Question

ath/grade-8/solve-a-system-of-equations-by-graphing
$2x - 3y = -18$
$y = \frac{1}{3}x + 5$
click to select points on the graph.
$2x - 3y = -18$
$y = \frac{1}{3}x + 5$

Explanation:

Step1: Rewrite first equation to slope-intercept form

Start with $2x - 3y = -18$. Isolate $y$:

$$\begin{align*} -3y &= -2x - 18 \\ y &= \frac{2}{3}x + 6 \end{align*}$$

Step2: Find intersection of the two lines

Set the two equations equal:

$$ \frac{1}{3}x + 5 = \frac{2}{3}x + 6 $$

Solve for $x$:

$$\begin{align*} 5 - 6 &= \frac{2}{3}x - \frac{1}{3}x \\ -1 &= \frac{1}{3}x \\ x &= -3 \end{align*}$$

Step3: Find y-value for x=-3

Substitute $x=-3$ into $y = \frac{1}{3}x + 5$:

$$ y = \frac{1}{3}(-3) + 5 = -1 + 5 = 4 $$

Step4: Verify with the first equation

Substitute $x=-3, y=4$ into $2x - 3y = -18$:

$$ 2(-3) - 3(4) = -6 -12 = -18 $$

This confirms the point is valid.

Step5: Identify points for graphing $y=\frac{2}{3}x+6$

• y-intercept: when $x=0$, $y=6$ (point $(0,6)$)
• Another point: when $x=3$, $y=\frac{2}{3}(3)+6=8$ (point $(3,8)$)

Answer:

The solution to the system is the point $(-3, 4)$. To graph $2x-3y=-18$, plot the points $(0,6)$ and $(3,8)$ and draw a line through them; the intersection of this line with $y=\frac{1}{3}x+5$ is $(-3,4)$.