Question

Solve the system of equations.
–x+6y=13
4x–10y=–10 (1 point)

Answer 1

To solve the system of equations

1. \(-x + 6y = 13\)
2. \(4x - 10y = -10\)

we can use the substitution or elimination method. Here, I will use the substitution method.

First, let's solve the first equation for \(x\):

\[
-x + 6y = 13
\]

We can rewrite this as:

\[
x = 6y - 13
\]

Next, we will substitute this expression for \(x\) in the second equation:

\[
4(6y - 13) - 10y = -10
\]

Now, distribute \(4\) in the equation:

\[
24y - 52 - 10y = -10
\]

Combine like terms:

\[
(24y - 10y) - 52 = -10
\]

This simplifies to:

\[
14y - 52 = -10
\]

Now, add \(52\) to both sides:

\[
14y = -10 + 52
\]

\[
14y = 42
\]

Now, divide by \(14\):

\[
y = \frac{42}{14} = 3
\]

Now that we have \(y = 3\), we can substitute this value back into the equation we derived for \(x\):

\[
x = 6(3) - 13
\]

\[
x = 18 - 13
\]

\[
x = 5
\]

Thus, the solution to the system of equations is:

\[
\boxed{(5, 3)}
\]

To verify, we can substitute \(x = 5\) and \(y = 3\) back into the original equations:

1. For the equation \(-x + 6y = 13\):

\[
-5 + 6(3) = -5 + 18 = 13 \quad \text{(True)}
\]

2. For the equation \(4x - 10y = -10\):

\[
4(5) - 10(3) = 20 - 30 = -10 \quad \text{(True)}
\]

The solution \( (5, 3) \) satisfies both equations.