Question

Solve: $\frac{1}{2x}-\frac{y}{3}=-2$ and $\frac{x}{2}+\frac{y}{4}=\frac{1}{2}$

• A. $-2,\frac{1}{3}$
• B. $3,-3$
• C. $-\frac{1}{2},3$
• D. None

Answer 1

Answer: (C)

$-\frac{1}{2},3$

Given equations are
$\frac{1}{2x}-\frac{y}{3}=-2$ --- (1)
And $\frac{x}{2}+\frac{y}{4}=\frac{1}{2}$ --- (2)

Multiplying equation $\left(1\right)$ with $\frac{1}{4}$ we get, $\frac{1}{8x}-\frac{y}{12}=-\frac{1}{2}$ ----- equation $\left(3\right)$

Multiplying equation $\left(2\right)$ with $\frac{1}{3}$ we get, $\frac{x}{6}+\frac{y}{12}=\frac{1}{6}$ ----- equation $\left(4\right)$

Adding equation $\left(3\right)$ and
$\left(4\right)$, we get $\frac{1}{8x}+\frac{x}{6}=-\frac{1}{3}=>x=-\frac{1}{2}$

Substituting $x=-\frac{1}{2}$
in the equation $\left(1\right)$, we get $\frac{1}{-2\left(\frac{1}{2}\right)}-\frac{y}{3}=-2=>y=3$