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 is \(-3\) and the y - intercept is \( 4 \). 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 \( y\geq3x - 2 \)

Next, graph the line \( y = 3x - 2 \). The slope is \( 3\) and the y - intercept is \(-2\). 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. 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

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 could also use the other equation), we get \( y=3(1)-2 = 1\). So the intersection point of the two lines is \((1,1)\). The overlapping region is bounded by the dashed line \( y=-3x + 4\) (above the line \( y = 3x - 2\) and below the line \( y=-3x + 4\)) and the solid line \( y = 3x - 2\), with the vertex at \((1,1)\).

Answer:

The solution is the region that is above the solid line \( y = 3x - 2\), below the dashed line \( y=-3x + 4\), and includes the intersection point \((1,1)\) (and the line \( y = 3x - 2\) but not the line \( y=-3x + 4\)). To graph it, plot the solid line \( y = 3x - 2\) (with slope 3, y - intercept - 2), the dashed line \( y=-3x + 4\) (with slope - 3, y - intercept 4), shade the overlapping region (above \( y = 3x - 2\) and below \( y=-3x + 4\)).