Question

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

Rearrange $3x + y = -4$ to $y = -3x -4$.

Step2: Find points for first line

For $y=\frac{1}{2}x+3$:

• When $x=0$, $y=3$ → $(0, 3)$
• When $x=2$, $y=\frac{1}{2}(2)+3=4$ → $(2, 4)$

Step3: Find points for second line

For $y=-3x-4$:

• When $x=0$, $y=-4$ → $(0, -4)$
• When $x=-2$, $y=-3(-2)-4=2$ → $(-2, 2)$

Step4: Find intersection algebraically

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 $y=\frac{1}{2}x+3$: $y=\frac{1}{2}(-2)+3=2$

Answer:

The solution (intersection point) is $(-2, 2)$. To graph:

1. Plot $(0, 3)$ and $(2, 4)$, draw the line for $y=\frac{1}{2}x+3$.
2. Plot $(0, -4)$ and $(-2, 2)$, draw the line for $3x+y=-4$.
3. The lines cross at $(-2, 2)$.