graph the solution to the following system of inequalities. \ny < -3x + 4 \ny ≥ 3x - 2

Explanation:

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

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

• When $x=0$, $y=4$ (y-intercept: $(0,4)$)
• When $y=0$, $0=-3x+4 \implies x=\frac{4}{3}\approx1.33$ (x-intercept: $(\frac{4}{3},0)$) Draw this as a dashed line (since the inequality is strict $<$). Shade the region below this line (because $y$ is less than the expression).

Step2: Graph boundary of $y\geq3x-2$

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

• When $x=0$, $y=-2$ (y-intercept: $(0,-2)$)
• When $y=0$, $0=3x-2 \implies x=\frac{2}{3}\approx0.67$ (x-intercept: $(\frac{2}{3},0)$) Draw this as a solid line (since the inequality is non-strict $\geq$). Shade the region above this line (because $y$ is greater than or equal to the expression).

Step3: Find intersection of lines

Set $-3x+4=3x-2$ to find where the boundaries cross: $$ \begin{align*} -3x+4&=3x-2\ 4+2&=3x+3x\ 6&=6x\ x&=1 \end{align*} $$ Substitute $x=1$ into $y=3x-2$: $y=3(1)-2=1$. The intersection point is $(1,1)$.

Step4: Identify overlapping shaded region

The solution to the system is the area that is shaded under the dashed line $y=-3x+4$ and above the solid line $y=3x-2$, bounded on the left/right by the lines meeting at $(1,1)$.

Answer:

1. Draw a dashed line through $(0,4)$ and $(\frac{4}{3},0)$ for $y=-3x+4$, shade below it.
2. Draw a solid line through $(0,-2)$ and $(\frac{2}{3},0)$ for $y=3x-2$, shade above it.
3. The overlapping shaded region (including the solid line, excluding the dashed line) is the solution, with the boundary lines intersecting at $(1,1)$.