To determine which ordered pair is in the solution set of the given system of inequalities, we can analyze the inequalities and the corresponding regions on the coordinate plane:
- Inequalities:
- \( y > \frac{3}{2}x - 1 \): This represents the area above the line, which has a positive slope and a y-intercept of -1.
- \( y < \frac{3}{2}x - 1 \): This represents the area below the same line.
Since both inequalities reference the same line, but one is above and one is below, there is no region where both inequalities are satisfied simultaneously. This means there is "no solution" for the system of inequalities.
Now let's evaluate the given ordered pairs to confirm:
-
(–5, 2):
- Substitute \( x = -5 \) into \( y > \frac{3}{2}(-5) - 1 \).
- \( y > -\frac{15}{2} - 1 \rightarrow y > -\frac{17}{2} \). For \( 2 > -\frac{17}{2} \) this is true. The second inequality \( y < \frac{3}{2}(-5) - 1 \) gives \( y < -\frac{15}{2} - 1 \rightarrow y < -\frac{17}{2} \), and \( 2 < -\frac{17}{2} \) is false.
-
(2, 2):
- Substitute \( x = 2 \) into \( y > \frac{3}{2}(2) - 1\).
- \( y > 3 - 1 \rightarrow y > 2 \), so \( 2 > 2 \) is false. The second inequality \( y < \frac{3}{2}(2) - 1 \) gives \( y < 2 \), \( 2 < 2 \) is false.
-
(5, 2):
- Substitute \( x = 5 \) into \( y > \frac{3}{2}(5) - 1\).
- \( y > \frac{15}{2} - 1 \rightarrow y > \frac{13}{2} \). Here, \( 2 > \frac{13}{2} \) is false. The second inequality \( y < \frac{3}{2}(5) - 1 \) gives \( y < \frac{15}{2} - 1 \rightarrow y < \frac{13}{2} \), so \( 2 < \frac{13}{2} \) is true, but since the first inequality is false, this does not count.
After evaluating each pair, none satisfy both conditions; therefore, the conclusion is:
There is no solution.