Question

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

Explanation:

Step1: Find equation of blue line

The blue line passes through \((0,1)\) (y-intercept \(b = 1\)) and has a slope \(m=\frac{8 - 1}{4 - 0}=\frac{7}{4}\)? Wait, no, looking at the graph, when \(x = 0\), \(y = 1\)? Wait, no, let's recheck. Wait, the blue line: when \(x = 0\), \(y = 1\)? Wait, no, maybe I made a mistake. Wait, the blue line: let's take two points. When \(x = 0\), \(y = 1\)? Wait, no, looking at the grid, the blue line passes through \((0,1)\) and \((1,3)\)? Wait, no, maybe better to use slope-intercept form \(y=mx + b\). For the blue line: when \(x = 0\), \(y = 1\)? Wait, no, the blue line seems to pass through \((0,1)\) and when \(x = 1\), \(y = 3\)? Wait, no, let's check the red line first.

Step2: Find equation of red line

The red line passes through \((0, - 2)\) (y-intercept \(b=-2\)) and when \(x = 3\), \(y = 0\)? Wait, no, when \(x = 3\), \(y = 0\)? Wait, the red line: from \((0,-2)\) to \((3,0)\), slope \(m=\frac{0 - (-2)}{3 - 0}=\frac{2}{3}\)? Wait, no, looking at the graph, the red line goes from \((0,-2)\) and when \(x = 3\), \(y = 0\)? Wait, no, maybe another approach. Wait, the blue line: let's take two points. When \(x = 0\), \(y = 1\) (wait, no, the blue line at \(x = 0\) is at \(y = 1\)? Wait, no, the grid lines: each square is 1 unit. So the blue line: when \(x = 0\), \(y = 1\)? Wait, no, maybe I messed up. Wait, the blue line: let's see, when \(x = 0\), \(y = 1\), and when \(x = 1\), \(y = 3\), so slope \(m = 2\). Wait, because from \((0,1)\) to \((1,3)\), the change in \(y\) is \(2\), change in \(x\) is \(1\), so \(m = 2\). So blue line: \(y = 2x+1\)? Wait, no, when \(x = 0\), \(y = 1\), and when \(x = 1\), \(y = 3\), so \(m = 2\), so \(y = 2x + 1\)? Wait, but when \(x = 2\), \(y = 5\), which matches the blue line.

For the red line: it passes through \((0,-2)\) (y-intercept \(b=-2\)) and when \(x = 3\), \(y = 0\)? Wait, no, when \(x = 3\), \(y = 0\)? Wait, no, the red line: from \((0,-2)\) to \((3,0)\), slope \(m=\frac{0 - (-2)}{3 - 0}=\frac{2}{3}\)? Wait, no, when \(x = 3\), \(y = 0\), so \(0=\frac{2}{3}(3)+b\), \(0 = 2 + b\), \(b=-2\), so \(y=\frac{2}{3}x-2\)? Wait, but when \(x = 6\), \(y = 2\), which matches the red line.

Wait, no, maybe the blue line: let's take two points. When \(x = 0\), \(y = 1\), and when \(x = 1\), \(y = 3\), so slope \(m = 2\), so \(y = 2x+1\). Wait, but the blue line at \(x = 0\) is \(y = 1\), and at \(x = 2\), \(y = 5\), which is correct.

The red line: passes through \((0,-2)\) and \((3,0)\), slope \(m=\frac{0 - (-2)}{3 - 0}=\frac{2}{3}\), so \(y=\frac{2}{3}x-2\). Wait, but when \(x = 3\), \(y = 0\), correct. When \(x = 6\), \(y = 2\), correct.

Wait, but maybe I made a mistake. Let's re-express:

For the blue line:

- Y-intercept (\(b\)): when \(x = 0\), \(y = 1\) (from the graph, the blue line crosses the y-axis at \(y = 1\)).

- Slope (\(m\)): take two points on blue line, say \((0,1)\) and \((1,3)\). Then \(m=\frac{3 - 1}{1 - 0}=2\). So equation: \(y = 2x+1\).

For the red line:

- Y-intercept (\(b\)): when \(x = 0\), \(y=-2\) (crosses y-axis at \(y = - 2\)).

- Slope (\(m\)): take two points, say \((0,-2)\) and \((3,0)\). Then \(m=\frac{0 - (-2)}{3 - 0}=\frac{2}{3}\). So equation: \(y=\frac{2}{3}x-2\).

Wait, but maybe the blue line is \(y = 2x + 1\) and red line is \(y=\frac{2}{3}x-2\). Wait, but let's check the graph again. Alternatively, maybe the blue line has slope 2 and y-intercept 1, red line has slope \(\frac{2}{3}\) and y-intercept -2.

Wait, but maybe I made a mistake in the blue line's y-intercept. Let's check the blue line: when \(x = 0\), the blue line is at \(y = 1\)? Wait, the grid: the y-axis is marked with 8,6,4,2,0,-2,-4,-6,-8. So at \(x = 0\), the blue line is at \(y = 1\)? Wait, no, the blue line passes through (0,1)? Wait, the blue line is above the red line. Wait, the red line passes through (0,-2) and (3,0), so slope 2/3. The blue line passes through (0,1) and (1,3), slope 2. So equations:

Blue line: \(y = 2x + 1\)

Red line: \(y=\frac{2}{3}x-2\)

Wait, but maybe the blue line is \(y = 2x + 1\) and red line is \(y=\frac{1}{2}x-2\)? Wait, no, let's take the red line: from (0,-2) to (4,0), slope (0 - (-2))/(4 - 0) = 2/4 = 1/2. Oh! Maybe I messed up the x-coordinate. Let's see, the red line: when x = 4, y = 0? Wait, the red line goes from (0,-2) to (4,0), so slope (0 - (-2))/(4 - 0) = 2/4 = 1/2. Ah, that's better. So red line: y-intercept b = -2, slope m = 1/2. So equation: \(y=\frac{1}{2}x-2\).

Now the blue line: when x = 0, y = 1? Wait, no, when x = 0, the blue line is at y = 1? Wait, no, the blue line passes through (0,1) and (1,3), slope 2. So equation: \(y = 2x + 1\). Wait, but when x = 1, y = 3, which is correct. When x = 2, y = 5, which is on the blue line.

Wait, but let's verify with the graph. The red line: from (0,-2) to (4,0), so when x = 4, y = 0. The blue line: from (0,1) to (1,3), so when x = 1, y = 3.

Alternatively, maybe the blue line has y-intercept 1 and slope 2, red line has y-intercept -2 and slope 1/2. So the system is:

\(y = 2x + 1\)

\(y=\frac{1}{2}x-2\)

Wait, but maybe I made a mistake. Let's re-express:

For the red line:

Points: (0, -2) and (4, 0). So slope \(m=\frac{0 - (-2)}{4 - 0}=\frac{2}{4}=\frac{1}{2}\). So equation: \(y=\frac{1}{2}x-2\).

For the blue line:

Points: (0, 1) and (1, 3). Slope \(m=\frac{3 - 1}{1 - 0}=2\). So equation: \(y = 2x + 1\).

Yes, that makes sense. So the two equations are \(y = 2x + 1\) and \(y=\frac{1}{2}x-2\).

Answer:

\(y = 2x + 1\)

\(y=\frac{1}{2}x-2\)