Question

write a system of linear equations for the graph below.
(the graph shows two lines on a coordinate plane with grid lines. one line passes through (0,7) and (4, -1), the other passes through (0, -2) and (4, -1). there is a system of equations box on the right with two y = blank lines.)

Explanation:

Step1: Find the equation of the steeper line

The steeper line passes through \((0, 7)\) and \((4, 0)\)? Wait, no, looking at the graph, the steeper line passes through \((0, 7)\) and \((4, -1)\)? Wait, no, let's check the intercepts. Wait, the steeper line crosses the y - axis at \((0, 7)\) and the x - axis? Wait, no, let's take two points. The steeper line goes through \((0, 7)\) and \((4, -1)\)? Wait, no, when \(x = 0\), \(y=7\); when \(x = 4\), \(y=-1\)? Wait, no, looking at the grid, the steeper line passes through \((0,7)\) and \((4, -1)\)? Wait, no, the two points on the steeper line: when \(x = 0\), \(y = 7\); when \(x=4\), \(y=-1\)? Wait, the slope \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{-1 - 7}{4 - 0}=\frac{-8}{4}=-2\). So the equation is \(y=-2x + 7\).

Step2: Find the equation of the shallower line

The shallower line passes through \((0, -2)\) and \((4, 0)\)? Wait, no, when \(x = 0\), \(y=-2\); when \(x = 4\), \(y = 0\)? Wait, no, let's take two points: \((0,-2)\) and \((4,0)\)? Wait, the slope \(m=\frac{0-(-2)}{4 - 0}=\frac{2}{4}=\frac{1}{2}\). Wait, no, another point: when \(x = 8\), \(y = 0\)? Wait, no, the shallower line passes through \((0,-2)\) and \((4,0)\)? Wait, no, let's check the points. The shallower line goes through \((0, -2)\) and \((4, 0)\)? Wait, the slope \(m=\frac{0 - (-2)}{4-0}=\frac{2}{4}=\frac{1}{2}\)? Wait, no, when \(x = 0\), \(y=-2\); when \(x = 4\), \(y = 0\), so the slope is \(\frac{0+2}{4 - 0}=\frac{2}{4}=\frac{1}{2}\). Wait, but when \(x = 8\), \(y = 2\)? Wait, no, let's take two points: \((0,-2)\) and \((8, 2)\). The slope \(m=\frac{2-(-2)}{8-0}=\frac{4}{8}=\frac{1}{2}\). So the equation is \(y=\frac{1}{2}x-2\). Wait, let's verify with \(x = 4\), \(y=\frac{1}{2}(4)-2=2 - 2=0\), which matches. And when \(x = 0\), \(y=-2\), which matches.

Wait, but let's check the intersection point. The two lines intersect at \((4, -1)\)? Wait, no, when \(x = 4\), for the first line \(y=-2(4)+7=-8 + 7=-1\); for the second line \(y=\frac{1}{2}(4)-2=2 - 2=0\)? Wait, no, that's a mistake. Wait, let's re - examine the graph. The two lines intersect at \((4, -1)\)? Wait, no, the steeper line: when \(x = 4\), \(y=-2(4)+7=-1\). The shallower line: when \(x = 4\), \(y=\frac{1}{2}(4)-2=0\)? That's not correct. Wait, maybe I took the wrong points. Let's take the intersection point. The two lines intersect at \((4, -1)\)? Wait, no, looking at the graph, the intersection point is at \((4, -1)\)? Wait, no, the steeper line: when \(x = 0\), \(y = 7\); when \(x = 4\), \(y=-1\) (since \(y=-2x + 7\), when \(x = 4\), \(y=-8 + 7=-1\)). The shallower line: when \(x = 0\), \(y=-2\); when \(x = 4\), \(y = 0\)? No, that's not intersecting at \((4, -1)\). Wait, maybe the shallower line passes through \((0,-2)\) and \((4,0)\) is wrong. Let's take the two points on the shallower line: \((0,-2)\) and \((8, 2)\). The slope is \(\frac{2-(-2)}{8 - 0}=\frac{4}{8}=\frac{1}{2}\). Then the equation is \(y=\frac{1}{2}x-2\). The steeper line: passes through \((0,7)\) and \((4, -1)\), slope \(m=\frac{-1 - 7}{4-0}=\frac{-8}{4}=-2\), equation \(y=-2x + 7\). Now, let's check the intersection: set \(-2x + 7=\frac{1}{2}x-2\). Multiply both sides by 2: \(-4x+14=x - 4\), \(-5x=-18\), \(x=\frac{18}{5}=3.6\)? No, that's not 4. Wait, I must have taken the wrong points. Let's look at the graph again. The steeper line: when \(x = 0\), \(y = 7\); when \(x = 4\), \(y=-1\) (correct, as per the grid). The shallower line: when \(x = 0\), \(y=-2\); when \(x = 4\), \(y = 0\) is wrong. Wait, the shallower line passes through \((0,-2)\) and \((4,0)\) is incorrect. Let's take the intersection point: the two lines intersect at \((4, -1)\)? Wait, no, the steeper line at \(x = 4\) is \(y=-1\), and the shallower line at \(x = 4\) should also be \(y=-1\). So for the shallower line, when \(x = 4\), \(y=-1\), and when \(x = 0\), \(y=-2\). So the slope \(m=\frac{-1-(-2)}{4 - 0}=\frac{1}{4}\)? Wait, no, \(\frac{-1+2}{4}=\frac{1}{4}\). Then the equation is \(y=\frac{1}{4}x-2\). Let's check: when \(x = 4\), \(y=\frac{1}{4}(4)-2=1 - 2=-1\), correct. When \(x = 8\), \(y=\frac{1}{4}(8)-2=2 - 2=0\), correct. Ah, there we go. So the shallower line has slope \(\frac{1}{4}\) and y - intercept \(-2\), so equation \(y=\frac{1}{4}x-2\). The steeper line has slope \(-2\) and y - intercept \(7\), so equation \(y=-2x + 7\).

Wait, let's verify with the graph. The steeper line: when \(x = 0\), \(y = 7\) (correct, as per the dot at \((0,7)\)). When \(x = 4\), \(y=-2(4)+7=-1\) (correct, the intersection point). The shallower line: when \(x = 0\), \(y=-2\) (correct, the dot at \((0,-2)\)). When \(x = 4\), \(y=\frac{1}{4}(4)-2=1 - 2=-1\) (correct, intersection point). When \(x = 8\), \(y=\frac{1}{4}(8)-2=2 - 2=0\) (correct, as per the graph).

So the two equations are \(y=-2x + 7\) and \(y=\frac{1}{4}x-2\).

Answer:

\(

$$\begin{cases}y=-2x + 7\\y=\frac{1}{4}x-2\end{cases}$$

\)