Question

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

Explanation:

Step1: Graph \( y < -3x + 4 \)

First, graph the line \( y = -3x + 4 \). The slope \( m = -3 \) and the y - intercept \( b = 4 \). Since the inequality is \( y < -3x + 4 \), the line should be dashed (because the inequality is strict, \( y\) is not equal to \( -3x + 4 \)). Then, shade the region below the line. We can test a point, for example, the origin \((0,0)\): \( 0 < - 3(0)+4\) which is \( 0 < 4 \), a true statement, so we shade the region containing the origin.

Step2: Graph \( y \geq 3x - 2 \)

Next, graph the line \( y = 3x - 2 \). The slope \( m = 3 \) and the y - intercept \( b=-2\). Since the inequality is \( y\geq3x - 2 \), the line should be solid (because \( y\) can be equal to \( 3x - 2 \)). Then, shade the region above the line. We can test the origin \((0,0)\): \( 0\geq3(0)-2\) which is \( 0\geq - 2 \), a true statement, so we shade the region containing the origin.

Step3: Find the Intersection

The solution to the system of inequalities is the region where the two shaded regions overlap. To find the intersection point of the two lines \( y=-3x + 4 \) and \( y = 3x-2 \), set them equal to each other:
\[

$$\begin{align*} -3x + 4&=3x-2\\ 4 + 2&=3x+3x\\ 6&=6x\\ x&=1 \end{align*}$$

\]
Substitute \( x = 1 \) into \( y = 3x-2 \), we get \( y=3(1)-2 = 1 \). So the intersection point is \((1,1)\). The overlapping region is bounded by the dashed line \( y=-3x + 4 \), the solid line \( y = 3x-2 \), and the area that satisfies both inequalities.

(Note: Since the problem asks to graph the solution, the key steps are graphing each inequality and identifying the overlapping region. The final graph will have a dashed line for \( y=-3x + 4 \) with shading below it, a solid line for \( y = 3x-2 \) with shading above it, and the overlapping region as the solution set.)

Answer:

The solution is the region that is below the dashed line \( y=-3x + 4 \) and above the solid line \( y = 3x-2 \), with the intersection point at \((1,1)\). (To graph it, follow the steps of graphing each inequality and shading the overlapping region.)