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..

Answer 1

To find the solution to the system of equations given by:

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

we will solve the system using the substitution or elimination method. Let’s use substitution here.

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

From the first equation: \[ x - 5y = 7 \]

We can express \( x \) in terms of \( y \): \[ x = 5y + 7 \]

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

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

Distributing \( 3 \): \[ 15y + 21 - 2y = -4 \]

Now combine like terms: \[ (15y - 2y) + 21 = -4 \] \[ 13y + 21 = -4 \]

Step 3: Isolate \( y \)

Subtract \( 21 \) from both sides: \[ 13y = -4 - 21 \] \[ 13y = -25 \]

Now divide by \( 13 \): \[ y = -\frac{25}{13} \]

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

Now substitute \( y \) back into the equation for \( x \): \[ x = 5\left(-\frac{25}{13}\right) + 7 \] \[ x = -\frac{125}{13} + 7 \] Convert \( 7 \) to a fraction with a denominator of \( 13 \): \[ 7 = \frac{91}{13} \] So, \[ x = -\frac{125}{13} + \frac{91}{13} \] \[ x = \frac{-125 + 91}{13} \] \[ x = \frac{-34}{13} \]

Step 5: Write the solution

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

This point gives the \( x \) and \( y \) coordinates where the two lines intersect.