Question

Solve the following system of equations. 2x-5y = 6 -2x+10y = -16

Answer 1

To solve the system of equations:

\[
\begin{align*}
1) & \quad 2x - 5y = 6 \quad \text{(Equation 1)} \\
2) & \quad -2x + 10y = -16 \quad \text{(Equation 2)}
\end{align*}
\]

We can start by manipulating these equations. Notice that Equation 2 can be simplified to align its structure with Equation 1. If we divide all terms of Equation 2 by -2, we get:

\[
\frac{-2x}{-2} + \frac{10y}{-2} = \frac{-16}{-2} \implies x - 5y = 8 \quad \text{(Equation 3)}
\]

Now we have a new system of equations:

\[
\begin{align*}
1) & \quad 2x - 5y = 6 \\
3) & \quad x - 5y = 8
\end{align*}
\]

Next, we can substitute Equation 3 into Equation 1. From Equation 3, we express \( x \) in terms of \( y \):

\[
x = 8 + 5y
\]

Now substitute \( x \) in Equation 1:

\[
2(8 + 5y) - 5y = 6
\]

Simplifying this gives:

\[
16 + 10y - 5y = 6 \\
16 + 5y = 6
\]

Now we isolate \( y \):

\[
5y = 6 - 16 \\
5y = -10 \\
y = -2
\]

Now that we have \( y \), we can substitute \( y = -2 \) back into Equation 3 to find \( x \):

\[
x - 5(-2) = 8 \\
x + 10 = 8 \\
x = 8 - 10 \\
x = -2
\]

Thus, the solution to the system of equations is:

\[
(x, y) = (-2, -2)
\]

To verify, we can substitute \( x = -2 \) and \( y = -2 \) back into the original equations:

For Equation 1:

\[
2(-2) - 5(-2) = -4 + 10 = 6 \quad \text{(True)}
\]

For Equation 2:

\[
-2(-2) + 10(-2) = 4 - 20 = -16 \quad \text{(True)}
\]

Both equations are satisfied, confirming that our solution is correct.

Finally, the solution is:

\[
\boxed{(-2, -2)}
\]