Question

Solve the following pair of linear equations by the elimination method and the substitution method :
(i) x + y = 5 and 2x – 3y = 4
(ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
(iv)$\frac{x}{2} + \frac{2y}{3} = -1$ and $x- \frac{y}{3} = 3$

Answer 1

(i) By elimination method
x + y = 5 …………….(1)
2x – 3y = 4 ………….(2)

Multiplying equation (1) by 2, we obtain
2x + 2y = 10 .............(3)

Subtracting equation (2) from equation (3), we obtain
5y = 6
y = $\frac{6}{5}$

Substituting the value in equation (1), we obtain
x = 5 - $\frac{6}{5}$ = $\frac{19}{5}$
∴ x = $\frac{19}{5}$ , y =$\frac{6}{5}$

By substitution method
From equation (1), we obtain
x = 5-y ...............(4)

Putting this value in equation (2), we obtain
2(5-y) - 3y = 4
-5y = -6
y= $\frac{6}{5}$

Substituting the value in equation (4), we obtain
x = 5 -$\frac{6}{5}$ = $\frac{19}{5}$

∴ x = $\frac{19}{5}$ , y = $\frac{6}{5}$

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(ii) By elimination method
3x + 4y = 10 .......(1)
2x - 2y + 2............(2)

Multiplying equation (2) by 2, we obtain
4x - 4y = 4 .............(3)

Adding equation (1) and (3), we obtain
7x = 14
x =2

Substituting in equation (1), we obtain
6+ 4y = 10
4y = 4
y=1

Hence, x = 2, y = 1

**By substitution method **
From equation (2), we obtain
x= 1+y .........(4)

Putting this value in equation (1), we obtain
3(1+y) +4y = 10
7y = 7
y=1

Substituting the value in equation (4), we obtain
x = 1 +1 = 2
∴ x = 2, y =1

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(iii) By elimination method
3x - 5y -4 = 0 .............(1)
and 9x = 2y + 7
9x -2y -7 = 0...............(2)

Multiplying equation (1) by 3, we obtain
9x - 15y -12 =0............(3)

Subtracting equation (3) from equation (2), we obtain
13y = -5
y= $-\frac{5}{13}$
Substituting in equation (1), we obtain
3x +$\frac{25}{13}$ -4 =0
3x = 27 /13
x= $\frac{9}{13}$

∴ x = $\frac{9}{13}$, y = $-\frac{5}{13}$

By substitution method
From equation (1), we obtain
x = $\frac{5y +4}{ 3}$.............(4)

Putting this value in equation (2), we obtain
9($\frac{5y +4}{ 3}$) -2y -7 =0
13y = -5
y= $-\frac{5}{13}$

Substituting the value in equation (4), we obtain
x =$\frac{5 (-\frac{5}{13}) + 4 }{3}$
x = $\frac{9}{13}$

∴ x = $\frac{9}{13}$, y = $-\frac{5}{13}$

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**(iv) By elimination method **
$\frac{x}{2} + \frac{2y} {3} = -1$
$3x + 4y = -6$.............(1)
and
$x- \frac{y}{3} = 3$
$3x -y = 9$.............(2)

Subtracting equation (2) from equation (1), we obtain
5y = -15
y= -3 ............(3)

Substituting this value in equation (1), we obtain
3x - 12 = -6
3x= 6
x = 2

Hence, x = 2, y = −3

By substitution method
From equation (2), we obtain
$x = \frac{y+9}{3}$ .............(4)

Putting this value in equation (1), we obtain
3($\frac{y+9}{3}$) + 4y + -6
5y = −15
y = -3

Substituting the value in equation (4), we obtain
x =$\frac{-3 + 9}{3}$ = 2

∴ x = 2, y = −3