Use the image to answer the question.

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 (
).
Remaining Attempts : 3

Answer 1

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

1. First equation: \( x - 5y = 7 \)

   We can solve for \( x \): \[ x = 7 + 5y \]

2. Second equation: \( 3x - 2y = -4 \)

   Now, we substitute \( x \) from the first equation into the second equation: \[ 3(7 + 5y) - 2y = -4 \]

   Expanding this: \[ 21 + 15y - 2y = -4 \] This simplifies to: \[ 21 + 13y = -4 \]

3. Solving for \( y \): \[ 13y = -4 - 21 \] \[ 13y = -25 \] \[ y = -\frac{25}{13} \]

4. Finding \( x \): Now substitute \( y \) back into the expression for \( x \): \[ x = 7 + 5\left(-\frac{25}{13}\right) \] \[ x = 7 - \frac{125}{13} \] We convert 7 into a fraction: \[ 7 = \frac{91}{13} \] So: \[ x = \frac{91}{13} - \frac{125}{13} = \frac{91 - 125}{13} = \frac{-34}{13} \]

Thus, the solution to the system of equations (the intersection point) is: \[ \left( -\frac{34}{13}, -\frac{25}{13} \right) \]

The final answer for the intersection point (to fill in the answer box) is: \[ \left( -\frac{34}{13}, -\frac{25}{13} \right) \]