Question

Use the image to answer the question.

A coordinate plane with four quadrants shows the x- and y-axes ranging from negative 5 to 5 in increments of 1. A solid line and a dotted line intersect each other. The equation of the solid line is x minus 5 y equals 3. The equation of the dotted line is 3 x minus 2 y equals negative 4. The intersection of both lines is shown at negative 2 on the x-axis and negative 1 on the y-axis in quadrant 3.

Review the graphs of a system of two linear equations in two variables: x−5y=7
and 3x−2y=−4
. Find the solution to both equations.

(1 point)
The intersection point is (
).

Answer 1

To find the solution to the system of equations given by the lines \( x - 5y = 7 \) and \( 3x - 2y = -4 \), we need to solve the equations simultaneously.

1. **Set the equations**:

The equations are:
\[
x - 5y = 7 \quad (1)
\]
\[
3x - 2y = -4 \quad (2)
\]

2. **Solve equation (1) for \( x \)**:
\[
x = 5y + 7
\]

3. **Substitute \( x \) into equation (2)**:
Substitute \( x \) in equation (2):
\[
3(5y + 7) - 2y = -4
\]
Expand and simplify:
\[
15y + 21 - 2y = -4
\]
\[
13y + 21 = -4
\]
Subtract 21 from both sides:
\[
13y = -25
\]
Divide by 13:
\[
y = -\frac{25}{13}
\]

4. **Substitute \( y \) back to find \( x \)**:
Now substitute \( y \) back into \( x = 5y + 7 \):
\[
x = 5\left(-\frac{25}{13}\right) + 7
\]
\[
x = -\frac{125}{13} + \frac{91}{13}
\]
\[
x = -\frac{125 - 91}{13}
\]
\[
x = -\frac{34}{13}
\]

5. **Final intersection point**:
The intersection point of the two lines (the solution to the system of equations) is:
\[
\left(-\frac{34}{13}, -\frac{25}{13}\right)
\]

Thus, the intersection point is:

\[
\left(-\frac{34}{13}, -\frac{25}{13}\right)
\]