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..

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.