91Ó°ÊÓ

We are given a system of equations:

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

and we need to find its solution by graphing.

Solve the equation $x-3y=-3$for $y$we get,

$\begin{array}{rcl}x-3y-x& =& -3-x\\ -3y& =& -x-3\\ \frac{-3y}{-3}& =& \frac{-x-3}{-3}\\ y& =& \frac{1}{3}x+1\end{array}$

So, on comparing with the slope-intercept form, the slope of the line is $\frac{1}{3}$and its $y$intercept is 1. Using this the graph of the equation is given as,

Solving the equation $x+y=5$for $y$we get.

localid="1644574421992" $\begin{array}{rcl}x+y-x& =& 5-x\\ y& =& -x+5\end{array}$

So the slope of the line is $-1$and $y$intercept is 5. So its graph can be drawn on the same plane as,

From the graph, it can be seen that the two lines intersect at the point $(3,2)$.

So $(3,2)$is the solution to the given system.

Substitute $3$for $x$and $2$for $y$in the first equation.

$\begin{array}{rcl}x-3y& =& -3\\ 3-3·2& =& -3\\ 3-6& =& -3\\ -3& =& -3\end{array}$

It is a true statement.

Again, substitute the values in the second equation.

$\begin{array}{rcl}x+y& =& 5\\ 3+2& =& 5\\ 5& =& 5\end{array}$

This is also a true statement.

So the point satisfies both the equations.