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

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

The given linear equations are

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

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 the second equation by -3

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

2) now, the equations are

localid="1644387875506" $$\begin{gathered} x+\frac{y}{3}=-1 \\ -x-\frac{3}{2}y=-3 \end{gathered}$$

just add the above $2$equations.

$x+\frac{y}{3}=-1-x-\frac{3}{2}y=-3-\frac{7}{6}y=-4$

$$\begin{gathered} \frac{7}{6}y=4 \\ y=\frac{24}{7} \end{gathered}$$

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

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

put $y=\frac{24}{7}$

$$\begin{gathered} x+\frac{24}{21}=-1 \\ x+\frac{8}{7}=-1 \\ x=-1-\frac{8}{7} \\ x=-\frac{15}{7} \end{gathered}$$

The solution of the above linear equation is$\left(\frac{-15}{7},\frac{24}{7}\right)$

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

$$\begin{gathered} \begin{array}{l}x+\frac{1}{3}y=-1\\ -\frac{15}{7}+\frac{1}{3}\left(\frac{24}{7}\right)=-1\\ -\frac{15}{7}+\frac{8}{7}=-1\\ \frac{-15+8}{7}=-1\end{array} \\ \frac{-7}{7}=-1 \\ -1=-1 \\ \begin{array}{l}\frac{1}{3}x+\frac{1}{2}y=1\\ \frac{1}{3}\left(\frac{-15}{7}\right)+\frac{1}{2}\left(\frac{24}{7}\right)=1\\ -\frac{5}{7}+\frac{12}{7}=1\\ \frac{-5+12}{7}=1\end{array} \\ \frac{7}{7}=1 \\ 1=1 \end{gathered}$$

This is true, hence the solution is correct.