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

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

To do: To solve: $\frac{x}{2}+\frac{2y}{3}=-1$ and $x-\frac{y}{3}=3$.

Solution:

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

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

Let us multiply $( ii)$ by $2$,

$2x-\frac{2y}{3}=6\ ......\ ( iii)$

Add $( i)$ and $( iii)$

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

$\Rightarrow \frac{5x}{2}=5$

$\Rightarrow x=2$, Put this value in $( i)$

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

$\Rightarrow 1+\frac{2y}{3}=-1$

$\Rightarrow \frac{2y}{3}=-1-1$

$\Rightarrow \frac{2y}{3}=-2$

$y=-3$

Thus, $x=2$ and $y=-3$.

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Updated on: 10-Oct-2022

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