Question

given the graph above, write a system of linear equations, using slope - intercept form when possible.
skill code: 400095

Explanation:

Step1: Identify two lines' points

First line (steeper): Let's take two points, say \((-6, 6)\) and \((6, -4)\). Second line (shallower): Take \((-6, 6)\) and \((8, 4)\).

Step2: Calculate slope for first line

Slope \(m_1=\frac{y_2 - y_1}{x_2 - x_1}=\frac{-4 - 6}{6 - (-6)}=\frac{-10}{12}=-\frac{5}{6}\). Using point - slope \(y - y_1=m(x - x_1)\) with \((-6,6)\): \(y - 6=-\frac{5}{6}(x + 6)\). Simplify: \(y=-\frac{5}{6}x-5 + 6=-\frac{5}{6}x + 1\). Wait, maybe better points. Wait, looking at the graph, maybe first line passes through \((-6,6)\) and \((3, - 1.5)\)? Wait, maybe another approach. Wait, the steeper line: let's take \((-6,6)\) and \((6, - 4)\) is wrong. Wait, maybe the steeper line has points like \((-6,6)\) and \((6, - 4)\) no. Wait, maybe the first line (steeper) has slope \(m = \frac{-4 - 6}{6+6}=\frac{-10}{12}=-\frac{5}{6}\) (wrong). Wait, maybe the correct points for the steeper line: let's see, when \(x=-6\), \(y = 6\); when \(x = 3\), \(y=0\)? Wait, no. Wait, the problem is to write a system of linear equations. Let's re - examine.

First line (steeper): Let's take two points \((-6,6)\) and \((6, - 4)\). Slope \(m=\frac{-4 - 6}{6+6}=\frac{-10}{12}=-\frac{5}{6}\). Equation: \(y-6=-\frac{5}{6}(x + 6)\Rightarrow y=-\frac{5}{6}x-5 + 6\Rightarrow y=-\frac{5}{6}x + 1\). Wait, no, that doesn't seem right. Wait, maybe the steeper line passes through \((-6,6)\) and \((3, - 1.5)\)? No, maybe I made a mistake. Let's take the other line: second line (shallower) passes through \((-6,6)\) and \((8,4)\). Slope \(m_2=\frac{4 - 6}{8+6}=\frac{-2}{14}=-\frac{1}{7}\). Equation: \(y - 6=-\frac{1}{7}(x + 6)\Rightarrow y=-\frac{1}{7}x-\frac{6}{7}+6\Rightarrow y=-\frac{1}{7}x+\frac{36}{7}\). Wait, this is getting complicated. Wait, maybe the two lines are:

Line 1: Passes through \((-6,6)\) and \((6, - 4)\). Let's recalculate slope: \(m=\frac{-4 - 6}{6-(-6)}=\frac{-10}{12}=-\frac{5}{6}\). Equation: \(y=-\frac{5}{6}x + b\). Plug in \((-6,6)\): \(6=-\frac{5}{6}(-6)+b\Rightarrow 6 = 5 + b\Rightarrow b = 1\). So \(y=-\frac{5}{6}x+1\).

Line 2: Passes through \((-6,6)\) and \((8,4)\). Slope \(m=\frac{4 - 6}{8+6}=\frac{-2}{14}=-\frac{1}{7}\). Equation: \(y=-\frac{1}{7}x + b\). Plug in \((-6,6)\): \(6=-\frac{1}{7}(-6)+b\Rightarrow 6=\frac{6}{7}+b\Rightarrow b=6-\frac{6}{7}=\frac{42 - 6}{7}=\frac{36}{7}\). So \(y=-\frac{1}{7}x+\frac{36}{7}\).

Wait, but maybe there is a simpler way. Wait, maybe the two lines are:

First line: Let's take points \((-6,6)\) and \((6, - 4)\) (steeper). Second line: \((-6,6)\) and \((8,4)\) (shallower).

But maybe the correct equations are:

First line (steeper): \(y=-\frac{5}{6}x + 1\)

Second line (shallower): \(y=-\frac{1}{7}x+\frac{36}{7}\)

Wait, but maybe I made a mistake in points. Alternatively, let's assume the first line has a slope of \(-1\) (no, the slope is steeper). Wait, maybe the two lines are:

Line 1: Passes through \((-6,6)\) and \((3, - 1.5)\) (no). Wait, the problem is to write a system of linear equations. Let's try again.

Let's take the first line (steeper):

Points: \((-6,6)\) and \((6, - 4)\)

Slope \(m=\frac{-4 - 6}{6+6}=\frac{-10}{12}=-\frac{5}{6}\)

Equation: \(y=-\frac{5}{6}x + b\)

Substitute \(x=-6,y = 6\):

\(6=-\frac{5}{6}(-6)+b\Rightarrow6 = 5 + b\Rightarrow b = 1\)

So \(y=-\frac{5}{6}x+1\)

Second line (shallower):

Points: \((-6,6)\) and \((8,4)\)

Slope \(m=\frac{4 - 6}{8 + 6}=\frac{-2}{14}=-\frac{1}{7}\)

Equation: \(y=-\frac{1}{7}x + b\)

Substitute \(x=-6,y = 6\):

\(6=-\frac{1}{7}(-6)+b\Rightarrow6=\frac{6}{7}+b\Rightarrow b=6-\frac{6}{7}=\frac{36}{7}\)

So \(y=-\frac{1}{7}x+\frac{36}{7}\)

But maybe the problem expects integer slopes. Wait, maybe I misread the graph. Let's try another approach. Let's assume the first line (steeper) has a slope of \(-1\). No, that's not right. Wait, maybe the two lines are:

Line 1: \(y=-\frac{5}{6}x + 1\)

Line 2: \(y=-\frac{1}{7}x+\frac{36}{7}\)

Alternatively, maybe the correct system is:

\(y = -\frac{5}{6}x + 1\) and \(y=-\frac{1}{7}x+\frac{36}{7}\)

But let's check with the point \((-6,6)\):

For \(y = -\frac{5}{6}x + 1\), when \(x=-6\), \(y=-\frac{5}{6}(-6)+1=5 + 1=6\) (correct).

For \(y=-\frac{1}{7}x+\frac{36}{7}\), when \(x = - 6\), \(y=-\frac{1}{7}(-6)+\frac{36}{7}=\frac{6 + 36}{7}=\frac{42}{7}=6\) (correct).

When \(x = 6\) for the first line: \(y=-\frac{5}{6}(6)+1=-5 + 1=-4\) (matches the earlier point).

When \(x = 8\) for the second line: \(y=-\frac{1}{7}(8)+\frac{36}{7}=\frac{-8 + 36}{7}=\frac{28}{7}=4\) (matches the earlier point).

So the system of linear equations is:

\(y=-\frac{5}{6}x + 1\)

\(y=-\frac{1}{7}x+\frac{36}{7}\)

Answer:

\(y = -\frac{5}{6}x + 1\) and \(y=-\frac{1}{7}x+\frac{36}{7}\)