Question
For solving pair of equation, in this exercise use the method of elimination by equating coefficients :$\frac{5 y}{2}-\frac{x}{3}=8; \frac{y}{2}+\frac{5 x}{3}=12$

Answer

The given pair of linear equations are
$\frac{5 y}{2}-\frac{x}{3}=8$
$\Rightarrow-\frac{x}{3}+\frac{5 y}{2}=8\dots.....(1) [$ On Similifying $]$
$\frac{y}{2}+\frac{5 x}{3}=12$
$\Rightarrow \frac{5 x}{3}+\frac{y}{2}=12\dots.....(2) [$ On Similifying $]$
Multiply equation $(1)$ by $5 ,$ we get
$-\frac{5 x}{3}+\frac{25 y}{2}=40\dots......(3)$
Adding equation $(3)$ and $(2)$
$-\frac{5 x}{3}+\frac{25 y}{2}=40$
$ + \frac{5 x}{3}+\frac{y}{2}=12$
$\frac{26 y}{2}=52$
$ \Rightarrow 13 y=52$
$ \Rightarrow \mathrm{y}=4$
Substituting $y=4$ in equation $(1)$, We get
$-\frac{x}{3}+\frac{5(4)}{2}=8$
$ \Rightarrow-\frac{x}{3}=8-10$
$ \Rightarrow \mathrm{x}=6$
$\therefore$ Solution is $\mathrm{x}=6$ and $\mathrm{y}=4$.

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