Question

A system of linear equations is shown on the graph.

The graph shows two intersecting lines. One line passes through the points (negative 3, negative 4), (0, negative 2), and (3, 0). The other line passes through the points (negative 3, negative 4), (negative 1, 0), (0, 2), and (1, 4).

What is the solution to the system of equations? Type the coordinates in the boxes below.

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Answer 1

To find the solution to the system of equations represented by the two intersecting lines, we first need to determine the equations of both lines based on the points provided.

### Line 1
The first line passes through the points (−3, −4), (0, −2), and (3, 0).

1. We can find the slope (m) using the points (0, -2) and (3, 0):
\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - (-2)}{3 - 0} = \frac{2}{3}
\]

2. Now, we can use the point-slope form of the line equation \(y - y_1 = m(x - x_1)\) with point (0, -2):
\[
y - (-2) = \frac{2}{3}(x - 0) \implies y + 2 = \frac{2}{3}x \implies y = \frac{2}{3}x - 2
\]

### Line 2
The second line passes through the points (−3, −4), (−1, 0), (0, 2), and (1, 4).

1. We can find the slope (m) using the points (−3, -4) and (−1, 0):
\[
m = \frac{0 - (-4)}{-1 - (-3)} = \frac{4}{2} = 2
\]

2. Using the point-slope form with the point (-3, -4):
\[
y - (-4) = 2(x + 3) \implies y + 4 = 2x + 6 \implies y = 2x + 2
\]

### Finding the Intersection
Now, we have two equations:
1. \(y = \frac{2}{3}x - 2\) (Line 1)
2. \(y = 2x + 2\) (Line 2)

To find the intersection, we set the two equations equal:
\[
\frac{2}{3}x - 2 = 2x + 2
\]

To eliminate the fraction, multiply through by 3:
\[
2x - 6 = 6x + 6
\]

Now, move all terms involving x to one side and constant terms to the other:
\[
2x - 6x = 6 + 6 \implies -4x = 12 \implies x = -3
\]

Now substitute \(x = -3\) back into either equation to find \(y\):
Using Line 1:
\[
y = \frac{2}{3}(-3) - 2 = -2 - 2 = -4
\]

Thus the solution to the system of equations is:
\[
(-3, -4)
\]

The coordinates are:
- \(x = -3\)
- \(y = -4\)

Final answer:
**(-3, -4)**