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.

Question

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 intersection point of the two equations \(x - 5y = 7\) and \(3x - 2y = -4\), we can solve the system of equations.

Step 1: Rewrite the first equation. From \(x - 5y = 7\): \[ x = 5y + 7 \]

Step 2: Substitute \(x\) in the second equation. Now, substitute \(x\) in the second equation \(3x - 2y = -4\): \[ 3(5y + 7) - 2y = -4 \]

Step 3: Distribute and combine like terms. \[ 15y + 21 - 2y = -4 \] Combine the \(y\) terms: \[ 13y + 21 = -4 \]

Step 4: Solve for \(y\). Subtract 21 from both sides: \[ 13y = -4 - 21 \] \[ 13y = -25 \] Divide by 13: \[ y = -\frac{25}{13} \]

Step 5: Substitute \(y\) back to find \(x\). Now substitute \(y\) back into the equation \(x = 5y + 7\): \[ x = 5\left(-\frac{25}{13}\right) + 7 \] \[ x = -\frac{125}{13} + 7 \] Convert 7 to a fraction: \[ x = -\frac{125}{13} + \frac{91}{13} \] Combine: \[ x = -\frac{125 - 91}{13} \] \[ x = -\frac{34}{13} \]

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