Question

graph the solution to the following system of inequalities.
$y\leq -3x + 4$
$y > 2x - 3$

Explanation:

Step1: Graph boundary of $y \leq -3x+4$

First, identify the line $y=-3x+4$. Find intercepts:

• When $x=0$, $y=4$ (y-intercept: $(0,4)$)
• When $y=0$, $0=-3x+4 \implies x=\frac{4}{3} \approx 1.33$ (x-intercept: $(\frac{4}{3},0)$)

Draw a solid line (since inequality is $\leq$) through these points. Shade the region below the line (because $y$ is less than or equal to the line).

Step2: Graph boundary of $y > 2x-3$

Identify the line $y=2x-3$. Find intercepts:

• When $x=0$, $y=-3$ (y-intercept: $(0,-3)$)
• When $y=0$, $0=2x-3 \implies x=\frac{3}{2}=1.5$ (x-intercept: $(\frac{3}{2},0)$)

Draw a dashed line (since inequality is $>$) through these points. Shade the region above the line (because $y$ is greater than the line).

Step3: Find overlapping shaded region

The solution is the area that is shaded by both inequalities. To confirm the intersection of the two lines:
Set $-3x+4=2x-3$
$4+3=2x+3x$
$7=5x \implies x=\frac{7}{5}=1.4$
Substitute $x=1.4$ into $y=2x-3$: $y=2(1.4)-3=2.8-3=-0.2$
Intersection point: $(1.4, -0.2)$

Answer:

1. Draw a solid line through $(0,4)$ and $(\frac{4}{3},0)$, shade below it.
2. Draw a dashed line through $(0,-3)$ and $(\frac{3}{2},0)$, shade above it.
3. The overlapping shaded region (including the solid line, excluding the dashed line) is the solution, bounded by the intersection point $(1.4, -0.2)$.