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: 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)
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\(\begin{array}{r}32 x+20 y-44=0 \\
15 x-20 y-50=0 \\
\hline 47 x-94=0
\end{array}\)
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.
2. 2x + 3y – 7 = 0, 3x + 2y -8=0
Solution: 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.
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: 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: 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
3. x + y = 48, x+4= \(\frac{5}{2}(y+4)\)
Solution: 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 \(\frac{x}{2}+\frac{y}{3}=8\)
\(\begin{aligned}& \frac{9 x}{2}+3 y=72 \\
& \frac{5 x}{4}-3 y=-3
\end{aligned}\)
Class 9 Maths WB Board
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: \(3 x-\frac{2}{y}=5\) …(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.
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
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.
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, \(\begin{aligned}\frac{2}{y-2}-\frac{3}{y-2}=1
\end{aligned}\)
& \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
& \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}\)
Class 9 Math Chapter 5 WBBSE
∴ \(x=1 \frac{1}{4}\)
∴ \(x=\frac{5}{4} \& y=1\)
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
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, \(\quad b y=\frac{b^2}{a^2+b^2}\)
\(or, \quad y=\frac{b^2}{b\left(a^2+b^2\right)}or, \quad y=\frac{b}{a^2+b^2}
therefore x=\frac{a}{a^2+b^2} \& y=\frac{b}{a^2+b^2}\)
16. (7x-y-6)2+(14x+2y-16)2=0
Solution: 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
