Question

Solve a system of linear equations by graphing.

$\left\{\begin{array}{l}-x+y=2\\ 2x+y=-4\end{array}\right.$

Short answer

The solution to the given linear system of equations is $(-2,0)$.

Step-by-step solution

Step 1. Given information

The given linear system of equations is as follows.

$\left\{\begin{array}{l}-x+y=2\\ 2x+y=-4\end{array}\right.$

The given linear system of equations.

The standard form of slope-intercept form is given by $y=mx+b$, where $m$denotes the slope and $b$denotes the $y$-intercept.

Assume the first equation $-x+y=2$.

$y=x+2$

So, the slope $m$is $1$and the $y$-intercept $b$is $2$.

Assume the second equation$2x+y=-4$.

$y=-2x-4$

So, the slope $m$is $-2$and the $y$-intercept $b$is $-4$.

Step 3. Graph the two given lines on the same rectangular coordinate system.

The graph of the two given lines is shown below:

Step 4. Determine the point of intersection by using the graph.

From the graph, observe that the point of intersection is $(-2,0)$.

Step 5. Check whether the point of intersection (-2,0) is a solution to both equations.

Substitute the values of the variables into each equation to check whether a point is a solution to a given system of two equations.

If the point makes both equations true, it is a solution to the given system.

Substitute $-2$for $x$and $0$for $y$into $-x+y=2$.

$\begin{array}{rcl}-(-2)+0& =& 2\\ 2& =& 2\left(\text{True}\right)\end{array}$

Substitute $-2$for $x$and $0$for $y$into $\begin{array}{rcl}2x+y& =& -4\end{array}$.

localid="1647605432145" $\begin{array}{rcl}2(-2)+0& =& -4\\ -4& =& -4\left(\text{True}\right)\end{array}$

Conclude that the point $(-2,0)$made both equations true.

Hence, $(-2,0)$is a solution.