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

$\left\{\begin{array}{l}\frac{1}{3}x-y=-3\\ \frac{2}{3}x+\frac{5}{2}y=3\end{array}\right.$

The given linear equations are :

$$\begin{gathered} \frac{1}{3}x-y=-3 \\ \frac{2}{3}x+\frac{5}{2}y=3 \end{gathered}$$

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

1) Multiply first equation with -2

localid="1644750728669" $$\begin{gathered} -2\left(\frac{1}{3}x-y=-3\right) \\ \frac{-2}{3}x+2y=6 \end{gathered}$$

2) now, the equations are

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

simply add the above two linear equations

$\frac{-2}{3}x+2y=6\frac{2}{3}x+\frac{5}{2}y=3\frac{9}{2}y=9$

$$\begin{gathered} \frac{1}{2}y=1 \\ y=2 \end{gathered}$$

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

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

Let us take the equation

$\frac{1}{3}x-y=-3$

put y=2

$$\begin{gathered} \frac{1}{3}x-2=-3 \\ \frac{1}{3}x=-1 \\ x=-3 \end{gathered}$$

The solution of the above linear equation is: $\left(-3,2\right)$

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

$$\begin{gathered} \frac{1}{3}x-y=-3 \\ \frac{1}{3}(-3)-2=-3 \\ -1-2=-3 \\ -3=-3 \\ \frac{2}{3}x+\frac{5}{2}y=3 \\ \frac{2}{3}(-3)+\frac{5}{2}\left(2\right)=3 \\ -2+5=3 \\ 3=3 \end{gathered}$$

This is true, hence the solution is correct.