To solve the system of inequalities graphically, we need to consider the following inequalities:
1. [tex]\( y < -\frac{3}{2} x - 7 \)[/tex]
2. [tex]\( y < \frac{1}{2} x + 1 \)[/tex]
Here's the step-by-step solution:
1. Graph the boundary lines:
- For the inequality [tex]\( y < -\frac{3}{2} x - 7 \)[/tex], we first draw the line [tex]\( y = -\frac{3}{2} x - 7 \)[/tex]. This line has a slope of [tex]\(-\frac{3}{2}\)[/tex] and a y-intercept of [tex]\(-7\)[/tex].
- For the inequality [tex]\( y < \frac{1}{2} x + 1 \)[/tex], we first draw the line [tex]\( y = \frac{1}{2} x + 1 \)[/tex]. This line has a slope of [tex]\(\frac{1}{2}\)[/tex] and a y-intercept of [tex]\(1\)[/tex].
2. Determine the region for each inequality:
- The region below the line [tex]\( y = -\frac{3}{2} x - 7 \)[/tex] represents the values where [tex]\( y < -\frac{3}{2} x - 7 \)[/tex].
- The region below the line [tex]\( y = \frac{1}{2} x + 1 \)[/tex] represents the values where [tex]\( y < \frac{1}{2} x + 1 \)[/tex].
3. Find the intersection point:
- To find where these two lines intersect, solve the equations simultaneously:
[tex]\[
-\frac{3}{2}x - 7 = \frac{1}{2}x + 1
\][/tex]
[tex]\[
-\frac{3}{2}x - \frac{1}{2}x = 1 + 7
\][/tex]
[tex]\[
-2x = 8
\][/tex]
[tex]\[
x = -4
\][/tex]
Substitute [tex]\( x = -4 \)[/tex] into one of the equations to find [tex]\( y \)[/tex]:
[tex]\[
y = \frac{1}{2}(-4) + 1
\][/tex]
[tex]\[
y = -2 + 1
\][/tex]
[tex]\[
y = -1
\][/tex]
So, the intersection point is [tex]\((-4, -1)\)[/tex].
4. Identify a point in the solution region:
- Now we should identify a point that lies below both lines. A point in the solution set must satisfy both inequalities. A good choice can be slightly below and to the left of the intersection point [tex]\((-4, -1)\)[/tex].
Let's choose the point [tex]\((-5, -2)\)[/tex].
5. Verify the point:
- Check to ensure [tex]\((-5, -2)\)[/tex] satisfies both inequalities:
- For [tex]\( y < -\frac{3}{2} x - 7 \)[/tex]:
[tex]\[
-2 < -\frac{3}{2}(-5) - 7
\][/tex]
[tex]\[
-2 < 7.5 - 7
\][/tex]
[tex]\[
-2 < 0.5 \quad \text{(True)}
\][/tex]
- For [tex]\( y < \frac{1}{2} x + 1 \)[/tex]:
[tex]\[
-2 < \frac{1}{2}(-5) + 1
\][/tex]
[tex]\[
-2 < -2.5 + 1
\][/tex]
[tex]\[
-2 < -1.5 \quad \text{(True)}
\][/tex]
Since the point [tex]\((-5, -2)\)[/tex] satisfies both inequalities, it is in the solution set.
Thus, the coordinates of a point in the solution set are:
[tex]\[
\boxed{(-5, -2)}
\][/tex]
