Question

question
solve the following system of inequalities graphically on the set of axes below. state the coordinates of a point in the solution set.
$y < \frac{5}{2}x - 3$
$y \geq -x + 4$
(there is a coordinate plane image with shaded region and a dashed line, along with change line and change shade buttons. also, a section for entering a point coordinate with point: ( , ) and a submit answer button.)

Explanation:

Step1: Analyze the first inequality \( y < \frac{5}{2}x - 3 \)

This is a linear inequality with a slope of \( \frac{5}{2} \) and a y - intercept of - 3. The line is dashed (because of the \( < \) sign) and we shade the region below the line.

Step2: Analyze the second inequality \( y \geq -x + 4 \)

This is a linear inequality with a slope of - 1 and a y - intercept of 4. The line is solid (because of the \( \geq \) sign) and we shade the region above the line.

Step3: Find the intersection of the two shaded regions

The solution set is the region that is shaded for both inequalities. To find a point in the solution set, we can test a point. Let's test the point \( (4,2) \).
For \( y < \frac{5}{2}x - 3 \): Substitute \( x = 4,y = 2 \). \( \frac{5}{2}(4)-3=10 - 3 = 7 \), and \( 2<7 \), so the point satisfies the first inequality.
For \( y\geq - x + 4 \): Substitute \( x = 4,y = 2 \). \( - 4 + 4=0 \), and \( 2\geq0 \), so the point satisfies the second inequality. Another valid point could be \( (3,1) \), but let's check the graph. From the graph, we can also see that the point \( (4,2) \) lies in the overlapping shaded region. Also, let's check the point \( (2,2) \) (the one in the input box). For \( y < \frac{5}{2}x-3 \), when \( x = 2 \), \( \frac{5}{2}(2)-3=5 - 3 = 2 \), but \( y = 2 \) is not less than 2, so \( (2,2) \) is not valid. Let's find a correct point. Let's solve the system of equations \( y=\frac{5}{2}x - 3 \) and \( y=-x + 4 \) to find the intersection of the two lines.
Set \( \frac{5}{2}x-3=-x + 4 \)
\( \frac{5}{2}x+x=4 + 3 \)
\( \frac{5x + 2x}{2}=7 \)
\( \frac{7x}{2}=7 \)
\( x = 2 \)
Substitute \( x = 2 \) into \( y=-x + 4 \), we get \( y=-2 + 4=2 \). But the first inequality is strict (\( y < \frac{5}{2}x-3 \)), so the intersection point of the lines is not in the solution set (since at \( x = 2,y = 2 \), \( y=\frac{5}{2}x-3 \)). Let's take \( x = 3 \). For \( y\geq - x + 4 \), when \( x = 3 \), \( y\geq1 \). For \( y < \frac{5}{2}x-3 \), when \( x = 3 \), \( \frac{5}{2}(3)-3=\frac{15}{2}-3=\frac{15 - 6}{2}=\frac{9}{2}=4.5 \). So a point like \( (3,2) \): For \( y < \frac{5}{2}x-3 \), \( \frac{5}{2}(3)-3 = 4.5 \), \( 2<4.5 \). For \( y\geq - x + 4 \), \( - 3+4 = 1 \), \( 2\geq1 \). So \( (3,2) \) is a valid point. Also, let's check the point \( (4,2) \) again. As we saw earlier, it satisfies both inequalities.

Answer:

A point in the solution set is \( (4,2) \) (other valid points like \( (3,2) \) also exist)