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: Determine 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, so the inequality is either \(y<-\frac{3}{2}x + 3\) or \(y>-\frac{3}{2}x+3\). The shaded region is above the line (we can test a point in the shaded region, for example, \((0,4)\): \(4\) vs \(-\frac{3}{2}(0)+3 = 3\), and \(4>3\)), so the inequality is \(y\geq-\frac{3}{2}x + 3\)? Wait, no, the line is dashed, so it should be \(y>-\frac{3}{2}x+3\)? Wait, let's re - check. Wait, the line passes through \((0,3)\) and \((2,0)\). Let's take a point in the shaded area, say \((-2,4)\). Plug into \(y=-\frac{3}{2}x + 3\): RHS is \(-\frac{3}{2}(-2)+3=3 + 3=6\), and \(4<6\). Wait, maybe I made a mistake in the direction. Wait, the shaded area is to the left of the line? Wait, no, let's look at the graph again. The line is dashed, and the shaded region: let's take the point \((0,4)\), when \(x = 0\), the line has \(y = 3\), and \(4>3\), but when we take \(x=-2\), the line at \(x=-2\) is \(y=-\frac{3}{2}(-2)+3=3 + 3=6\), and the shaded point \((-2,4)\) has \(y = 4<6\). Wait, maybe the slope calculation is wrong. Wait, another way: from \((0,3)\) to \((2,0)\), the rise is \(0 - 3=-3\), run is \(2-0 = 2\), so slope is \(-\frac{3}{2}\), correct. Let's take the point \((-1,4)\): plug into \(y=-\frac{3}{2}x+3\), RHS is \(-\frac{3}{2}(-1)+3=\frac{3}{2}+3=\frac{9}{2}=4.5\), and \(4<4.5\). Wait, the shaded area seems to be below the line? No, the graph shows the green area is on the left - upper side. Wait, maybe I messed up the slope. Wait, let's check the two points again. The y - intercept is at \((0,3)\), and it crosses the x - axis at \((2,0)\). So the slope is \(\frac{0 - 3}{2-0}=-\frac{3}{2}\), correct. Now, the inequality: 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 the point \((0,4)\): \(4\) compared to \(-\frac{3}{2}(0)+3 = 3\), \(4>3\), but when we take \(x = 1\), the line at \(x = 1\) is \(y=-\frac{3}{2}(1)+3=\frac{3}{2}=1.5\), and a point in the shaded area at \(x = 1\) would be, say, \((1,2)\), and \(2>1.5\). Wait, maybe the initial thought was correct. Wait, maybe the error was in the first test point. Let's take \((0,4)\): \(y = 4\), line at \(x = 0\) is \(y = 3\), \(4>3\), so the inequality is \(y>-\frac{3}{2}x+3\)? But when we take \(x=-2\), the line at \(x=-2\) is \(y=-\frac{3}{2}(-2)+3=3 + 3=6\), and the shaded point \((-2,4)\) has \(y = 4<6\), which contradicts. Wait, maybe the line is \(y=-\frac{3}{2}x + 3\), and the shaded region is \(y\geq-\frac{3}{2}x+3\)? But the line is dashed, so it should be \(y>-\frac{3}{2}x+3\) or \(y<-\frac{3}{2}x+3\). Wait, maybe I made a mistake in the slope. Wait, let's recalculate the slope between \((0,3)\) and \((2,0)\): \(m=\frac{0 - 3}{2-0}=-\frac{3}{2}\), correct. Let's take the point \((-1,5)\): plug into \(y=-\frac{3}{2}x+3\), RHS is \(-\frac{3}{2}(-1)+3=\frac{3}{2}+3=\frac{9}{2}=4.5\), and \(5>4.5\). Ah, so when \(x=-1\), \(y = 5\) is in the shaded area and \(5>4.5\). When \(x = 0\), \(y = 4\) (shaded) is greater than \(3\). When \(x = 2\), the line has \(y = 0\), and the shaded area at \(x = 2\) would be above \(y = 0\)? Wait, no, the line crosses the x - axis at \((2,0)\). The shaded area: let's look at the graph, the green area is above the line? Wait, the line is going from \((0,3)\) down to \((2,0)\), so the region above the line (in the sense of \(y\) - values) would be where \(y>-\frac{3}{2}x + 3\). Wait, maybe my first test point \((-2,4)\) was wrong. Let's take \((-2,6)\): plug into \(y=-\frac{3}{2}x+3\), RHS is \(-\frac{3}{2}(-2)+3=3 + 3=6\), so \(y = 6\) is on the line, but the line is dashed, so the shaded area should be \(y>-\frac{3}{2}x+3\) (since the line is dashed, not including the line itself) and the shaded area is above the line. Wait, but when \(x=-2\), the line has \(y = 6\), and if we take \(y = 7\) at \(x=-2\), \(7>6\), which would be in the shaded area. So the correct inequality is \(y>-\frac{3}{2}x+3\)? Wait, no, let's check the slope - intercept form again. Wait, the line equation is \(y=-\frac{3}{2}x + 3\). The shaded region: let's use the origin \((0,0)\), plug into the inequality. If we assume the inequality is \(y<-\frac{3}{2}x+3\), then \(0<-\frac{3}{2}(0)+3=3\), which is true, but the origin is not in the shaded area. So the inequality should be \(y\geq-\frac{3}{2}x+3\)? But the line is dashed, so it's \(y>-\frac{3}{2}x+3\). Wait, the key is: dashed line means the inequality is strict (\(>\) or \(<\)), and the shaded region is where the inequality holds. Since the origin \((0,0)\) is not in the shaded area, and \(0<-\frac{3}{2}(0)+3 = 3\) is true, but the origin is not shaded, so the inequality should be the opposite, \(y>-\frac{3}{2}x+3\) (because when we plug in \((0,0)\) into \(y>-\frac{3}{2}x+3\), we get \(0>3\), which is false, and the origin is not in the shaded area, which matches). So the correct inequality is \(y>-\frac{3}{2}x + 3\). Wait, but let's re - check the slope. Wait, another way: the line has a slope of \(-\frac{3}{2}\) and y - intercept \(3\), so the equation is \(y=-\frac{3}{2}x+3\). The shaded region is above the line (since the line is dashed and the shaded area is on the side where \(y\) is greater than the line's \(y\) - value for a given \(x\)), so the inequality is \(y>-\frac{3}{2}x+3\).

Answer:

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