Question

this is the graph of a linear inequality. write the inequality in slope - intercept form.
write your answer with y first, followed by an inequality symbol. use integers, proper fractions, and improper fractions in simplest form.

Explanation:

Step1: Find the slope of the line

The line passes through the points \((0, 3)\) and \((2, 0)\). The slope \(m\) is calculated as \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{0 - 3}{2 - 0}=\frac{-3}{2}\).

Step2: Find the y - intercept

The y - intercept \(b\) is the value of \(y\) when \(x = 0\). From the point \((0, 3)\), we know that \(b = 3\). So the equation of the line in slope - intercept form \(y=mx + b\) is \(y=-\frac{3}{2}x+3\).

Step3: Determine the inequality symbol

The line is dashed, which means the inequality is either \(y<-\frac{3}{2}x + 3\) or \(y>-\frac{3}{2}x+3\). We test a point in the shaded region. Let's take the point \((0,0)\). Substitute \(x = 0\) and \(y = 0\) into the inequality. If we use \(y<-\frac{3}{2}x + 3\), then \(0<-\frac{3}{2}(0)+3\), which simplifies to \(0 < 3\), and this is true. So the inequality is \(y\leq-\frac{3}{2}x + 3\)? Wait, no, the line is dashed, so it should be a strict inequality. Wait, let's re - check the shaded region. Looking at the graph, the shaded region includes the area below the line? Wait, when \(x = 0\), the line is at \(y = 3\), and the shaded region at \(x = 0\) is below \(y = 3\) (since at \(x=0\), the shaded region has \(y\) values less than 3? Wait, no, the point \((0,0)\) is in the shaded region. Let's re - calculate the slope. Wait, the two points are \((0,3)\) and \((2,0)\). The slope is \(\frac{0 - 3}{2-0}=-\frac{3}{2}\). The equation of the line is \(y=-\frac{3}{2}x + 3\). Now, the line is dashed, so the inequality is either \(y<-\frac{3}{2}x + 3\) or \(y>-\frac{3}{2}x+3\). Let's take a point in the shaded region, say \((-2,0)\). Substitute into \(y=-\frac{3}{2}x + 3\): \(y=-\frac{3}{2}(-2)+3=3 + 3=6\). The \(y\) - value of the point \((-2,0)\) is \(0\), and \(0<6\), so the inequality is \(y<-\frac{3}{2}x + 3\)? Wait, no, wait the graph: when \(x = 0\), the line is at \(y = 3\), and the shaded region is above or below? Wait, the green shaded region: looking at the graph, when \(x\) increases, the line goes down. Let's take another point, say \((-2,4)\). Wait, no, the grid: each square is 1 unit. Wait, the line passes through \((0,3)\) and \((2,0)\). Let's check the direction of the inequality. The general form of a linear inequality is \(y>mx + b\) (above the line) or \(y<mx + b\) (below the line) for a line \(y = mx + b\). If the line is dashed, it's a strict inequality. Let's take the point \((0,0)\): \(0<-\frac{3}{2}(0)+3\) (since \(0<3\)) is true. So the inequality is \(y<-\frac{3}{2}x + 3\)? Wait, but when we look at the graph, the shaded region at \(x=-2\) is above the line? Wait, no, maybe I made a mistake in the slope. Wait, the two points: \((0,3)\) and \((2,0)\). The slope \(m=\frac{0 - 3}{2-0}=-\frac{3}{2}\). The equation of the line is \(y=-\frac{3}{2}x+3\). Let's take \(x=-2\), then \(y=-\frac{3}{2}(-2)+3=3 + 3=6\). The point \((-2,7)\): if we substitute into \(y>-\frac{3}{2}x + 3\), then \(7>-\frac{3}{2}(-2)+3=6\), which is true. Wait, maybe the shaded region is above the line. Let's check the point \((0,4)\). If we use \(y>-\frac{3}{2}x + 3\), then \(4>-\frac{3}{2}(0)+3\), which is \(4>3\), true. And \((0,4)\) is in the shaded region (the green region at the top - left). Oh! I made a mistake earlier. The shaded region is above the line. So let's re - test. Take the point \((0,4)\): \(4>-\frac{3}{2}(0)+3\) ( \(4>3\) is true). Take the point \((0,0)\): \(0>-\frac{3}{2}(0)+3\) ( \(0>3\) is false). So the correct inequality is \(y>-\frac{3}{2}x + 3\)? Wait, no, the line is dashed. Wait, the two points are \((0,3)\) and \((2,0)\). Let's re - examine the graph. The green shaded region is above the line? Wait, the graph shows that at \(x = 0\), the line is at \(y = 3\), and the green region is above \(y = 3\) (towards the top - left). Wait, maybe I mixed up the points. Let's list the coordinates again. The line passes through \((0,3)\) (on the y - axis) and \((2,0)\) (on the x - axis). The slope is \(\frac{0 - 3}{2-0}=-\frac{3}{2}\). The equation of the line is \(y=-\frac{3}{2}x + 3\). Now, the shaded region: if we take a point in the green area, say \((-2,6)\). Substitute into \(y\) and the line equation: \(y=6\), and the line at \(x=-2\) is \(y=-\frac{3}{2}(-2)+3=3 + 3=6\)? Wait, no, \(y=-\frac{3}{2}(-2)+3 = 3 + 3=6\). So the point \((-2,6)\) is on the line? But the line is dashed. Wait, maybe the two points are \((0,3)\) and \((2,0)\), and the slope is \(-\frac{3}{2}\). Let's check the inequality symbol again. The line is dashed, so it's a strict inequality. Let's take the point \((0,4)\): \(4>-\frac{3}{2}(0)+3\) ( \(4>3\) is true). Take the point \((0,2)\): \(2>-\frac{3}{2}(0)+3\) ( \(2>3\) is false). So the shaded region is where \(y>-\frac{3}{2}x + 3\)? Wait, but when \(x = 2\), the line is at \(y = 0\), and the shaded region at \(x = 2\) is below \(y = 0\)? No, that can't be. Wait, maybe I have the slope wrong. Let's calculate the slope again. The two points are \((0,3)\) and \((2,0)\). The change in \(y\) is \(0 - 3=-3\), the change in \(x\) is \(2 - 0 = 2\), so slope \(m=\frac{-3}{2}\). The equation of the line is \(y=-\frac{3}{2}x+3\). Now, the shaded region: looking at the graph, the green area is on the left - hand side. Let's use the test point \((-4,0)\). Substitute into \(y\) and the line equation: \(y = 0\), and the line at \(x=-4\) is \(y=-\frac{3}{2}(-4)+3=6 + 3=9\). So \(0<9\), which would mean \(y<-\frac{3}{2}x + 3\). But \((-4,0)\) is in the shaded region. Wait, now I'm confused. Let's look at the graph again. The line is dashed, and the shaded region is below the line? Wait, when \(x = 0\), the line is at \(y = 3\), and the shaded region at \(x = 0\) is below \(y = 3\) (since the point \((0,0)\) is in the shaded region). When \(x = 2\), the line is at \(y = 0\), and the shaded region at \(x = 2\) is below \(y = 0\) (the point \((2,-1)\) would be in the shaded region? Wait, the graph's y - axis goes from - 8 to 8. Let's re - evaluate. The correct way:

1. Find the equation of the boundary line:
- The line passes through \((0,3)\) (y - intercept \(b = 3\)) and \((2,0)\).
- Slope \(m=\frac{0 - 3}{2-0}=-\frac{3}{2}\).
- So the equation of the line is \(y=-\frac{3}{2}x + 3\).
2. Determine the inequality symbol:
- The line is dashed, so the inequality is either \(y<-\frac{3}{2}x + 3\) or \(y>-\frac{3}{2}x + 3\).
- Test a point in the shaded region. Let's choose \((0,0)\).
- Substitute \(x = 0\) and \(y = 0\) into the inequality \(y<-\frac{3}{2}x + 3\): \(0<-\frac{3}{2}(0)+3\) which simplifies to \(0 < 3\), which is true.
- If we test \(y>-\frac{3}{2}x + 3\) with \((0,0)\), we get \(0>3\), which is false.

So the inequality in slope - intercept form is \(y<-\frac{3}{2}x + 3\). Wait, but when I look at the graph, the green region is also above the line in some parts. Wait, maybe the correct inequality is \(y\leq-\frac{3}{2}x + 3\)? No, the line is dashed, so it's a strict inequality. Wait, perhaps I made a mistake in the direction of the inequality. Let's check the slope again. The two points are \((0,3)\) and \((2,0)\). The slope is \(-\frac{3}{2}\), so the line is decreasing. The shaded region: if we move from left to right, the line goes down. The shaded region is below the line. So the inequality is \(y<-\frac{3}{2}x + 3\).

Answer:

\(y<-\frac{3}{2}x + 3\)