Question

solve the following system of inequalities graphically on the set of axes below. state the coordinates of a point in the solution set.
$y \geq -2x - 6$
$y < \frac{1}{3}x + 8$

Explanation:

Step1: Analyze the first inequality \( y \geq -2x - 6 \)

This is a linear inequality. The boundary line is \( y = -2x - 6 \), which has a slope of \(-2\) and a y - intercept of \(-6\). Since the inequality is \( \geq \), we draw a solid line for \( y=-2x - 6 \) and shade the region above the line (because for a point \((x,y)\) to satisfy \( y\geq - 2x-6\), \(y\) values are greater than or equal to the values on the line \(y = - 2x-6\)).

Step2: Analyze the second inequality \( y<\frac{1}{3}x + 8 \)

The boundary line is \( y=\frac{1}{3}x + 8 \), with a slope of \(\frac{1}{3}\) and a y - intercept of \(8\). Since the inequality is \(<\), we draw a dashed line for \( y = \frac{1}{3}x+8 \) and shade the region below the line (because \(y\) values are less than the values on the line \(y=\frac{1}{3}x + 8\)).

Step3: Find the intersection of the two shaded regions

The solution set of the system of inequalities is the region that is shaded by both inequalities. To find a point in the solution set, we can pick a point and check if it satisfies both inequalities. Let's try the point \((0,0)\):

• For \( y\geq - 2x-6 \): Substitute \(x = 0,y = 0\) into \(y\geq - 2x-6\), we get \(0\geq-2(0)-6= - 6\), which is true.
• For \( y<\frac{1}{3}x + 8 \): Substitute \(x = 0,y = 0\) into \(y<\frac{1}{3}x + 8\), we get \(0<\frac{1}{3}(0)+8 = 8\), which is true.

So \((0,0)\) is a point in the solution set. We can also find the intersection point of the two lines \(y=-2x - 6\) and \(y=\frac{1}{3}x + 8\) by setting \(-2x-6=\frac{1}{3}x + 8\).

Multiply through by 3 to eliminate the fraction: \(-6x-18=x + 24\)

Subtract \(x\) from both sides: \(-7x-18 = 24\)

Add 18 to both sides: \(-7x=42\)

Divide by \(-7\): \(x=-6\)

Substitute \(x = - 6\) into \(y=-2x-6\), we get \(y=-2(-6)-6=12 - 6 = 6\). So the intersection point of the two lines is \((-6,6)\). The solution region is bounded by the two lines (with one solid and one dashed) and the overlapping shaded area.

Answer:

A point in the solution set is \((0,0)\) (other valid points like \((-3,0)\), \((3,5)\) etc. also work. The intersection point of the two lines is \((-6,6)\), but since the second line is dashed, \((-6,6)\) is not included in the solution set as it does not satisfy \(y<\frac{1}{3}x + 8\) (it satisfies \(y=\frac{1}{3}x + 8\) when \(x=-6\), \(y = 6=\frac{1}{3}(-6)+8=-2 + 8=6\))).