Question

Find the value of $x$ and $y$ using elimination method:
$\frac{x}{2}+\frac{y}{3}=1$ and $\frac{x}{4}+\frac{y}{7}=2$

• A. $(44,-63)$
• B. $(-41,63)$
• C. $(44,63)$
• D. $(-44,-63)$

Answer 1

The correct option is A $(44,-63)$

$\frac{x}{2}+\frac{y}{3}=1\left(Given\right)$

$\Rightarrow$ $3x+2y=6$ $⟶{eq}^n\left(i\right)$

$\frac{x}{4}+\frac{y}{7}=2\left(Given\right)$

$\Rightarrow$ $7x+4y=56$ $⟶{eq}^n\left(ii\right)$

Multiplying ${eq}^n\left(i\right)$ by $2$, we get

$6x+4y=12$ $⟶{eq}^n\left(i\right)$

On subtracting ${eq}^n\left(iii\right)$ from $\left(ii\right)$, we have

$7x+4y-6x-4y=56-12$

$\Rightarrow$ $x=44$

On substituting the value of $x$ in ${eq}^n\left(i\right)$, we get

$3\times$ $44+2y=6$

$\Rightarrow$ $2y=6-132$

$\Rightarrow$ $y=-63$