Question

solve the following system of inequalities graphically on the set of axes below. state the coordinates of a point in the solution set.
$y < -dfrac{2}{3}x + 5$
$y geq dfrac{4}{3}x - 1$
answer
point: ( , )

Explanation:

Step1: Find intersection of two lines

First, find the intersection point of \( y = -\frac{2}{3}x + 5 \) and \( y=\frac{4}{3}x - 1 \). Set them equal:
\( -\frac{2}{3}x + 5=\frac{4}{3}x - 1 \)
Add \( \frac{2}{3}x \) to both sides: \( 5 = 2x - 1 \)
Add 1 to both sides: \( 6 = 2x \), so \( x = 3 \)
Substitute \( x = 3 \) into \( y=\frac{4}{3}x - 1 \): \( y=\frac{4}{3}(3)-1 = 4 - 1 = 3 \). Intersection is \( (3,3) \).

Step2: Analyze the inequalities

For \( y<-\frac{2}{3}x + 5 \), the line is dashed (since \( < \)) and shade below.
For \( y\geq\frac{4}{3}x - 1 \), the line is solid (since \( \geq \)) and shade above.
The solution set is the overlapping region. A point in this region: test \( (0,0) \)? No, \( 0\geq - 1 \) (good for second inequality), but \( 0<5 \) (good for first), wait no: \( y\geq\frac{4}{3}x - 1 \) at \( x = 0 \), \( y\geq - 1 \), and \( y<-\frac{2}{3}(0)+5 = 5 \). But better to use the intersection or a point in the overlap. Let's take \( (3,3) \): check \( 3<-\frac{2}{3}(3)+5= - 2 + 5 = 3 \)? No, \( 3<3 \) is false. Take \( (0,3) \): \( 3<-\frac{2}{3}(0)+5 = 5 \) (true), \( 3\geq\frac{4}{3}(0)-1=-1 \) (true). Wait, \( (3,3) \) is on the boundary of the second inequality but not the first. Let's take \( (3,2) \): \( 2<-\frac{2}{3}(3)+5 = 3 \) (true), \( 2\geq\frac{4}{3}(3)-1 = 3 \)? No. Wait, \( (0,0) \): \( 0<5 \) (true), \( 0\geq - 1 \) (true). Wait, no, \( y\geq\frac{4}{3}x - 1 \) at \( x = 0 \), \( y\geq - 1 \), which is true, and \( y<5 \) is true. But maybe a better point: let's solve for the region. The overlap is between the two lines, from \( x \) where the lines meet (and beyond? Wait, the lines intersect at \( (3,3) \), and the first line has y-intercept 5, second has y-intercept -1. The solution region is between the two lines, above \( y=\frac{4}{3}x - 1 \) and below \( y = -\frac{2}{3}x + 5 \). Let's take \( (3,3) \) is on the boundary of the second inequality, but \( y<-\frac{2}{3}x + 5 \) at \( x = 3 \) is \( y<3 \), so \( (3,2) \): \( 2<3 \) (true), \( 2\geq\frac{4}{3}(3)-1 = 3 \)? No. Wait, \( (0,3) \): \( 3<5 \) (true), \( 3\geq - 1 \) (true). Yes, \( (0,3) \) is in the solution set? Wait, \( y\geq\frac{4}{3}x - 1 \) at \( x = 0 \), \( y\geq - 1 \), which is true, and \( y<5 \) is true. Wait, but maybe \( (3,3) \) is not in the first inequality (since \( < \)), so a point like \( (3,2) \): no, \( 2\geq\frac{4}{3}(3)-1 = 3 \)? No. Wait, I made a mistake. Let's re - evaluate. The intersection is \( (3,3) \). The first inequality is \( y<-\frac{2}{3}x + 5 \) (below the dashed line), the second is \( y\geq\frac{4}{3}x - 1 \) (above the solid line). So the overlapping region is where \( \frac{4}{3}x - 1\leq y<-\frac{2}{3}x + 5 \). Let's find a point: when \( x = 3 \), \( y \) must be less than 3 and greater than or equal to 3? No, wait, the lines intersect at \( (3,3) \), so the region is between the two lines, with \( y \) above the solid line and below the dashed line. So a point like \( (3,3) \) is on the solid line (so included in the second inequality) but not on the dashed line (so \( y<3 \) is false). Wait, maybe \( (0,0) \): \( 0\geq - 1 \) (yes), \( 0<5 \) (yes). Wait, but \( y\geq\frac{4}{3}x - 1 \) at \( x = 0 \), \( y\geq - 1 \), which is true, and \( y<5 \) is true. But is \( (0,0) \) in the overlap? Let's check the graph: the solid line \( y=\frac{4}{3}x - 1 \) at \( x = 0 \) is \( y=-1 \), so above that line (including) and below \( y = -\frac{2}{3}x + 5 \) (which at \( x = 0 \) is \( y = 5 \)). So \( (0,0) \) is in the overlap. Wait, but earlier mistake: \( (0,0) \): \( 0\geq\frac{4}{3}(0)-1=-1 \) (true), \( 0<-\frac{2}{3}(0)+5 = 5 \) (true). So \( (0,0) \) is a solution? Wait, no, the second inequality is \( y\geq\frac{4}{3}x - 1 \), so \( y \) must be greater than or equal to that line. At \( x = 0 \), the line is \( y=-1 \), so \( y = 0 \) is above, good. And \( y = 0 \) is below \( y = 5 \), good. Wait, but maybe the intended point is \( (3,3) \) but no, because \( 3<3 \) is false. Wait, let's take \( (3,2) \): \( 2<-\frac{2}{3}(3)+5 = 3 \) (true), \( 2\geq\frac{4}{3}(3)-1 = 3 \) (false). Oops. Take \( (0,3) \): \( 3<5 \) (true), \( 3\geq - 1 \) (true). Yes, \( (0,3) \): check \( y\geq\frac{4}{3}x - 1 \) at \( x = 0 \), \( 3\geq - 1 \) (true), \( y<-\frac{2}{3}x + 5 \) at \( x = 0 \), \( 3<5 \) (true). So \( (0,3) \) is in the solution set. Wait, but maybe the intersection point's neighbor. Alternatively, \( (3,3) \) is on the solid line, so \( y\geq\frac{4}{3}x - 1 \) is true (since \( 3\geq3 \)), but \( y<-\frac{2}{3}x + 5 \) is \( 3<3 \), which is false. So \( (3,3) \) is not in the solution set. Wait, I think I messed up the inequality directions. Let's re - graph mentally:

- \( y = -\frac{2}{3}x + 5 \): slope - 2/3, y - intercept 5 (goes down from (0,5) to (3,3) to (7.5,0)).
- \( y=\frac{4}{3}x - 1 \): slope 4/3, y - intercept - 1 (goes up from (0,-1) to (3,3) to (6,7)).

So the region below \( y = -\frac{2}{3}x + 5 \) (dashed) and above \( y=\frac{4}{3}x - 1 \) (solid) is the area between the two lines, above the solid line and below the dashed line. So a point in this region: let's take \( (3,3) \) is on the solid line (so included in the second inequality) but not the dashed line. Wait, no, the dashed line means the boundary is not included. So a point like \( (3,2) \): below dashed line (2 < 3) and above solid line (2 ≥ 3? No, 2 < 3). Wait, this is confusing. Wait, at \( x = 3 \), the solid line is \( y = 3 \), dashed line is \( y = 3 \). So the overlapping region is where \( \frac{4}{3}x - 1\leq y<-\frac{2}{3}x + 5 \). So for \( x = 3 \), \( 3\leq y<3 \), which is only \( y = 3 \), but \( y<3 \) excludes \( y = 3 \). So the solution set is the region where \( x<3 \) (since the lines meet at \( x = 3 \))? Wait, no, when \( x>3 \), the dashed line is below the solid line, so no overlap. When \( x<3 \), the dashed line is above the solid line, so overlap is above solid line and below dashed line. So take \( x = 0 \), \( y \) between - 1 and 5, so \( (0,0) \): \( 0\geq - 1 \) (yes), \( 0<5 \) (yes). \( (0,3) \): \( 3\geq - 1 \) (yes), \( 3<5 \) (yes). So \( (0,3) \) is in the solution set. Wait, but let's check \( (3,3) \): \( 3\geq3 \) (yes for second inequality), \( 3<3 \) (no for first). So \( (3,3) \) is not in. Let's take \( (2,2) \): \( 2<-\frac{2}{3}(2)+5=-\frac{4}{3}+5=\frac{11}{3}\approx3.666 \) (true), \( 2\geq\frac{4}{3}(2)-1=\frac{8}{3}-1=\frac{5}{3}\approx1.666 \) (true). So \( (2,2) \) is in the solution set.

Answer:

(2, 2) (or other valid points like (0, 0), (3, 2) is invalid, (0, 3) is valid, (2, 3) is valid as 3 < 5 and 3 ≥ 4/3*2 - 1 = 8/3 - 1 = 5/3 ≈ 1.666. Let's confirm (2,3): 3 < -4/3 + 5 = 11/3 ≈ 3.666 (true), 3 ≥ 8/3 - 1 = 5/3 ≈ 1.666 (true). So (2,3) is also valid. But a simple one is (3,3) is on the boundary of the second inequality but not the first, so (2,2) is safe.)