WBBSE Solutions For Class 9 Maths Chapter 5 Linear Simultaneous Equations Exercise 5.4

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

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

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),

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

Class 9 Math Chapter 5 WBBSE

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.

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.

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.

Class IX Maths Solutions WBBSE 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)

Class IX Maths Solutions WBBSE

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

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

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).

\(or, \frac{1}{x}=\frac{5-2}{6}
or, \frac{1}{x}=\frac{3}{6}
or, \quad 3 x=6
or, \quad x=\frac{6}{3}
or, \quad x=2
therefore   x=2 \)

Class 9 Maths WB Board

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

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