Question

Solve using substitution.

10x − 6y = –10
–10x + 7y = 20

(
,
)

Answer 1

To solve the system of equations using substitution, we will first solve one of the equations for one variable and then substitute it into the other equation.

Given the equations:
1. \( 10x - 6y = -10 \)
2. \( -10x + 7y = 20 \)

Let's solve the first equation for \( y \):

1. \( 10x - 6y = -10 \)

Rearranging to solve for \( y \):

\[
-6y = -10 - 10x
\]
\[
6y = 10 + 10x
\]
\[
y = \frac{10 + 10x}{6}
\]
\[
y = \frac{5}{3} + \frac{5}{3}x
\]

Now that we have \( y \) expressed in terms of \( x \), we can substitute this expression into the second equation.

Substituting into the second equation:

2. \( -10x + 7y = 20 \)

\[
-10x + 7\left(\frac{5}{3} + \frac{5}{3}x\right) = 20
\]

Now we simplify and solve for \( x \):

\[
-10x + \frac{35}{3} + \frac{35}{3}x = 20
\]

To eliminate the fraction, multiply the entire equation by 3:

\[
-30x + 35 + 35x = 60
\]

Combine like terms:

\[
5x + 35 = 60
\]

Isolating \( x \):

\[
5x = 60 - 35
\]
\[
5x = 25
\]
\[
x = 5
\]

Now that we have \( x \), we can substitute this back into our expression for \( y \):

\[
y = \frac{5}{3} + \frac{5}{3}(5)
\]
\[
y = \frac{5}{3} + \frac{25}{3}
\]
\[
y = \frac{30}{3}
\]
\[
y = 10
\]

Thus, the solution to the system of equations is:

\((x, y) = (5, 10)\)

So the solution is:

\((5, 10)\)