \quad x=2
therefore   x=2 \)

Wbbse Class 9 Maths Methods To Solve Linear Simultaneous Equations Exercise 5.4

9. \(\frac{x+y}{x y}=2, \frac{x-y}{x y}=1\)

Solution:

Given

\(\frac{x+y}{x y}=2\) …(1)

\(\frac{x-y}{x y} = 1\) …(2)

From equation (1) \(\frac{x}{x y}+\frac{y}{x y}=2\)

or,\(\frac{1}{y}+\frac{1}{x}=2\)

or, \(\frac{1}{y}=2-\frac{1}{x}\) ….(3)

From equation (2)=\(\frac{x}{x y}-\frac{y}{x y}=1\)

or, \(\frac{1}{y}-\frac{1}{x}=1\)

or, \(\frac{1}{y}=1+\frac{1}{x}\)

Comparing the value of x from equations (3) & (4),

\(or, \quad 2-\frac{1}{x}=1+\frac{1}{x}
or, \quad 2-1=\frac{1}{x}+\frac{1}{x}
or, \quad 1=\frac{1+1}{x}
or, 1=\frac{2}{x}
or, \quad x=2\)

 

Putting the value of y in equation (3),

\(or, \frac{1}{y}=\frac{4-1}{2}
or, \quad \frac{1}{y}=\frac{3}{2}
or, 3 y=2
or, \quad y=\frac{2}{3}\)

Class 9 Maths WB Board

∴ x =2, \(y=\frac{2}{3}\)

10. \(\frac{x+y}{5}+\frac{x-y}{4}=5, \frac{x+y}{4}+\frac{x-y}{5}=5 \frac{4}{5}\)

Solution:

Given

\(\frac{x+y}{5}+\frac{x-y}{4}=5\) …(1)

\(\frac{x+y}{4}+\frac{x-y}{5}=5 \frac{4}{5}\) …(2)

 

From equation (1)=\(\frac{x+y}{5}+\frac{x-y}{4}=5\)

\(or, \quad \frac{4 x+4 y+5 x-5 y}{20}=5
or, \quad 9 x-v=100
or, \quad 9 x=100+y\)

 

or, \(x=\frac{100+y}{9}\) …(3)

From equation (2) \(\frac{x+y}{4}+\frac{x-y}{5}=\frac{29}{5}\)

\(or, \quad \frac{5 x+5 y+4 x-4 y}{20}=\frac{29}{5}
or, \quad 9 x+y=\frac{29}{5} \times 20
or, \quad 9 x+y=116
or, \quad 9 x=116-y\)

\(x=\frac{16-y}{9}\)….(4)

Class 9 Maths WB Board

Comparing the value of x from equations (3) & (4),

\(or, \quad \frac{100+y}{9}=\frac{116-y}{9}
or, \quad 100+y=116-y
or, y+y=116-100
or, 2 y=16
or, y=\frac{16}{2}
or, y=8\)

 

Putting the value of y in equation (3),

\(\begin{aligned}
& x=\frac{100+8}{9} \\
& x=\frac{108}{9}
\end{aligned}\)

 

∴ x = 12
y = 8

Class 9 Wbbse Linear Equations Chapter 5.4 Solved Exercises

11. \(\frac{4}{x}-\frac{y}{2}=-1, \frac{8}{x}+2 y=10\)

Solution:

Given

\(\frac{4}{x}-\frac{y}{2}=-1\)….(1)

\(\frac{8}{x}+2y=10\) …(2)

 

From equation (1)\(-\frac{y}{2}=-\frac{4}{x}-1\)

or, \(\frac{y}{2}=\frac{4}{x}+1\)

\(y=2\left(\frac{4}{x}+1\right)\)…(3)

 

From equation (2) 2y = \(=10-\frac{8}{x}\)

or, \(y=\frac{10}{2}-\frac{8}{2 x}\)

or, \(y=5-\frac{4}{x}\) …(4)

Comparing the value of x from equations (3) & (4),

\(2\left(\frac{4}{x}+1\right)=5-\frac{4}{x}\)

 

\(or, \frac{8}{x}+2=5-\frac{4}{x}
or, \frac{8}{x}+\frac{4}{x}=5-2\)

 

\(or, \quad \frac{8+4}{x}=3
or, \quad \frac{12}{x}=3\)

 

or, 3x = 12

\(x=\frac{12}{3}\)

 

or, x = 4

Putting the value of y in equation (4),

\(y=5-\frac{4}{4}\)

or, y=5-1

y = 4
∴ x= 4 , y = 4

12. 2-2(3x-y)=10(4-y)- 5x = 4(y-x)

Solution:

Given

2- 2(3x-y) = 4(y-x) …..(1)

10(4-y)-5x = 4(y-x) …(2)

From equation (1) 2-6x+2y= 4y – 4x
or, – 6x+4x=4y-2y-2
or, -2x =-2y-2
or, 2x=2-2y

or, \(x=\frac{2-2 y}{2}\) …(3)

From equation (2) 40-10y-5x = 4y – 4x
or, – 5x + 4x = 4y+ 10y – 40
or, -x=14y-40
or, X = 40-14y

Comparing the value of x from equations (3) & (4),

\(\frac{2-2 y}{2}=40-14 y\)

 

or, \(\frac{2}{2}-\frac{2 y}{2}=40-14y\)

or, 1-y=40-14y
or, -y+14y=40-1
or, 13y=39

or, y = \(\frac{39}{13}\)

y = 3

Putting the value of y in equation (4)
x=40-14 x 3

or, X = 40-42

∴ x=-2,y =3

 

Class IX Maths Solutions WBBSE Chapter 5 Linear Simultaneous Equations Exercise 5.3

Question 1. Let us solve the following linear equations in two variables by elimination method and check them graphically:

1. 8x+5y-11=0,3x-4y-10=0

Solution:

Given

8x+5y = 11 …(1)
3x-4y=10 …(2)

To eliminate y we multiply equation (1) by 4 and equation (2) by 5; we get

32x+20y-44 =0…(3)

Adding, eq(3)+(2)

Read and Learn More WBBSE Solutions For Class 9 Maths

\(\begin{array}{r}
32 x+20 y-44=0 \\
15 x-20 y-50=0 \\
\hline 47 x-94=0
\end{array}\)

Class 9 Mathematics West Bengal Board

∴ 47x=94

∴ X = \(\frac{94}{47}\) = 2

Putting, x = 2 in equation (1) we get
8×2+5y-11=0
=>5y=11-16
5y=-5

∴ \(y=-\frac{5}{5}=-1\)

∴ The required solution is x = 2 &y=-1.

 

Wbbse Class 9 Maths Chapter 5.3 Linear Simultaneous Equations Solutions

2. 2x + 3y – 7 = 0, 3x + 2y -8=0

Solution:

Let

2x + 3y-7=0 …(1)

3x + 2y -8=0 …(2)

To eliminate x, we multiply equation (1) by 3 & equation (2) by 2, we get,

 

Class 9 Mathematics West Bengal Board

Subtracting we get

\(\begin{gathered}
6 x+9 y-21=0 \\
6 x+4 y-16=0 \\
(-)(-)(+) \\
5 y-5=0
\end{gathered}\)

∴ 5y = 5
∴ y = 1

Putting, y = 1 in equation (1) we get
2x + 3×1=7=0
2x+3-7=0
∴ 2x = 4

∴ \(x=\frac{4}{2}=2\)

The required solution is x = 2 & y = 1.

 

Wbbse Class 9 Linear Simultaneous Equations Exercise 5.3 Solutions

Question 2. To eliminate y, what number is multiplied with the equation 7x-5y + 2 = 0 and then added to the equation 2x + 15y+3=0?

Solution:  To eliminate y, 3 is multiplied with the equation.

Class 9 Mathematics West Bengal Board

Question 3. Let us write the least natural number by which we can multiply both the equations and get the equal co-efficients of x.

Solution: 4x-3y = 16 …(1)
6x+5y= 62….(2)

If we multiply equation (1) by 3 & equation (2) by 2, we will get the equal co- efficient of x.

Question 4. Let us solve the following linear equations in two variables by elimination method:

1. 3x+2y= 6, 2x-3y = 17

Solution:

Let

3x + 2y =-6 …(1)
2x-3y = 17 …(2)

To eliminate y, we multiply equation (1) by 3 and equation (2) by 2; we get

9x+6y=18 …(3)

Adding,eq(3)+(2)

\(\begin{aligned}
& 9 x+6 y=18 \\
& 4 x-6 y=34 \\
& \hline 13 x=52
\end{aligned}\)

Class 9 Mathematics West Bengal Board

∴ \(x=\frac{52}{13}\)

Putting, x = 4 in equation (1) we get
3×4+2y=6
2y=-6
∴ 2y=-6

∴ \(y=-\frac{6}{2}=-3\)

The required solution is x = 4 & y = – 3

2. 2x + 3y = 32, 11y-9x=3

Solution:

Given

2x + 3y = 32 …(1)
– 9x+11y = 3 …(2)

To eliminate x, we multiply equation (1) by 9 & equation (2) by 2, we get

18x+27y=288…(3)

Adding,eq(3)+(2)

\(\begin{array}{r}
18 x+27 y=288 \\
-18 x+22 y=\quad 6 \\
\hline 49 y=294
\end{array}\)

Class 9 Maths WB Board

∴ \(y=\frac{294}{49}=6\)

Putting, y 6 in equation (1) we get
2x + 3 x 6 = 32
or, 2x = 32-18
or, 2x = 14

or, \(x=\frac{14}{2}\)

∴ X=7.
The required solution is x = 7 & y = 6

Class 9 Wbbse Maths Linear Equations Exercise 5.3 Solved Problems

3. x + y = 48, x+4= \(\frac{5}{2}(y+4)\)

Solution:

Given

x + y = 48 …(1)

x+4=\(\frac{5}{2}(y+4)\)  …(2)

or, 2x+8=5y + 20
or, 2x-5y=20-8
or, 2x-5y = 12 …(3)

To eliminate y, we multiply equation (1) by 5 and equation (2) by 1, we get

5x+5y =240…(4)

Adding eq(4) +(3)

\(\begin{aligned}
& 5 x+5 y=240 \\
& 2 x-5 y=12 \\
& (-)(+) \quad(-) \\
& \hline 7 x=252
\end{aligned}\)

Class 9 Maths WB Board

Putting x = 36 in equation (1) we get
36+ y = 48
∴ y= 48 – 36 = 12

The required solution is x = 36 & y = 12.

4. \(\frac{x}{2}+\frac{y}{3}=8, \quad \frac{5 x}{4}-3 y=-3\)

Solution: \(\frac{x}{2}+\frac{y}{3}=8\)  ….(1)

\(\frac{5 x}{4}-3 y=-3\) …(2)

To eliminate y, we multiply equation (1) by & equation (2) by 1, we get \(\frac{x}{2}+\frac{y}{3}=8\)

\(\begin{aligned}
& \frac{9 x}{2}+3 y=72 \\
& \frac{5 x}{4}-3 y=-3
\end{aligned}\)

Wbbse Class 9 Maths Chapter 5.3 Linear Equations Notes

Adding,

\(\frac{9 x}{2}+\frac{5 x}{4}=69\)

 

⇒ \(\frac{18 x+5 x}{4}=69\)

23x = 69 Χ 4 …(3)

∴ \(x=\frac{69 \times 4}{23}=12\)

Putting, x = 12 in equation (1) we get,

⇒ \(\frac{12}{2}+\frac{y}{3}=8\)

∴ \(\frac{y}{3}=8-6=2\)

The required soluition is x = 12 & y = 6.

5. \(3 x-\frac{2}{y}=5, x+\frac{4}{y}=4\)

Solution:

Given

\(3 x-\frac{2}{y}=5\) …(1)

\(x+\frac{4}{y}=\) …(2)

To eliminate y we multiply equation no. (1) by 2 & equation no. (2) by 1, we get

\(6 x-\frac{4}{y}=10\) …(3)

Adding, eq (3) + eq(2)

\(
\begin{aligned}
& 6 x-\frac{4}{y}=10 \\
& x+\frac{4}{y}=4
\end{aligned}
7 x=14\)

Class 9 Math Chapter 5 WBBSE

or, x = 14/7
or, x = 2

Putting, x = 2 in equation no. (2) we get

\(
2+\frac{4}{y}=4
or, \frac{4}{y}=2
or, y=\frac{4}{2}
or, y=2\)

 

The required solution x = 2 & y = 2.

West Bengal Board Class 9 Linear Simultaneous Equations Exercise 5.3 Solutions

6. \(\frac{x}{2}+\frac{y}{3}=1, \frac{x}{3}+\frac{y}{2}=1\)

Solution:
\(\frac{x}{2}+\frac{y}{3}=1\) …(1)

\(\frac{x}{3}+\frac{y}{2}=1\) …(2)

The eliminate x, we multiply equation no. (1) by 1/3 & equation no. (2) by 1/2, we get

\(
\begin{aligned}
& \frac{x}{6}+\frac{y}{9}=\frac{1}{3} \\
& \frac{x}{6}+\frac{y}{4}=\frac{1}{2}
\end{aligned}\)

Class 9 Math Chapter 5 WBBSE

Subtracting \(\frac{y}{9}-\frac{y}{4}=\frac{1}{3}-\frac{1}{2}
\)

 

\(\begin{aligned}
& \Rightarrow \frac{4 y-9 y}{36}=\frac{2-3}{6} \\
& \Rightarrow \frac{-5 y}{36}=\frac{-1}{6} \\
& therefore   y=\frac{1}{6} \times \frac{36}{5}=\frac{6}{5}
\end{aligned}\)

 

Putting y = 6/5 in equation no. (1) we get

\(\begin{aligned}
& \Rightarrow \frac{x}{2}+\frac{6}{5} \times \frac{1}{3}=1 \\
& \Rightarrow \frac{x}{2}=1-\frac{2}{5}=\frac{3}{5} \\
& therefore x=\frac{6}{5}
\end{aligned}\)

 

∴ The required solution, x = 6/5 & y = 6/5

Wbbse Class 9 Linear Equations Exercise 5.3 Important Questions

7. \(\frac{x+y}{2}+\frac{3 x-5 y}{4}=2, \frac{x}{14}+\frac{y}{18}=1\)

Solution:

\(\frac{x+y}{2}+\frac{3 x-5 y}{4}=2\) …(1)

\(\frac{x}{14}+\frac{y}{18}=1\)….(2)

Class 9 Math Chapter 5 WBBSE

\(\frac{x+y}{2}+\frac{3 x-5 y}{4}=2\)

 

or, \(\frac{2 x+2 y+3 x-5 y}{4}=2\)

or, 5x-3y = 8 …(3)

\(\frac{x}{14}+\frac{y}{18}=1\)

 

or, \(\frac{9 x+7 y}{126}\) = 1

or, 9x + 7y = 126 ….(4)

To eliminate y, we multiply equation no. (1) by 7 & equation no. (2) by 3; we get

\(\begin{aligned}
& 35 x 121 y=56 \\
& 27 x+21 y=378 \\
& \text { Adding, } 62 x=434
\end{aligned}\)

Class 9 Math Chapter 5 WBBSE

or, \(x=\frac{434}{62}\)

∴ x = 7

Putting x = 7 in equation no. (2), we get
9×7+ 7y = 126
⇒126-63 = 63

or, y= \(\frac{63}{7}\)

∴ y = 9

The required solution, x = 7, & y = 9.

Wbbse Class 9 Maths Linear Equations Exercise 5.3

8. \(\frac{x y}{x+y}=\frac{1}{5}, \frac{x y}{x-y}=\frac{1}{9}\)

Solution: \(\frac{x y}{x+y}=\frac{1}{5}\) …(1)

⇒ x+y =5x-y

\(\frac{x y}{x-y}=\frac{1}{9}\) ….(2)

⇒ x + y = 9xy

Adding eq (1)  and eq (2) we get,

x + y = 5xy ….(1)
x – y = 9xy ….(2)

2x=14xy

or, 2 = 14y

\(or, \quad \mathrm{y}=\frac{2}{14}
or, y=\frac{1}{7}\)

 

Again, subtracting  eq (1) and (2)

x + y = 5xy
x – y = 9xy

we get 2y= -4xy
=> -4x=2

or, \(x=\frac{2}{-4}\)

∴ \(x=-\frac{1}{2}\)

∴ The required solution, x=-1/2 , y = 1/7

9. \(\frac{1}{x-1}+\frac{1}{y-2}=3, \frac{2}{x-1}+\frac{3}{y-2}=5\)

Solution:

\(\frac{1}{x-1}+\frac{1}{y-2}=3\) …(1)

\(\frac{2}{x-1}+\frac{3}{y-2}=5\) …(2)

To eliminate x, we multiply equation no. (1) by 2 & equation no. (2) by 1, we get

\(
\begin{aligned}
& \frac{2}{x-1}+\frac{2}{y-2}=6 \\
& \frac{2}{x-1}+\frac{3}{y-2}=5 \\
& (-) \quad(-) \quad(-)
\end{aligned}\)

Subtracting, \(\begin{aligned}\frac{2}{y-2}-\frac{3}{y-2}=1
\end{aligned}\)

\(\begin{aligned}
& \Rightarrow \quad \frac{2-3}{y-2}=1 \\
& \Rightarrow \quad \frac{-1}{y-2}=1
\end{aligned} \)

 

∴ y-2=-1
∴ y = 1
Putting, y = 1 in equation (1) we get

\(\begin{aligned}
& \frac{1}{x-1}+\frac{1}{1-2}=3 \\
& \frac{1}{x-1}=3+1 \\
& x-1=\frac{1}{4} \\
& x=\frac{5}{4}
\end{aligned}\)

Wbbse 9th Class Maths Linear Simultaneous Equations Step By Step Solutions Exercise 5.3

∴ \(x=1 \frac{1}{4}\)

∴ \(x=\frac{5}{4} \& y=1\)

10. \(\frac{14}{x+y}+\frac{3}{x-y}=5, \frac{21}{x+y}-\frac{1}{x-y}=2\)

Solution:

\(\frac{14}{x+y}+\frac{3}{x-y}=5\) …(1)

multiplying equation (1) by 1

\(\frac{21}{x+y}-\frac{1}{x-y}=2\) …(2)

and equation (2) by 3, we get,

\(\frac{63}{x+y}-\frac{3}{x-y}=6\) …(3)

 

eq(3) + eq (1)

\(\begin{aligned}
& \frac{63}{x+y}-\frac{3}{x-y}=6 \\
& \frac{14}{x+y}+\frac{3}{x-y}=5
\end{aligned}\)

Adding, \(\frac{77}{x+y} \quad=11\)

 

or, x + y = 7

or, x – y = 1 …..(5)

or, x+y = 7 ….(4)

Adding eq(5) + eq(4) we get

2x = 8
⇒ x =4 & y=7-4 = 3

∴ x= 4; y =3

11. \(\frac{x+y}{5}-\frac{x-y}{4}=\frac{7}{20}, \frac{x+y}{3}-\frac{x-y}{2}+\frac{5}{6}=0\)

Solution: \(\frac{x+y}{5}-\frac{x-y}{4}=\frac{7}{20} \ldots \ldots \times(1)\) …(1)

 

\(\frac{x+y}{3}-\frac{x-y}{2}=-\frac{5}{6} \ldots \ldots \times\left(\frac{1}{2}\right)\) …(2)

 

\(\begin{aligned}
& \frac{x+y}{5}-\frac{x-y}{4}=\frac{7}{20} \\
& \frac{x+y}{6}-\frac{x-y}{4}=-\frac{5}{12} \\
& -\quad+\quad +\quad \\
& \text { subracting, } \frac{x+y}{5}-\frac{x+y}{6}=\frac{7}{20}+\frac{5}{12} \\
& \end{aligned}\)

 

\(\begin{aligned}
& \Rightarrow \quad \frac{6(x+y)-5(x+y)}{30}=\frac{21+25}{60} \\
& \Rightarrow \quad \frac{x+y}{30}=\frac{46}{60} \\
& \Rightarrow \quad x+y=\frac{46 \times 30}{60} \\
& \Rightarrow \quad y=23
\end{aligned}\)

 

∴Putting, x + y = 23 in equation (1)

\(\begin{aligned}
& \Rightarrow \frac{23}{5}-\frac{x-y}{4}=\frac{7}{20} \\
& \Rightarrow \frac{-(x-y)}{4}=\frac{7}{20}-\frac{23}{5} \\
& \Rightarrow \frac{7-92}{20} \\
& \Rightarrow x-y=-\left(\frac{-85}{20} \times 4\right)=17
\end{aligned}\)

 

\(\begin{aligned}
&  therefore x+y=23 \\
&\frac{x-y=17}{2 x=40} \\
&\end{aligned}\)

 

∴ x = 20 & y = 23-20=3
∴ x 20, y = 3

Wbbse Class 9 Maths Methods To Solve Linear Simultaneous Equations Exercise 5.3

12. x + y = a + b, ax-by= \(a^2-b^2\)

Solution: x + y = a + b …(1) Χ b we get

ax-by= \(a^2-b^2\) …(2) by 1, we get

Multiplying equation (1) by b & equation (2) by 1, we get

\(b x+b y=a b+b^2\)

 

Adding eq(3)+eq(2)

\(\begin{aligned}
& b x+b y=a b+b^2 \\
& a x-b y=a^2-b^2
\end{aligned}\)

Adding, \(a x+b x=a^2 a b \)

or, (a + b) x = a(a+b)

or, \(x=\frac{a(a+b)}{(a+b)}\)

or x = a

Putting the value of x in equation (1)
a+y=a+b
or, y=a+b-a
or, y=b

∴ Solution is x = a ,y = b

13. \(\frac{x+a}{a}=\frac{y+b}{b}, a x-b y=a^2-b^2\)

Solution: \(\frac{x+a}{a}=\frac{y+b}{b}\)…(1)

\(a x-b y=a^2-b^2\)…(2)

 

From equation (1) \(\frac{x+a}{a}=\frac{y+b}{b}\)
or, bx+ab = ay + ab
or, bx-ay = ab- ab
or, bx – ay = 0 …(3)

Multiplying equation (2) by a & equation (3) by b

\(\begin{aligned}
& a^2 x-a b y=a\left(a^2-b^2\right) \\
& b^2 x-a b y=0 \\
& (-) \quad(+) \\
& \text {subtracting, } a^2 x-b^2 x=a\left(a^2-b^2\right)
\end{aligned}\)

 

or, \(\left(a^2-b^2\right) x=a\left(a^2-b^2\right)\)

or, \(x=\frac{a\left(a^2-b^2\right)}{\left(a^2-b^2\right)}\)

or, x = a

Putting the value of x in equation (3), we get
b.a – ay = 0
or, – ay = – ab

or, \(y=\frac{-a b}{-a}\)

or, y = b

∴ x=a, & y = b

14. ax + by = c, \(a^2 x+b^2 y=c^2\)

Solution:  ax + by = c …(1)

\(a^2 x+b^2 y=c^2\)…(2)

Multiplying equation (1) by a & equation (2) by 1,

\(a^2 x+a b y=a c\)…(3)

 

\(\begin{aligned}
& a^2 x+a b y=a c \\
& a^2 x+b^2 y=c^2 \\
& (-) \quad(-) \quad(-)
\end{aligned}\)

Subtracting, \(aby -b^2 y=a c-c^2\)

 

or, by(ab) = c(a – c)

or, \({y}=\frac{c(a-c)}{b(a-b)}\)

Putting the value of y in equation (1)

\(a x+b \cdot \frac{c(a-c)}{b(a-b)}=c\)

or, \(\quad a x=c-\frac{c(a-c)}{(a-b)}\)

or, \(\quad a x=\frac{a c-b c-a c+c^2}{a-b}\)

or, \(\quad x=\frac{c(c-b)}{a(a-b)}\)

therefore \(x=\frac{c(c-b)}{a(a-b)} \& y=\frac{c(a-c)}{b(a-b)}\)

 

15. ax + by = 1, \(b x+a y=\frac{(a+b)^2}{a^2+b^2}-1\)

Solution: ax + by = 1 ….(1)

\(b x+a y=\frac{(a+b)^2}{a^2+b^2}-1\)…(2)

 

or, \(b x+a y=\frac{a^2+2 a b+b^2-a^2-b^2}{a^2+b^2}\)

or, \(b x+a y=\frac{2 a b}{a^2+b^2}\) …(3)

Multiplying equation (1) by a & equation (3) by b, we get

\(
\begin{aligned}
& a^2 x+a b y=a \\
& b^2 x+a b y=\frac{2 a b^2}{a^2+b^2} \\
& (-) \quad(-) \quad(-)
\end{aligned}\)

Subtracting, \(a^2 x-b^2 x=a-\frac{2 a b^2}{a^2+b^2}\)

\(or, \quad\left(a^2-b^2\right) x=\frac{a^3+a b^2-2 a b^2}{a^2+b^2}\)

 

or, \(\quad\left(a^2-b^2\right) x=\frac{a^3-a b^2}{a^2+b^2}\)

or, \(\quad x=\frac{a\left(a^2-b^2\right)}{\left(a^2+b^2\right)\left(a^2-b^2\right)}\)

or, \(x=\frac{a}{a^2+b^2}\)

Putting the value of x in equation (1) we get

\(a. \frac{a}{a^2+b^2}+b y=1\)

 

or, \(\quad \frac{a^2}{a^2+b^2}+by=1 \)

or, \(\quad b y=1-\frac{a^2}{a^2+b^2}\)

or, \(\quad b y=\frac{a^2+b^2-a^2}{a^2+b^2}\)

or, \(\quad b y=\frac{b^2}{a^2+b^2}\)

\(or, \quad y=\frac{b^2}{b\left(a^2+b^2\right)}
or, \quad y=\frac{b}{a^2+b^2}
therefore  x=\frac{a}{a^2+b^2} \& y=\frac{b}{a^2+b^2}\)

Class 9 Wbbse Linear Equations Chapter 5.3 Solved Exercises

16. (7x-y-6)²+(14x+2y-16)²=0

Solution:

Given

7x-y-6 …(1)
14x+2y-16…..(2)

From equation (1) 7x-y=6 ….(3)
From equation (2) 14x + 2y = 16…(4)

Multiplying the equation (3) by 2 & equation (2) by 1

14x-2y= 12 ….(5)

Adding, eq(5) + eq(4)

\(\begin{aligned}
& 14 x-2 y=12 \\
& 14 x+2 y=16 \\
& \hline 28 x=28
\end{aligned}\)

 

or, \(x=\frac{28}{28}\)

or, x = 1

Putting the value of x in equation (4), we get
14 x 1 + 2y = 16

or, 2y=16-14
or,2y=2

or, \(y=\frac{2}{2}\)

∴ X=1&y= 1