Question

write a system of linear equations for the graph below.
\\(\

$$\begin{cases} \\ y = \\square \\\\ \\ y = \\square \\end{cases}$$

\\)

Explanation:

Step1: Find the first line's equation

The steeper line passes through \((0, 5)\) (y-intercept \(b = 5\)) and \((2, -4)\)? Wait, no, let's check points. Wait, when \(x = 0\), \(y = 5\); when \(x = 2\), let's see, the intersection point? Wait, maybe better to take two points. Let's take \((0, 5)\) and \((2, -4)\)? Wait, no, the steeper line: from \((0,5)\) to \((2, -4)\)? Wait, slope \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Let's take \((0,5)\) and \((2, -4)\)? Wait, no, when \(x = 0\), \(y = 5\); when \(x = 2\), let's see the graph. Wait, the steeper line: when \(x = 0\), \(y = 5\); when \(x = 2\), \(y = -4\)? Wait, no, maybe \((0,5)\) and \((1, 1)\)? Wait, no, let's recalculate. Wait, the steeper line: let's take two points. Let's say \((0, 5)\) and \((2, -4)\)? Wait, no, the slope between \((0,5)\) and \((2, -4)\) is \(\frac{-4 - 5}{2 - 0}=\frac{-9}{2}=-4.5\)? Wait, no, maybe I made a mistake. Wait, the other line: the shallower one passes through \((-9, 0)\) and \((0, -3)\). Let's do that first. For the shallower line: points \((-9, 0)\) and \((0, -3)\). Slope \(m=\frac{-3 - 0}{0 - (-9)}=\frac{-3}{9}=-\frac{1}{3}\). Y-intercept \(b = -3\) (since when \(x = 0\), \(y = -3\)). So equation is \(y = -\frac{1}{3}x - 3\).

Step2: Find the second line's equation

The steeper line: passes through \((0, 5)\) (y-intercept \(b = 5\)) and \((2, -4)\)? Wait, no, let's take \((0,5)\) and \((2, -4)\)? Wait, slope \(m=\frac{-4 - 5}{2 - 0}=\frac{-9}{2}\)? No, wait, when \(x = 0\), \(y = 5\); when \(x = 2\), let's see the intersection point. Wait, the intersection point is \((2, -4)\)? Wait, no, the two lines intersect at \((2, -4)\)? Wait, no, let's check the shallower line: when \(x = 2\), \(y = -\frac{1}{3}(2) - 3=-\frac{2}{3}-3=-\frac{11}{3}\approx -3.666\)? No, that's not right. Wait, maybe the steeper line: points \((0,5)\) and \((2, -4)\) is wrong. Wait, let's take \((0,5)\) and \((1, 1)\)? No, when \(x = 1\), \(y = 1\)? Wait, no, the steeper line: from \((0,5)\) to \((2, -4)\) is incorrect. Wait, maybe the steeper line has slope \(-4\)? Wait, when \(x = 0\), \(y = 5\); when \(x = 1\), \(y = 1\)? No, \(5 - 4 = 1\), so slope \(-4\). So equation \(y = -4x + 5\). Let's check: when \(x = 0\), \(y = 5\) (correct). When \(x = 2\), \(y = -8 + 5 = -3\)? No, that's not matching. Wait, maybe the intersection point is \((2, -3)\)? No, the graph shows the intersection at \((2, -4)\)? Wait, I think I messed up. Let's start over.

For the shallower line:
- Points: \((-9, 0)\) and \((0, -3)\).
- Slope \(m = \frac{-3 - 0}{0 - (-9)} = \frac{-3}{9} = -\frac{1}{3}\).
- Y-intercept \(b = -3\) (since it crosses y-axis at \((0, -3)\)).
- Equation: \(y = -\frac{1}{3}x - 3\).

For the steeper line:
- Points: \((0, 5)\) (y-intercept \(b = 5\)) and \((2, -4)\)? Wait, no, when \(x = 2\), what's \(y\)? Let's see, the steeper line: when \(x = 0\), \(y = 5\); when \(x = 2\), \(y = -4\)? Then slope \(m = \frac{-4 - 5}{2 - 0} = \frac{-9}{2} = -4.5\)? No, that can't be. Wait, maybe the steeper line passes through \((0, 5)\) and \((1, 1)\)? Then slope \(m = \frac{1 - 5}{1 - 0} = -4\). So equation \(y = -4x + 5\). Let's check when \(x = 2\), \(y = -8 + 5 = -3\), but the intersection point should be where both lines meet. Wait, maybe the intersection is at \((2, -3)\)? No, the graph shows the intersection at \((2, -4)\)? I think I made a mistake. Wait, let's look at the shallower line: when \(x = -9\), \(y = 0\); when \(x = 0\), \(y = -3\); when \(x = 3\), \(y = -4\). Ah! So the intersection point is \((3, -4)\). So for the steeper line: passes through \((0, 5)\) and \((3, -4)\). Then slope \(m = \frac{-4 - 5}{3 - 0} = \frac{-9}{3} = -3\). So equation: \(y = -3x + 5\) (since \(b = 5\) when \(x = 0\), \(y = 5\)). Let's check: when \(x = 3\), \(y = -9 + 5 = -4\) (correct, matches the intersection point). For the shallower line: passes through \((-9, 0)\) and \((3, -4)\). Slope \(m = \frac{-4 - 0}{3 - (-9)} = \frac{-4}{12} = -\frac{1}{3}\). Y-intercept: when \(x = 0\), \(y = -\frac{1}{3}(0) - 3 = -3\) (since using point \((0, -3)\)). So equation \(y = -\frac{1}{3}x - 3\). Let's verify at \(x = 3\): \(y = -\frac{1}{3}(3) - 3 = -1 - 3 = -4\) (correct, matches intersection).

So the two equations are \(y = -3x + 5\) and \(y = -\frac{1}{3}x - 3\).

Wait, let's confirm the steeper line: when \(x = 0\), \(y = 5\) (correct, as per graph). When \(x = 1\), \(y = -3(1) + 5 = 2\) (matches the graph? Let's see, the steeper line at \(x = 1\) is at \(y = 2\)? Yes, that looks right. The shallower line at \(x = 1\): \(y = -\frac{1}{3}(1) - 3 = -\frac{10}{3}\approx -3.333\), which matches the graph.

Step3: Write the system

So the system is:
\(y = -3x + 5\)
\(y = -\frac{1}{3}x - 3\)

Answer:

\(

$$\begin{cases} y = -3x + 5 \\ y = -\frac{1}{3}x - 3 \end{cases}$$

\)