Solve Systems of Equations by Graphing: example 5.6

Solve the system by graphing: ${\begin{matrix} y = 6 \\ 2 x + 3 y = 12 \end{matrix} .$

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: ${\begin{matrix} y = −1 \\ x + 3 y = 6 \end{matrix} .$

How did it go?


We know the first equation represents a horizontal line whose y-intercept is 6.
The second equation is most conveniently graphed using intercepts.
To find the intercepts, let x = 0 and then y = 0.
Graph the lines.
Determine the point of intersection.	The lines intersect at (−3, 6).
Check the solution to both equations.	$\begin{matrix} y & = & 6 & 2 x + 3 y & = & 12 \\ 6 & \overset{?}{=} & 6 ✓ & 2 (−3) + 3 (6) & \overset{?}{=} & 12 \\ 2 & = & 2 & −6 + 18 & \overset{?}{=} & 12 \\ 12 & = & 12 ✓ \end{matrix}$
	The solution is (−3, 6).


We know the first equation represents a horizontal line whose y-intercept is 6.
The second equation is most conveniently graphed using intercepts.
To find the intercepts, let x = 0 and then y = 0.
Graph the lines.
Determine the point of intersection.	The lines intersect at (−3, 6).
Check the solution to both equations.	$\begin{matrix} y & = & 6 & 2 x + 3 y & = & 12 \\ 6 & \overset{?}{=} & 6 ✓ & 2 (−3) + 3 (6) & \overset{?}{=} & 12 \\ 2 & = & 2 & −6 + 18 & \overset{?}{=} & 12 \\ 12 & = & 12 ✓ \end{matrix}$
	The solution is (−3, 6).