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) Elimination Method
Given equations:
x + y = 5 …………….(1)
2x – 3y = 4 ………….(2)
Multiply equation (1) by 2: 2x + 2y = 10 .....(3)
Subtract equation (2) from equation (3): 5y = 6 => y = \( \frac{6}{5} \)
Substitute y in equation (1): x = 5 - \( \frac{6}{5} \) = \( \frac{19}{5} \)
Solution: x = \( \frac{19}{5} \), y = \( \frac{6}{5} \)

Substitution Method
From equation (1): x = 5 - y ...............(4)
Substitute x in equation (2): 2(5-y) - 3y = 4 => 10 - 2y - 3y = 4 => -5y = -6 => y = \( \frac{6}{5} \)
Substitute y in equation (4): x = 5 - \( \frac{6}{5} \) = \( \frac{19}{5} \)
Solution: x = \( \frac{19}{5} \), y = \( \frac{6}{5} \)

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(ii) Elimination Method
Given equations:
3x + 4y = 10 .......(1)
2x - 2y + 2 = 0 => 2x - 2y = -2 .....(2)
Multiply equation (2) by 2: 4x - 4y = -4 .............(3)
Add equation (1) and (3): 7x = 6 => x = \( \frac{6}{7} \)
Substitute x in equation (1): 3(\( \frac{6}{7} \)) + 4y = 10 => \( \frac{18}{7} \) + 4y = 10 => 4y = 10 - \( \frac{18}{7} \) = \( \frac{52}{7} \) => y = \( \frac{13}{7} \)
Solution: x = \( \frac{6}{7} \), y = \( \frac{13}{7} \)

Substitution Method
From equation (2): 2x = 2y - 2 => x = y - 1 .....(4)
Substitute x in equation (1): 3(y - 1) + 4y = 10 => 3y - 3 + 4y = 10 => 7y = 13 => y = \( \frac{13}{7} \)
Substitute y in equation (4): x = \( \frac{13}{7} \) - 1 = \( \frac{6}{7} \)
Solution: x = \( \frac{6}{7} \), y = \( \frac{13}{7} \)

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(iii) Elimination Method
Given equations:
3x - 5y - 4 = 0 .............(1)
9x - 2y - 7 = 0...............(2)
Multiply equation (1) by 3: 9x - 15y - 12 = 0............(3)
Subtract equation (3) from equation (2): 13y + 5 = 0 => 13y = -5 => y = \( -\frac{5}{13} \)
Substitute y in equation (1): 3x - 5(-\( \frac{5}{13} \)) - 4 = 0 => 3x + \( \frac{25}{13} \) - 4 = 0 => 3x = 4 - \( \frac{25}{13} \) = \( \frac{27}{13} \) => x = \( \frac{9}{13} \)
Solution: x = \( \frac{9}{13} \), y = \( -\frac{5}{13} \)

Substitution Method
From equation (1): x = \( \frac{5y + 4}{3} \) .............(4)
Substitute x in equation (2): 9(\( \frac{5y + 4}{3} \)) - 2y - 7 = 0 => 3(5y + 4) - 2y - 7 = 0 => 15y + 12 - 2y - 7 = 0 => 13y = -5 => y = \( -\frac{5}{13} \)
Substitute y in equation (4): x = \( \frac{5 (-\frac{5}{13}) + 4 }{3} \) = \( \frac{-\frac{25}{13} + 4}{3} \) = \( \frac{\frac{27}{13}}{3} \) = \( \frac{9}{13} \)
Solution: x = \( \frac{9}{13} \), y = \( -\frac{5}{13} \)

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(iv) Elimination Method
Given equations:
\( \frac{x}{2} + \frac{2y} {3} = -1 \) => 3x + 4y = -6 .............(1)
\( x- \frac{y}{3} = 3 \) => 3x - y = 9 .............(2)
Subtract equation (2) from equation (1): 5y = -15 => y = -3 ............(3)
Substitute y in equation (1): 3x + 4(-3) = -6 => 3x - 12 = -6 => 3x = 6 => x = 2
Solution: x = 2, y = -3

Substitution Method
From equation (2): 3x = y + 9 => x = \( \frac{y+9}{3} \) .............(4)
Substitute x in equation (1): 3(\( \frac{y+9}{3} \)) + 4y = -6 => y + 9 + 4y = -6 => 5y = -15 => y = -3
Substitute y in equation (4): x = \( \frac{-3 + 9}{3} \) = \( \frac{6}{3} \) = 2
Solution: x = 2, y = -3