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 \((0, 0)\) and \((1, -1)\) (we can also use other points, but these are easy to identify). The slope \(m\) is calculated as \(m=\frac{y_2 - y_1}{x_2 - x_1}\). So \(m=\frac{-1 - 0}{1 - 0}=-1\).

Step2: Determine the equation of the boundary line

The slope - intercept form of a line is \(y = mx + b\). Since the line passes through the origin \((0,0)\), \(b = 0\). So the equation of the boundary line is \(y=-x\) (or \(y=-1x + 0\)).

Step3: Determine the inequality symbol

The line is dashed, so the inequality is either \(y\lt - x\) or \(y\gt - x\). We test a point in the shaded region. Let's take the point \((-1,0)\). Substitute into \(y\) and \(-x\): \(y = 0\), \(-x=-(-1)=1\). Since \(0\lt1\), the shaded region satisfies \(y\lt - x\)? Wait, no, wait. Wait, if we take \((-2,0)\): \(y = 0\), \(-x=-(-2) = 2\), \(0\lt2\). Wait, but let's check the direction of the shading. The shaded region is to the left of the line \(y=-x\). Wait, maybe I made a mistake in the slope. Wait, let's take another pair of points. Let's take \((0,0)\) and \((2,-2)\). Then \(m=\frac{-2 - 0}{2 - 0}=-1\), same slope. Now, let's take a point in the shaded area, say \((-3,0)\). Plug into \(y\) and \(-x\): \(y = 0\), \(-x=-(-3)=3\), \(0\lt3\). Wait, but the line is dashed, and the shading: let's see, when \(x = 0\), the line is \(y = 0\). The shaded region includes the area where, for example, when \(x=-1\), \(y\) can be 0 (which is in the shaded region). Wait, no, maybe I got the slope wrong. Wait, let's look at the graph again. The line goes from the origin down to the right, so for every 1 unit we move to the right (increase \(x\) by 1), we move down 1 unit (decrease \(y\) by 1), so slope is - 1. Now, the shaded region: let's take the point \((0,1)\), is it in the shaded region? No, the shaded region at \(x = 0\) is below \(y = 0\)? Wait, no, the graph shows that at \(x = 0\), the shaded region includes \(y\) values from \(-\infty\) up to... Wait, no, looking at the graph, the shaded area is above or below? Wait, the red shaded area: when \(x=-2\), the \(y\) values are from, say, - 8 up to, like, 6? Wait, maybe I messed up the slope. Wait, let's take two points on the dashed line: \((0,0)\) and \((3,-3)\). So slope \(m=\frac{-3 - 0}{3 - 0}=-1\). Now, the inequality: the line is dashed, so it's a strict inequality. Now, let's test the point \((-1,1)\). Wait, \((-1,1)\): \(y = 1\), \(-x=-(-1)=1\), so \(y=-x\) at that point, but the line is dashed, so that point is not on the line. Wait, maybe the shaded region is \(y\leq - x\)? No, the line is dashed, so it's \(y\lt - x\) or \(y\gt - x\). Wait, let's take the point \((0, - 1)\). \(y=-1\), \(-x = 0\), \(-1\lt0\), and \((0,-1)\) is in the shaded region. Take the point \((0,1)\), \(y = 1\), \(-x=0\), \(1\gt0\), and \((0,1)\) is not in the shaded region. So the inequality is \(y\lt - x\)? Wait, no, wait the slope: maybe I had the slope sign wrong. Wait, if we go from \((0,0)\) to \((-1,1)\), then \(m=\frac{1 - 0}{-1 - 0}=-1\), same as before. Wait, maybe the correct inequality is \(y\leq - x\)? No, the line is dashed, so it's a strict inequality. Wait, maybe I made a mistake in the direction of the shading. Let's look at the graph again. The shaded area is on the left - hand side of the line \(y=-x\). Let's take \(x=-2\), then \(y=-x = 2\). The shaded region at \(x=-2\) includes \(y\) values from, say, - 8 up to 6? Wait, no, the grid: each square is 1 unit. The dashed line goes from the top - left (around \((-6,6)\)) down to the bottom - right (around \((6,-6)\))? Wait, no, the endpoints of the dashed line: looking at the graph, the top end is around \((-6,6)\) and the bottom end is around \((6,-6)\). So the slope is \(\frac{-6 - 6}{6-(-6)}=\frac{-12}{12}=-1\), correct. Now, take a point in the shaded region, say \((-3,3)\): \(y = 3\), \(-x=-(-3)=3\), so \(y=-x\) at that point, but the line is dashed, so that point is not on the line. Wait, no, \((-3,3)\) is on the line \(y=-x\) (since \(3=-(-3)\)), but the line is dashed, so that point is not included. Wait, maybe the shaded region is \(y\geq - x\)? Let's test \((-1,0)\): \(y = 0\), \(-x = 1\), \(0\lt1\), so \(0\geq1\) is false. Wait, \((-1,2)\): \(y = 2\), \(-x = 1\), \(2\geq1\) is true. Ah! There we go. I was testing the wrong points. Let's take \((-1,2)\): it's in the shaded region. \(y = 2\), \(-x=-(-1)=1\), and \(2\gt1\) (or \(2\geq1\)). Wait, the line is dashed, so the inequality is \(y\gt - x\)? Wait, no, \((-1,2)\): \(y = 2\), \(-x = 1\), \(2\gt1\). Let's take \((0,1)\): \(y = 1\), \(-x = 0\), \(1\gt0\), and \((0,1)\) is in the shaded region. \((1,0)\): \(y = 0\), \(-x=-1\), \(0\gt - 1\), but \((1,0)\) is not in the shaded region (the shaded region at \(x = 1\) is below the line? Wait, no, the shaded region at \(x = 1\): the line at \(x = 1\) is \(y=-1\). The shaded region at \(x = 1\) includes \(y\) values from, say, - 8 up to - 1? No, that can't be. Wait, I think I messed up the direction of the slope. Wait, let's re - examine the graph. The dashed line goes from the upper left (where \(x\) is negative and \(y\) is positive) to the lower right (where \(x\) is positive and \(y\) is negative). So for a point in the shaded region, say \((-2,3)\): \(y = 3\), \(-x=-(-2)=2\), \(3\gt2\). Another point: \((-3,4)\): \(y = 4\), \(-x = 3\), \(4\gt3\). A point on the other side of the line, say \((2,-3)\): \(y=-3\), \(-x=-2\), \(-3\lt - 2\), and \((2,-3)\) is not in the shaded region. So the inequality should be \(y\gt - x\)? But wait, the line is dashed, so the inequality is \(y\gt - x\) or \(y\lt - x\). Wait, no, when \(x = 0\), the line is \(y = 0\). The shaded region at \(x = 0\) includes \(y\) values greater than 0? Wait, looking at the graph, at \(x = 0\), the shaded region is above the \(x\) - axis (for \(x\lt0\)) and below? No, the graph shows that the shaded area is a large triangle - like region. Wait, maybe the correct slope - intercept form with the inequality: the boundary line is \(y=-x\) (dashed), and the shading is above the line? Wait, no, when \(x=-1\), the line is \(y = 1\). The shaded region at \(x=-1\) includes \(y\) values above 1? Wait, the graph's shaded area: the top part is at \(x=-6,y = 6\) and the bottom part at \(x = 6,y=-6\). So the line is \(y=-x\), dashed. Now, let's take the point \((-1,2)\): it's in the shaded region. \(y = 2\), \(-x=-(-1)=1\), so \(2\gt1\), so \(y\gt - x\). Wait, but when \(x = 1\), the line is \(y=-1\). The shaded region at \(x = 1\) includes \(y\) values less than - 1? No, that doesn't make sense. Wait, I think I had the slope wrong. Wait, let's calculate the slope again. Let's take two points on the dashed line: \((-6,6)\) and \((6,-6)\). Then \(m=\frac{-6 - 6}{6-(-6)}=\frac{-12}{12}=-1\), correct. Now, the inequality: the shaded region is on the side of the line where \(y\geq - x\)? Wait, at \((-6,6)\), \(y = 6\), \(-x=-(-6)=6\), so \(y=-x\) at that point, but the line is dashed, so that point is not included. At \((-5,5)\), same thing. Wait, maybe the correct inequality is \(y\leq - x\)? No, because when \(x=-1\), \(y\leq1\), but the shaded region at \(x=-1\) has \(y\) values greater than 1? I'm confused. Wait, let's start over.

1. Find the equation of the boundary line:
- The line passes through \((0,0)\) and has a slope of \(-1\) (since for every 1 unit increase in \(x\), \(y\) decreases by 1). So the equation of the boundary line is \(y=-x\) (in slope - intercept form \(y = mx + b\), \(m=-1\), \(b = 0\)).
2. Determine the inequality symbol:
- The line is dashed, so the inequality is either \(y\gt - x\) or \(y\lt - x\).
- Test a point in the shaded region. Let's choose \((-2,0)\). Substitute into \(y\) and \(-x\):
- \(y = 0\), \(-x=-(-2)=2\).
- Since \(0\lt2\), does \(0\lt - x\) (i.e., \(0\lt2\))? Yes. Wait, but when \(x=-3\), \(y = 0\), \(-x = 3\), \(0\lt3\). When \(x=-1\), \(y = 0\), \(-x = 1\), \(0\lt1\). When \(x = 0\), \(y = 0\), \(-x = 0\), \(0 = 0\) (but the line is dashed, so the point \((0,0)\) is not included). When \(x = 1\), \(y=-2\), \(-x=-1\), \(-2\lt - 1\), and \((1,-2)\) is in the shaded region. Ah! Now I see. The shaded region is where \(y\lt - x\). Let's test \((1,-2)\): \(y=-2\), \(-x=-1\), \(-2\lt - 1\), which is true. Test \((-2,0)\): \(y = 0\), \(-x = 2\), \(0\lt2\), true. Test \((0,-1)\): \(y=-1\), \(-x = 0\), \(-1\lt0\), true. Test \((2,-3)\): \(y=-3\), \(-x=-2\), \(-3\lt - 2\), true. Test \((-1,1)\): \(y = 1\), \(-x = 1\), \(1 = 1\) (not in shaded region, since line is dashed), and \(1\lt1\) is false, which is correct because \((-1,1)\) is on the dashed line. So the inequality is \(y\lt - x\). Wait, but the slope - intercept form is \(y=mx + b\), and the inequality is \(y\lt - x\) (or \(y\leq - x\), but the line is dashed, so it's \(y\lt - x\)). Wait, no, the correct inequality is \(y\leq - x\) if the line is solid, but since it's dashed, it's \(y\lt - x\). Wait, but let's check the graph again. The shaded region: when \(x\) is positive, \(y\) is negative and less than \(-x\) (since for \(x = 2\), \(y\lt - 2\)), and when \(x\) is negative, \(y\) is less than \(-x\) (since for \(x=-2\), \(y\lt2\)). So the inequality in slope - intercept form is \(y\lt - x\). Wait, but let's write it as \(y\lt - x\) (or \(y\lt - 1x+0\)).

Answer:

\(y\lt - x\)