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.