Question

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solve the following system of equations graphically on the set of axes below.
$y = \frac{1}{2}x + 3$
$3x + y = -4$
plot two lines by clicking the graph.
click a line to delete it.

Explanation:

Step1: Rewrite 2nd equation to slope-intercept

$y = -3x - 4$

Step2: Find points for $y=\frac{1}{2}x+3$

When $x=0$, $y=3$; when $x=2$, $y=4$

Step3: Find points for $y=-3x-4$

When $x=0$, $y=-4$; when $x=-2$, $y=2$

Step4: Solve algebraically to find intersection

Set $\frac{1}{2}x + 3 = -3x - 4$
$\frac{1}{2}x + 3x = -4 - 3$
$\frac{7}{2}x = -7$
$x = -2$
Substitute $x=-2$ into $y=\frac{1}{2}x+3$:
$y=\frac{1}{2}(-2)+3 = -1 + 3 = 2$

Answer:

The solution to the system is the point $(-2, 2)$
(To graph: Plot $(0,3)$ and $(2,4)$ for $y=\frac{1}{2}x+3$, draw the line; plot $(0,-4)$ and $(-2,2)$ for $y=-3x-4$, draw the line. The lines intersect at $(-2,2)$)