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 (
).

Answer 1

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

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

2. Substitute \( x \) into the second equation: \[ 3(5y + 7) - 2y = -4 \] Simplifying this gives: \[ 15y + 21 - 2y = -4 \ 13y + 21 = -4 \] Now, isolate \( y \): \[ 13y = -4 - 21 \ 13y = -25 \ y = -\frac{25}{13} \]

3. Substitute \( y \) back into the expression for \( x \): \[ x = 5\left(-\frac{25}{13}\right) + 7 \ x = -\frac{125}{13} + 7 \ x = -\frac{125}{13} + \frac{91}{13} \ x = -\frac{125 - 91}{13} \ x = -\frac{34}{13} \]

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

So, the final answer is: \[ \text{The intersection point is } \left(-\frac{34}{13}, -\frac{25}{13}\right). \]

Answer 2

wrong

Answer 3

I apologize for any confusion caused by the earlier response. Let's solve the system of equations correctly.

Given:

1. \( x - 5y = 7 \)
2. \( 3x - 2y = -4 \)

Step 1: Solve for one variable in one equation

Let's express \( x \) in terms of \( y \) using the first equation:

\[ x = 5y + 7 \]

Step 2: Substitute into the second equation

Now substitute \( x = 5y + 7 \) into the second equation:

\[ 3(5y + 7) - 2y = -4 \]

Step 3: Simplify that equation

Distributing \( 3 \):

\[ 15y + 21 - 2y = -4 \]

Combine like terms:

\[ (15y - 2y) + 21 = -4 \ 13y + 21 = -4 \]

Now isolate \( y \):

\[ 13y = -4 - 21 \ 13y = -25 \ y = -\frac{25}{13} \]

Step 4: Substitute back to find \( x \)

Substituting \( y \) back into the equation for \( x \):

\[ x = 5\left(-\frac{25}{13}\right) + 7 \ x = -\frac{125}{13} + 7 \ x = -\frac{125}{13} + \frac{91}{13} \ x = -\frac{125 - 91}{13} \ x = -\frac{34}{13} \]

So the solution to the system of equations is:

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

Please check the parameters of the original equations given in your query; if they have any discrepancies, let me know, and I will fix them accordingly.

If instead the equation you were working with is \( x - 5y = 3 \) and \( 3x - 2y = -4 \) (as seen in the question about the image, with a solution given at (-2, -1)), let's use them too:

1. \( x - 5y = 3 \)
2. \( 3x - 2y = -4 \)

Let’s solve this new set step-by-step:

Step 1: Solve for \( x \) in the first equation:

\[ x = 5y + 3 \]

Step 2: Substitute \( x \) into the second equation:

\[ 3(5y + 3) - 2y = -4 \] \[ 15y + 9 - 2y = -4 \] \[ 13y + 9 = -4 \] \[ 13y = -4 - 9 \] \[ 13y = -13 \] \[ y = -1 \]

Step 3: Substitute back to find \( x \):

\[ x = 5(-1) + 3 \] \[ x = -5 + 3 \] \[ x = -2 \]

Thus, the correct intersection point for \( x - 5y = 3 \) and \( 3x - 2y = -4 \) is:

\[ (-2, -1) \]

Final Answer:

The intersection point is \( (-2, -1) \). Thank you for your patience!