Question

Below is the graph of a system of two linear equations.

A coordinate plane with two lines is shown. A line with positive slope crosses the x-axis at (negative 8, 0) and the y-axis at (0, 8). A line with a negative slope crosses the x-axis at (negative 2, 0) and the y-axis at (0, negative 2). The lines intersect at (negative 5, 3).

What system does it represent, and what is the solution of this system?

Answer 1

The graph you described represents a system of two linear equations.

Let's denote the two lines based on their intercepts:

1. The first line, which has a positive slope and crosses the x-axis at \((-8, 0)\) and the y-axis at \((0, 8)\), can be expressed in slope-intercept form \(y = mx + b\). The slope \(m\) can be calculated as follows:

\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 0}{0 - (-8)} = \frac{8}{8} = 1
\]

The equation of the first line is:

\[
y = 1x + 8 \Rightarrow y = x + 8
\]

2. The second line, which has a negative slope and crosses the x-axis at \((-2, 0)\) and the y-axis at \((0, -2)\), also can be expressed in slope-intercept form. The slope \(m\) is calculated as follows:

\[
m = \frac{-2 - 0}{0 - (-2)} = \frac{-2}{2} = -1
\]

The equation of the second line is:

\[
y = -1x - 2 \Rightarrow y = -x - 2
\]

Now, the system of equations can be represented as:

\[
\begin{align*}
1.\quad y &= x + 8 \\
2.\quad y &= -x - 2
\end{align*}
\]

To find the solution of this system, we substitute \(y\) from the first equation into the second equation:

\[
x + 8 = -x - 2
\]

Now, solve for \(x\):

\[
x + x = -2 - 8
\]
\[
2x = -10
\]
\[
x = -5
\]

Now substitute \(x = -5\) back into the first equation to find \(y\):

\[
y = -5 + 8 = 3
\]

Thus, the solution of this system of equations is:

\[
\boxed{(-5, 3)}
\]

This confirms that the lines intersect at the coordinate \((-5, 3)\), which is the solution to the system.