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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 (
).

1 answer

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:

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