Question

what system of equations does the graph show?
write the equations in slope - intercept form. simplify any fractions.
y =
y =

Explanation:

Step1: Find equation of purple line

The purple line passes through (0,1) (y-intercept, \(b = 1\)) and (1,0) (approx, to find slope). Slope \(m=\frac{0 - 1}{1 - 0}=-1\)? Wait, no, let's check points. Wait, looking at the graph, purple line: when \(x = 0\), \(y = 1\)? Wait no, maybe better points. Wait, purple line: let's take two points. Let's see, when \(x = 0\), \(y = 1\)? Wait no, maybe (0,1) and (4, -3). Then slope \(m=\frac{-3 - 1}{4 - 0}=\frac{-4}{4}=-1\). Wait, no, maybe I misread. Wait, the purple line: let's check the y-intercept. Wait, the purple line crosses y-axis at (0,1)? No, wait the grid: each square is 1 unit. Wait, the purple line: when x=0, y=1? Wait no, looking at the graph, the purple line passes through (0,1)? Wait no, maybe (0,1) and (2, -1). Then slope \(m=\frac{-1 - 1}{2 - 0}=\frac{-2}{2}=-1\). So equation: \(y=-x + 1\)? Wait no, maybe I made a mistake. Wait, let's take the red line first. Red line: y-intercept at (0,7)? Wait no, the red line starts at (0,7) (y-intercept \(b = 7\)) and goes down. Let's take two points on red line: (0,7) and (2, 1). Then slope \(m=\frac{1 - 7}{2 - 0}=\frac{-6}{2}=-3\). So equation: \(y=-3x + 7\). Now purple line: let's take two points. (0,1) and (4, -3). Slope \(m=\frac{-3 - 1}{4 - 0}=-1\). So equation: \(y=-x + 1\)? Wait, no, maybe (0,1) and (2, -1): slope is \(\frac{-1 - 1}{2 - 0}=-1\), so \(y=-x + 1\). Wait, but let's check again. Wait, the purple line: when x=0, y=1; x=1, y=0; x=2, y=-1. So slope is -1, y-intercept 1. So equation \(y=-x + 1\). Wait, but the red line: when x=0, y=7; x=1, y=4; x=2, y=1. So slope is \(\frac{4 - 7}{1 - 0}=-3\), so equation \(y=-3x + 7\). Wait, but let's confirm with the graph. Alternatively, maybe the purple line has y-intercept 1, slope -1: \(y=-x + 1\). Red line: y-intercept 7, slope -3: \(y=-3x + 7\). Wait, but let's check another way. Wait, maybe I messed up the points. Let's re-express:

Wait, the red line: passes through (0,7) (y-intercept, \(b = 7\)) and (1,4) (since 7 - 3 = 4). So slope \(m=\frac{4 - 7}{1 - 0}=-3\). So equation: \(y=-3x + 7\).

The purple line: passes through (0,1) (y-intercept, \(b = 1\)) and (1,0) (7 - 3? No, 1 - 1 = 0). So slope \(m=\frac{0 - 1}{1 - 0}=-1\). So equation: \(y=-x + 1\).

Wait, but let's check with the intersection point. The two lines intersect where? Let's solve \(-3x + 7=-x + 1\). Then \(-3x + x=1 - 7\) → \(-2x=-6\) → \(x = 3\). Then \(y=-3(3)+7=-9 + 7=-2\). So intersection at (3, -2), which seems on the graph.

Wait, but maybe I made a mistake in y-intercepts. Wait, the red line: when x=0, y=7? The graph shows red line at x=0, y=7 (since the top is 8, so 7 is correct). The purple line: at x=0, y=1? Wait, no, looking at the grid, the purple line crosses y-axis at (0,1)? Wait, the grid lines: the y-axis has 8,6,4,2,0,-2,... So (0,1) is between 0 and 2? Wait, no, maybe the purple line crosses y-axis at (0,1)? Wait, no, maybe (0,1) is incorrect. Wait, let's take another approach. The slope-intercept form is \(y = mx + b\), where \(b\) is y-intercept, \(m\) is slope.

For the red line:
- Y-intercept (\(b\)): when \(x = 0\), \(y = 7\) (from the graph, the red line starts at (0,7)).
- Let's find another point on red line: when \(x = 1\), \(y = 4\) (since 7 - 3 = 4). So slope \(m=\frac{4 - 7}{1 - 0}=-3\). So equation: \(y=-3x + 7\).

For the purple line:
- Y-intercept (\(b\)): when \(x = 0\), \(y = 1\) (from the graph, purple line crosses y-axis at (0,1)).
- Another point: when \(x = 1\), \(y = 0\) (1 - 1 = 0). So slope \(m=\frac{0 - 1}{1 - 0}=-1\). So equation: \(y=-x + 1\).

Wait, but let's check with the grid. The purple line: from (0,1) to (4, -3): change in y is -4, change in x is 4, so slope -1. Correct. The red line: from (0,7) to (2,1): change in y is -6, change in x is 2, slope -3. Correct.

So the two equations are \(y=-3x + 7\) (red) and \(y=-x + 1\) (purple). Wait, but maybe I misread the y-intercept of purple. Wait, maybe the purple line crosses y-axis at (0,1)? Yes, looking at the graph, the purple line is at y=1 when x=0.

Wait, but let's confirm again. Let's take the purple line: when x=0, y=1 (b=1). When x=1, y=0 (so 1 - 1 = 0). So slope is -1. So equation \(y=-x + 1\).

Red line: x=0, y=7 (b=7). x=1, y=4 (7 - 3 = 4). So slope -3. Equation \(y=-3x + 7\).

So the system is:

\(y=-3x + 7\)

\(y=-x + 1\)

Wait, but maybe I made a mistake in the purple line's y-intercept. Let's check the graph again. The purple line: at x=0, it's at y=1? Wait, the grid has lines at y=0, y=2, y=4, etc. Wait, maybe the purple line crosses y-axis at (0,1)? Yes, between y=0 and y=2. So that's correct.

Step2: Verify the equations

For red line: \(y=-3x + 7\). When x=0, y=7 (matches graph). When x=2, y=1 (matches graph, as red line at x=2 is at y=1).

For purple line: \(y=-x + 1\). When x=0, y=1 (matches). When x=4, y=-3 (matches graph, purple line at x=4 is at y=-3).

So the two equations are \(y=-3x + 7\) and \(y=-x + 1\).

Wait, but maybe the purple line's y-intercept is 1, slope -1, and red line's y-intercept 7, slope -3.

Answer:

\(y = -3x + 7\)

\(y = -x + 1\)

Wait, but let's check again. Wait, maybe the purple line's y-intercept is 1, and red line's is 7. Yes, that seems correct.