Question

Solve the following pair of equations by the elimination method and the substitution method:

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

• A. $x=1$ and $y=-3$
• B. $x=5$ and $y=2$
• C. $x=4$ and $y=6$
• D. $x=2$ and $y=-3$

Answer 1

The correct option is D $x=2$ and $y=-3$

Given Equations are:
$\frac{x}{2}+\frac{2y}{3}=-1$ .....(1)
$x-\frac{y}{3}=3$ .....(2)
From equation (1) will get
$\Rightarrow 3x+4y=-6$ ....(3)
From equation (2) will get
$\Rightarrow 3x-y=9$ ....(4)
$\text{Elimination Method}$
Now multiplying equation (3) by $1$ and equation (4) by $4$ and adding will get,
$\Rightarrow 1(3x+4y)+4(3x-y)=1\times -6+4\times 9$
$\Rightarrow 3x+4y+12x-4y=-6+36$
$\Rightarrow 15x=30$
$\Rightarrow x=2$
Now putting $x$ in equation (4) will get
$\Rightarrow 3\times 2-y=9$
$\Rightarrow 6-y=9$
$\Rightarrow -y=9-6$
$\Rightarrow y=-3$
Hence, $x=2$ and $y=-3$
$\text{Substitution Method}$
Now substituting $y$ from equation (3)
$\Rightarrow 4y=-6-3x$
$\Rightarrow y=\frac{-6-3x}{4}$
Now putting $y$ in equation (4) will get
$\Rightarrow 3x-\left(\frac{-6-3x}{4}\right)=9$
$\Rightarrow 12x+6+3x=36$
$\Rightarrow 15x=36-6$
$\Rightarrow x=2$
Now putting $x$ in equation (3) will get
$\Rightarrow 3\times 2+4y=-6$
$\Rightarrow 6+4y=-6$
$\Rightarrow 4y=-6-6$
$\Rightarrow 4y=-12$
$\Rightarrow y=-3$
Hence, $x=2$ and $y=-3$