Solve the system by graphing:$\left\{\begin{array}{l}-x+y=1\\ 3x+2y=12\end{array}\right.$

We are given a system of equations$\left\{\begin{array}{l}-x+y=1\\ 3x+2y=12\end{array}\right.$and we need to find its solution by graphing.

Solving the equation $-x+y=1$for $y$we get,

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

So the slope of the line is 1 and its $y$intercept is 1. Using this the graph of the equation is given as

Solving the equation $3x+2y=12$for $y$we get,

$\begin{array}{rcl}3x+2y-3x& =& 12-3x\\ 2y& =& -3x+12\\ \frac{2y}{2}& =& \frac{-3x+12}{2}\\ y& =& -1.5x+6\end{array}$

So the slope of the line is $-1.5$and its $y$intercept is 6.

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 $(2,3)$.

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

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

$\begin{array}{rcl}-x+y& =& 1\\ -2+3& =& 1\\ 1& =& 1\end{array}$

It is a true statement.

Again, substitute the values in the second equation.

$\begin{array}{rcl}3x+2y& =& 12\\ 3·2+2·3& =& 12\\ 6+6& =& 12\\ 12& =& 12\end{array}$

This is also a true statement.

So the point satisfies both the equations.