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