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. To determine which side to shade, we can test a point not on the line, for example, the origin \((0,0)\). Substitute \( x = 0 \) and \( y = 0 \) into the inequality: \( 0 < -3(0)+4\), which simplifies to \( 0 < 4 \), a true statement. So we shade the region that includes the origin (below the dashed line \( y=-3x + 4\)).

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 the inequality is non - strict, \( y\) can be equal to \( 3x - 2 \)). Then, shade the region above the line. Test the origin \((0,0)\) in the inequality: \( 0\geq3(0)-2\), which simplifies to \( 0\geq - 2 \), a true statement. So we shade the region that includes the origin (above the solid line \( y = 3x - 2\)).

Step3: Find the Intersection Region

The solution to the system of inequalities is the region that is shaded for both inequalities. So we look for the area that is below the dashed line \( y=-3x + 4\) and above the solid line \( y = 3x - 2\).

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 two lines intersect at the point \((1,1)\).

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 of the two lines at \((1,1)\). When graphing, the dashed line has a slope of - 3 and y - intercept of 4, the solid line has a slope of 3 and y - intercept of - 2, and the solution region is the overlapping shaded area of the two individual shaded regions.