Question

set.
$y \geq \frac{1}{6}x + 4$
$y < -\frac{4}{3}x - 5$
(graph of coordinate plane with x from -10 to 10 and y from -10 to 10)
answer attempt 1 out of 2
point: (\boxed{ }, \boxed{ })
submit answer

Explanation:

Step1: Find intersection of inequalities

To find a point in the solution set, we first find the intersection of the two inequalities \( y \geq \frac{1}{6}x + 4 \) and \( y < -\frac{4}{3}x - 5 \). We can solve the system of inequalities by finding a point that satisfies both. Let's try to find the intersection of the two lines (even though the inequalities are strict/non - strict, we can use the lines to guide us).

Set \( \frac{1}{6}x + 4=-\frac{4}{3}x - 5 \)

Multiply through by 6 to eliminate fractions: \( x + 24=-8x - 30 \)

Step2: Solve for x

Add \( 8x \) to both sides: \( x + 8x+24=-8x + 8x-30 \)
\( 9x+24=-30 \)

Subtract 24 from both sides: \( 9x+24 - 24=-30 - 24 \)
\( 9x=-54 \)

Divide by 9: \( x = - 6 \)

Step3: Solve for y (using one of the line equations)

Substitute \( x=-6 \) into \( y=\frac{1}{6}x + 4 \): \( y=\frac{1}{6}(-6)+4=-1 + 4 = 3 \)

Now we check if this point satisfies the second inequality \( y<-\frac{4}{3}x - 5 \). Substitute \( x = - 6 \) into \( -\frac{4}{3}x-5 \): \( -\frac{4}{3}(-6)-5=8 - 5 = 3 \). But our point has \( y = 3 \) and the inequality is \( y<3 \), so \( ( - 6,3) \) does not satisfy the second inequality. Let's try a value of \( x\) less than - 6 (since the slope of the second line is negative, for \( x < - 6\), let's try \( x=-12 \))

For \( x=-12 \), in \( y\geq\frac{1}{6}x + 4 \): \( y\geq\frac{1}{6}(-12)+4=-2 + 4=2 \)

In \( y<-\frac{4}{3}x - 5 \): \( y<-\frac{4}{3}(-12)-5 = 16 - 5=11 \)

Let's pick \( x=-12 \) and find a \( y \) that satisfies both. Let's take \( y = 3 \) (since \( 2\leq3<11 \)). Wait, but let's check with the first inequality when \( x=-12 \), \( \frac{1}{6}(-12)+4=-2 + 4 = 2 \), and \( 3\geq2 \), and for the second inequality, \( -\frac{4}{3}(-12)-5=16 - 5 = 11 \), and \( 3<11 \). Wait, but maybe a better approach is to look at the graph. Let's consider the region. The first line \( y=\frac{1}{6}x + 4 \) has a y - intercept of 4 and slope \( \frac{1}{6} \), the second line \( y = -\frac{4}{3}x-5 \) has a y - intercept of - 5 and slope \( -\frac{4}{3} \).

Let's try \( x=-6 \), \( y = 2 \). Check first inequality: \( 2\geq\frac{1}{6}(-6)+4=-1 + 4 = 3 \)? No, \( 2<3 \). Try \( x=-18 \)

For \( x=-18 \), \( y\geq\frac{1}{6}(-18)+4=-3 + 4 = 1 \)

For \( x=-18 \), \( y<-\frac{4}{3}(-18)-5=24 - 5 = 19 \)

Let's take \( y = 2 \). Check first inequality: \( 2\geq1 \), yes. Second inequality: \( 2<19 \), yes. But maybe a more straightforward point. Wait, let's solve the system again.

Wait, maybe I made a mistake earlier. Let's re - solve the equation \( \frac{1}{6}x + 4=-\frac{4}{3}x - 5 \)

\( \frac{1}{6}x+\frac{4}{3}x=-5 - 4 \)

\( \frac{1 + 8}{6}x=-9 \)

\( \frac{9}{6}x=-9 \)

\( \frac{3}{2}x=-9 \)

\( x=-9\times\frac{2}{3}=-6 \). So the intersection of the lines is at \( x = - 6,y = 3 \). But since \( y<-\frac{4}{3}x - 5 \) and at \( x=-6 \), \( -\frac{4}{3}x - 5=8 - 5 = 3 \), so \( y \) must be less than 3 and greater than or equal to \( \frac{1}{6}x + 4 \). At \( x=-6 \), \( \frac{1}{6}x + 4 = 3 \), so there is no solution at \( x=-6 \). Let's try \( x=-12 \)

\( \frac{1}{6}(-12)+4=-2 + 4 = 2 \), \( -\frac{4}{3}(-12)-5=16 - 5 = 11 \)

So a point like \( (-12,3) \): \( 3\geq2 \) (satisfies first inequality) and \( 3<11 \) (satisfies second inequality). Wait, but let's check the graph. The first line at \( x=-12 \) has \( y = 2 \), so \( y = 3\) is above the first line (since \( 3\geq2 \)) and below the second line (since \( 3<11 \)). Another way: let's find a point in the region.

Let's take \( x=-6 \), \( y = 2 \): no, as \( 2<3 \) (first inequality fails). \( x=-18 \), \( y = 2 \): \( 2\geq\frac{1}{6}(-18)+4=-3 + 4 = 1 \) (good), \( 2<-\frac{4}{3}(-18)-5=24 - 5 = 19 \) (good). But maybe the problem expects us to find the intersection of the two lines (even though it's not in the solution set) and then adjust. Wait, actually, the system of inequalities \( y\geq\frac{1}{6}x + 4 \) and \( y<-\frac{4}{3}x - 5 \) may have no solution? Wait, let's check the slopes. The first line has slope \( \frac{1}{6}\), the second has slope \( -\frac{4}{3}\). The first line is increasing, the second is decreasing. Let's check the y - values at a common x. Let's take \( x = 0 \): first inequality \( y\geq4 \), second inequality \( y<-5 \). No overlap. At \( x=-6 \): first \( y\geq3 \), second \( y<3 \). No overlap. Wait, this means the solution set is empty? But that can't be. Wait, maybe I made a mistake in calculation.

Wait, let's re - solve \( \frac{1}{6}x + 4=-\frac{4}{3}x - 5 \)

\( \frac{1}{6}x+\frac{4}{3}x=-5 - 4 \)

\( \frac{1 + 8}{6}x=-9 \)

\( \frac{9}{6}x=-9 \)

\( x=-9\times\frac{6}{9}=-6 \). So at \( x = - 6 \), \( y = 3 \). So the first inequality is \( y\geq3 \), the second is \( y<3 \). So there is no solution? But the problem says "Point: (□,□)", so maybe there is a mistake in my approach. Wait, maybe the inequalities are written wrong? Or maybe I misread. Wait, the first inequality is \( y\geq\frac{1}{6}x + 4 \), the second is \( y<-\frac{4}{3}x - 5 \). Let's check the y - intercepts. The first line is above \( y = 4 \) when \( x = 0 \), the second line is below \( y=-5 \) when \( x = 0 \). So there is no overlap between the two regions. So the solution set is empty. But that seems odd. Wait, maybe the user made a typo, but assuming the problem is correct, maybe we made a mistake.

Wait, let's check with \( x=-12 \), \( y = 5 \). First inequality: \( 5\geq\frac{1}{6}(-12)+4=-2 + 4 = 2 \) (good). Second inequality: \( 5<-\frac{4}{3}(-12)-5=16 - 5 = 11 \) (good). Wait, but when \( x=-12 \), the first line gives \( y = 2 \), so \( y = 5\) is above the first line (satisfies \( y\geq2 \)) and below the second line (satisfies \( y<11 \)). So \( (-12,5) \) is a point in the solution set.

Answer:

\((-12,5)\) (Note: There are other possible points, this is one example. If the system has no solution, but based on the calculation of non - overlapping at \( x=-6 \), but for \( x<-6 \) (more negative), the first line's \( y\) value decreases (since slope is positive? Wait, no, slope of first line is \( \frac{1}{6}>0 \), so as \( x\) decreases (becomes more negative), \( y\) decreases. The second line has slope \( -\frac{4}{3}<0 \), so as \( x\) decreases (more negative), \( y\) increases. So at some \( x\) very negative, the first line's \( y\) (decreasing) and the second line's \( y\) (increasing) will cross. Wait, when \( x\) approaches \( -\infty \), first line \( y\) approaches \( -\infty \), second line \( y\) approaches \( +\infty \). So there must be a point where they cross. Wait, our earlier solution for the intersection of the lines was \( x=-6,y = 3 \). But for \( x<-6 \), let's take \( x=-18 \)

First line: \( y=\frac{1}{6}(-18)+4=-3 + 4 = 1 \)

Second line: \( y=-\frac{4}{3}(-18)-5=24 - 5 = 19 \)

So \( y = 2 \) (between 1 and 19) satisfies both \( y\geq1 \) and \( y<19 \). So \( (-18,2) \) is also a solution. So the answer can be any point like \((-12,5)\), \((-18,2)\) etc. A simple one is \((-12,5)\)

\((-12,5)\)