Solve Systems of Equations by Graphing: example 5.5

Solve the system by graphing: ${\begin{matrix} x + y = 2 \\ x - y = 4 \end{matrix} .$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Solve each system by graphing: ${\begin{matrix} x + y = 6 \\ x - y = 2 \end{matrix} .$

How did it go?


To find the intercepts, let x = 0 and solve for y, then let y = 0 and solve for x.	$\begin{matrix} \begin{matrix} x + y & = & 2 \\ 0 + y & = & 2 \\ y & = & 2 \end{matrix} & \begin{matrix} x + y & = & 2 \\ x + 0 & = & 2 \\ x & = & 2 \end{matrix} \end{matrix}$

To find the intercepts, let x = 0 then let y = 0.	$\begin{matrix} \begin{matrix} x - y & = & 4 \\ 0 - y & = & 4 \\ - y & = & 4 \\ y & = & - 4 \end{matrix} & \begin{matrix} x - y & = & 4 \\ x - 0 & = & 4 \\ x & = & 4 \end{matrix} \end{matrix}$
Graph the line.
Determine the point of intersection.	The lines intersect at (3, −1).
Check the solution in both equations.	$\begin{matrix} x + y & = & 2 & x - y & = & 4 \\ 3 + (−1) & \overset{?}{=} & 2 & 3 - (−1) & \overset{?}{=} & 4 \\ 2 & = & 2 ✓ & 4 & = & 4 ✓ \end{matrix}$ The solution is (3, −1).


To find the intercepts, let x = 0 and solve for y, then let y = 0 and solve for x.	$\begin{matrix} \begin{matrix} x + y & = & 2 \\ 0 + y & = & 2 \\ y & = & 2 \end{matrix} & \begin{matrix} x + y & = & 2 \\ x + 0 & = & 2 \\ x & = & 2 \end{matrix} \end{matrix}$

To find the intercepts, let x = 0 then let y = 0.	$\begin{matrix} \begin{matrix} x - y & = & 4 \\ 0 - y & = & 4 \\ - y & = & 4 \\ y & = & - 4 \end{matrix} & \begin{matrix} x - y & = & 4 \\ x - 0 & = & 4 \\ x & = & 4 \end{matrix} \end{matrix}$
Graph the line.
Determine the point of intersection.	The lines intersect at (3, −1).
Check the solution in both equations.	$\begin{matrix} x + y & = & 2 & x - y & = & 4 \\ 3 + (−1) & \overset{?}{=} & 2 & 3 - (−1) & \overset{?}{=} & 4 \\ 2 & = & 2 ✓ & 4 & = & 4 ✓ \end{matrix}$ The solution is (3, −1).