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 can solve the equations using the substitution or elimination method. Let's solve it using the substitution method.

### Step 1: Solve for \( x \) in terms of \( y \) from the first equation

From the first equation \( x - 5y = 7 \):

\[
x = 5y + 7
\]

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

Now substitute \( x \) in the second equation:

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

### Step 3: Simplify and solve for \( y \)

Expanding this we get:

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

Combine like terms:

\[
13y + 21 = -4
\]

Now isolate \( y \):

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

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

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

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

Calculating \( 5\left(-\frac{25}{13}\right) \):

\[
x = -\frac{125}{13} + 7
\]
Convert 7 to a fraction with a denominator of 13:

\[
7 = \frac{91}{13}
\]

Now, combine the fractions:

\[
x = -\frac{125}{13} + \frac{91}{13}
\]
\[
x = \frac{-125 + 91}{13}
\]
\[
x = \frac{-34}{13}
\]

### Final Result

The solution to the system of equations is:

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

This point is the intersection of the two lines represented by the equations.