Question

Question 1 (iv)
Solve the following pair of linear equations by the elimination method and the substitution method:
$\frac{x}{2}+\frac{2y}{3}=-1$ and $x-\frac{y}{3}=3$

Answer 1

$\frac{x}{2}+\frac{2y}{3}=-1$ and $x-\frac{y}{3}=3$

By elimination method:

$\frac{x}{2}+\frac{2y}{3}=-1$ ... (i)
$x-\frac{y}{3}=3$ ... (ii)

Multiplying equation (i) by 2, we get
$x+\frac{4y}{3}=-2$ ... (iii)
$x-\frac{y}{3}=3$ ... (ii)
Subtracting equation (ii) from equation (iii), we get

$\frac{5y}{3}=-5$

Dividing by 5 and multiplying by 3, we get

$y=\frac{-15}{5}$
y = - 3

Putting this value in equation (ii), we get

$x-\frac{y}{3}=3$ ... (ii)
$x-\frac{(-3)}{3}=3$

x + 1 = 3
x = 2

Hence our answer is x = 2 and y = -3.

By substitution method

$x-\frac{y}{3}=3$ ... (ii)

Add y/3 both side, we get

$x=3+\frac{y}{3}$... (iv)

Putting this value in equation (i), we get

$\frac{x}{2}+\frac{2y}{3}=-1$ ... (i)

$\frac{(3+\frac{y}{3})}{2}+\frac{2y}{3}=-1$

$\frac{3}{2}+\frac{y}{6}+\frac{2y}{3}=-1$

Multiplying by 6, we get

9 + y + 4y = - 6

5y = -15

y = - 3

Hence, our answer is x = 2 and y = -3.