In the following exercise, solve the systems of equations by elimination.

$\left\{\begin{array}{l}x+\frac{y}{2}=\frac{3}{2}\\ \frac{x}{5}-\frac{y}{5}=3\end{array}\right.$

The given equations are

$$\begin{gathered} x+\frac{y}{2}=\frac{3}{2} \\ \frac{x}{5}-\frac{y}{5}=3 \end{gathered}$$

To calculate the value of $y$, we have to eliminate the $x$term. For which we will do the following changes in the equation, so that $x$term can be easily eliminated.

1) Multiply the 2^nd linear equation by -5

$$\begin{gathered} -5\left(\frac{x}{5}-\frac{y}{5}=3\right) \\ -x+y=-15 \end{gathered}$$

2) now the equations are

$$\begin{gathered} x+\frac{y}{2}=\frac{3}{2} \\ -x+y=-15 \end{gathered}$$

Now, simply add the equations

$x+\frac{y}{2}=\frac{3}{2}-x+y=-15\frac{3y}{2}=-\frac{27}{2}$

$$\begin{gathered} 3y=-27 \\ y=-9 \end{gathered}$$

As we have the value of $y$now, we can use this value to calculate the value of $x$

We will put the value of $y$in any equation and then calculate the value of $x$.

Let us take the equation.

$x+\frac{y}{2}=\frac{3}{2}$

put y= -9

$$\begin{gathered} x+\frac{(-9)}{2}=\frac{3}{2} \\ x=\frac{3}{2}+\frac{9}{2} \\ x=6 \end{gathered}$$

The solution of the above linear equation by elimination is $(6,-9)$.

Checking the solution by putting the value of $x,y$in the equation, we get

$$\begin{gathered} x+\frac{y}{2}=\frac{3}{2} \\ 6+\frac{-9}{2}=\frac{3}{2} \\ \frac{12-9}{2}=\frac{3}{2} \\ \frac{3}{2}=\frac{3}{2} \\ \frac{x}{5}-\frac{y}{5}=3 \\ \frac{6}{5}-\frac{-9}{5}=3 \\ \frac{6+9}{5}=3 \\ \frac{15}{5}=3 \\ 3=3 \end{gathered}$$

This is true, hence the solution is correct.