Question

solve the following system of inequalities graphically on the set of axes below. state the coordinates of a point in the solution set.
$y > \frac{3}{2}x - 6$
$y \leq -x - 1$
(graph of coordinate axes is shown)
answer attempt 1 out of 2
point: (1, blank) submit answer

Explanation:

Step1: Find intersection of inequalities

First, find the intersection of \( y > \frac{3}{2}x - 6 \) and \( y \leq -x - 1 \). Let's solve the system of equations \( y=\frac{3}{2}x - 6 \) and \( y=-x - 1 \) to find the intersection point of the two lines.
Set \( \frac{3}{2}x - 6=-x - 1 \), add \( x \) to both sides: \( \frac{3}{2}x+x - 6=- 1 \), which is \( \frac{5}{2}x - 6=-1 \). Add 6 to both sides: \( \frac{5}{2}x=5 \), multiply both sides by \( \frac{2}{5} \): \( x = 2 \). Then substitute \( x = 2 \) into \( y=-x - 1 \), we get \( y=-3 \). So the intersection point of the two lines is \( (2,-3) \).

Step2: Test a point in the solution region

We can test a point that satisfies both inequalities. Let's try \( x = 0 \). For \( y > \frac{3}{2}(0)-6=-6 \) and \( y \leq -0 - 1=-1 \). So a point like \( (0,-2) \): check \( -2 > - 6 \) (true) and \( -2 \leq -1 \) (true). Another point: let's take \( x = 2 \), but \( y=-3 \) satisfies \( y \leq -x - 1 \) (since \( -3\leq - 3 \)) and \( y > \frac{3}{2}(2)-6=3 - 6=-3 \)? No, \( -3\) is not greater than \( -3 \). Let's take \( x = 0 \), \( y=-2 \) as above, or \( x = 1 \). For \( x = 1 \), \( y > \frac{3}{2}(1)-6=\frac{3}{2}-6=-\frac{9}{2}=-4.5 \) and \( y \leq -1 - 1=-2 \). So \( y=-2 \) when \( x = 1 \): \( -2 > - 4.5 \) (true) and \( -2 \leq -2 \)? Wait, \( y \leq -x - 1 \) when \( x = 1 \) is \( y \leq -2 \). So \( y=-2 \) is valid. Wait, \( (1,-2) \): check \( -2 > \frac{3}{2}(1)-6=\frac{3}{2}-6=-\frac{9}{2}=-4.5 \) (true) and \( -2 \leq -1 - 1=-2 \) (true, since it's equal). Wait, \( y \leq -x - 1 \) allows equality. So \( (1,-2) \) is valid? Wait, \( y \leq -x - 1 \) at \( x = 1 \) is \( y \leq -2 \), so \( y=-2 \) is okay. And \( y > \frac{3}{2}x - 6 \) at \( x = 1 \) is \( y > -4.5 \), which \( -2 \) is. Alternatively, let's take \( (0,-1) \): check \( -1 > -6 \) (true) and \( -1 \leq -1 \) (true, since it's equal). Wait, \( y=-1 \) when \( x = 0 \): \( y \leq -0 - 1=-1 \) (true) and \( y > -6 \) (true). So \( (0,-1) \) is also valid. But let's confirm with the graph. The region is where \( y \) is above the line \( y=\frac{3}{2}x - 6 \) (dashed line) and below or on the line \( y=-x - 1 \) (solid line). So a point like \( (0,-1) \), \( (0,-2) \), \( (1,-2) \), etc. Let's take \( (0,-1) \):

Check \( y > \frac{3}{2}x - 6 \): \( -1 > 0 - 6=-6 \) (true).

Check \( y \leq -x - 1 \): \( -1 \leq 0 - 1=-1 \) (true, since it's equal).

So \( (0,-1) \) is in the solution set.

Answer:

\((0, -1)\) (other valid points like \((0, -2)\), \((1, -2)\) are also correct)