Which system is equivalent to

\[
\left\{
\begin{array}{l}
5x^2 + 6y^2 = 50 \\
7x^2 + 2y^2 = 10
\end{array}
\right.
\]

A.
\[
\left\{
\begin{aligned}
5x^2 + 6y^2 & = 50 \\
-21x^2 - 6y^2 & = 10
\end{aligned}
\right.
\]

B.
\[
\left\{
\begin{aligned}
5x^2 + 6y^2 & = 50 \\
-21x^2 - 6y^2 & = 30
\end{aligned}
\right.
\]

C.
\[
\left\{
\begin{aligned}
35x^2 + 42y^2 & = 250 \\
-35x^2 - 10y^2 & = -50
\end{aligned}
\right.
\]

D.
\[
\left\{
\begin{aligned}
35x^2 + 42y^2 & = 350 \\
-35x^2 - 10y^2 & = -50
\end{aligned}
\right.
\]

We start with the original system:
[tex]\[
\begin{cases}
5x^2 + 6y^2 = 50, \\
7x^2 + 2y^2 = 10.
\end{cases}
\][/tex]

Step 1. Multiply the first equation by 7

Multiplying the equation
[tex]\[
5x^2 + 6y^2 = 50
\][/tex]
by 7 gives:
[tex]\[
7(5x^2 + 6y^2) = 7 \cdot 50,
\][/tex]
which simplifies to:
[tex]\[
35x^2 + 42y^2 = 350.
\][/tex]

Step 2. Multiply the second equation by 5

Multiplying the equation
[tex]\[
7x^2 + 2y^2 = 10
\][/tex]
by 5 gives:
[tex]\[
5(7x^2 + 2y^2) = 5 \cdot 10,
\][/tex]
which simplifies to:
[tex]\[
35x^2 + 10y^2 = 50.
\][/tex]

Step 3. Multiply the result of Step 2 by [tex]$-1$[/tex]

To obtain a system where the equations have opposite [tex]$x^2$[/tex] terms, we multiply the second equation by [tex]$-1$[/tex]:
[tex]\[
-1(35x^2 + 10y^2) = -1 \cdot 50,
\][/tex]
which simplifies to:
[tex]\[
-35x^2 - 10y^2 = -50.
\][/tex]

Now, we have transformed the original system into the following equivalent system:
[tex]\[
\begin{cases}
35x^2 + 42y^2 = 350, \\
-35x^2 - 10y^2 = -50.
\end{cases}
\][/tex]

This corresponds to the fourth option in the question.