Question

which ordered pairs are solutions to the equation? select all that apply. (2, -2) (0, -4) (-2, 6) (6, -5) (-6, 4) (5, 0)

Explanation:

Step1: Find the equation of the line

First, identify two points on the line. From the graph, we can see that the line passes through \((0, - 0.5)\)? Wait, no, let's re - examine. Wait, when \(x = 0\), the \(y\) - intercept: looking at the graph, when \(x = 0\), \(y=-0.5\)? No, maybe I made a mistake. Wait, let's take two clear points. Let's take \((-2,1)\) and \((2, - 2)\). Wait, the slope \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{-2 - 1}{2-(-2)}=\frac{-3}{4}=- 0.75\)? Wait, no, maybe another pair. Wait, when \(x = 2\), \(y=-2\); when \(x=-2\), \(y = 1\). Wait, the slope \(m=\frac{-2 - 1}{2+2}=\frac{-3}{4}\). Wait, but let's check the ordered pairs.

For an ordered pair \((x,y)\) to be a solution, it must lie on the line. Let's check each ordered pair:

- For \((2,-2)\): From the graph, we can see that the point \((2,-2)\) lies on the line.
- For \((0,-4)\): Does \((0,-4)\) lie on the line? Looking at the graph, when \(x = 0\), the \(y\) - value is not \(-4\). Wait, maybe my initial point selection is wrong. Let's re - evaluate the line. Let's take two points: when \(x=-2\), \(y = 1\); when \(x = 2\), \(y=-2\). The slope \(m=\frac{-2 - 1}{2-(-2)}=\frac{-3}{4}\). The equation of the line using point - slope form \(y - y_1=m(x - x_1)\) with \((x_1,y_1)=(-2,1)\) is \(y - 1=-\frac{3}{4}(x + 2)\). Simplifying, \(y=-\frac{3}{4}x-\frac{3}{2}+1=-\frac{3}{4}x-\frac{1}{2}\).

Now check each ordered pair:

1. \((2,-2)\): Substitute \(x = 2\) into \(y=-\frac{3}{4}x-\frac{1}{2}\), \(y=-\frac{3}{4}(2)-\frac{1}{2}=- \frac{3}{2}-\frac{1}{2}=-2\). So \((2,-2)\) is a solution.
2. \((0,-4)\): Substitute \(x = 0\) into \(y=-\frac{3}{4}x-\frac{1}{2}\), \(y=-\frac{1}{2} eq - 4\). So \((0,-4)\) is not a solution. Wait, but the original check in the problem has a checkmark. Maybe my equation is wrong. Let's try another approach. Let's look at the line again. Let's take \(x=-6\), \(y = 4\): Substitute into the line. If the line has a slope of \(-1\)? Wait, maybe the slope is \(-1\). Let's check: If \(x=-2\), \(y = 1\); \(x=-1\), \(y = 0\); \(x = 0\), \(y=-1\); \(x = 1\), \(y=-2\); \(x = 2\), \(y=-3\)? No, that's not matching. Wait, the user's original checkmarks: Let's check the graph again. The line passes through \((-2,1)\), \((0,-0.5)\)? No, maybe the correct way is to check if the point is on the line by visual inspection.

Looking at the graph:

- \((2,-2)\): Lies on the line.
- \((-2,6)\): Wait, \((-2,6)\) is far from the line. Wait, maybe the user made a mistake in the original checkmarks, but we have to go by the graph.

Wait, let's re - list the ordered pairs:

1. \((2,-2)\): From the graph, the line passes through \((2,-2)\), so it's a solution.
2. \((0,-4)\): The line at \(x = 0\) has \(y\approx - 0.5\), so \((0,-4)\) is not on the line.
3. \((-2,6)\): The line at \(x=-2\) has \(y\approx1\), so \((-2,6)\) is not on the line.
4. \((6,-5)\): Let's see, if we follow the slope. From \(x = 2\) (\(y=-2\)) to \(x = 6\), \(x\) increases by 4, \(y\) should decrease by \(4\times\)slope. If slope is \(-1\), \(y\) decreases by 4, so \(y=-2 - 4=-6\), but \((6,-5)\) is close? Wait, maybe the correct equation is \(y=-\frac{1}{2}x - 1\)? Let's check: For \(x = 2\), \(y=-\frac{1}{2}(2)-1=-2\), correct. For \(x=-2\), \(y=-\frac{1}{2}(-2)-1 = 1 - 1=0\)? No. Wait, maybe \(y=-\frac{3}{4}x-\frac{1}{2}\) is wrong. Let's take two points: \((-6,4)\) and \((2,-2)\). Slope \(m=\frac{-2 - 4}{2 + 6}=\frac{-6}{8}=-\frac{3}{4}\). Equation: \(y-4=-\frac{3}{4}(x + 6)\), \(y=-\frac{3}{4}x-\frac{18}{4}+4=-\frac{3}{4}x-\frac{9}{2}+4=-\frac{3}{4}x-\frac{1}{2}\). For \(x=-6\), \(y=-\frac{3}{4}(-6)-\frac{1}{2}=\frac{18}{4}-\frac{1}{2}=\frac{9}{2}-\frac{1}{2}=4\), so \((-6,4)\) is on the line.

For \(x = 5\), \(y=-\frac{3}{4}(5)-\frac{1}{2}=-\frac{15}{4}-\frac{2}{4}=-\frac{17}{4}=-4.25 eq0\), so \((5,0)\) is not on the line.

For \(x = 6\), \(y=-\frac{3}{4}(6)-\frac{1}{2}=-\frac{18}{4}-\frac{2}{4}=-\frac{20}{4}=-5\), so \((6,-5)\) is on the line.

For \(x=-6\), \(y = 4\), so \((-6,4)\) is on the line.

For \(x = 2\), \(y=-2\), so \((2,-2)\) is on the line.

So the correct solutions are \((2,-2)\), \((6,-5)\), \((-6,4)\).

Wait, let's verify each:

- \((2,-2)\): Substitute into \(y=-\frac{3}{4}x-\frac{1}{2}\), \(y=-\frac{3}{4}(2)-\frac{1}{2}=- \frac{3}{2}-\frac{1}{2}=-2\), correct.
- \((6,-5)\): \(y=-\frac{3}{4}(6)-\frac{1}{2}=-\frac{18}{4}-\frac{2}{4}=-\frac{20}{4}=-5\), correct.
- \((-6,4)\): \(y=-\frac{3}{4}(-6)-\frac{1}{2}=\frac{18}{4}-\frac{2}{4}=\frac{16}{4}=4\), correct.

The other points:

- \((0,-4)\): \(y=-\frac{3}{4}(0)-\frac{1}{2}=-\frac{1}{2} eq - 4\), not a solution.
- \((-2,6)\): \(y=-\frac{3}{4}(-2)-\frac{1}{2}=\frac{6}{4}-\frac{2}{4}=\frac{4}{4}=1 eq6\), not a solution.
- \((5,0)\): \(y=-\frac{3}{4}(5)-\frac{1}{2}=-\frac{15}{4}-\frac{2}{4}=-\frac{17}{4} eq0\), not a solution.

Answer:

The ordered pairs that are solutions are \((2, - 2)\), \((6, - 5)\), \((-6, 4)\)