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

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

To determine which system is equivalent to the given system of equations:

1. Start with the original system:
[tex]\[
\begin{array}{l}
5x^2 + 6y^2 = 50 \quad \text{(Equation 1)} \\
7x^2 + 2y^2 = 10 \quad \text{(Equation 2)}
\end{array}
\][/tex]

2. To eliminate the coefficients of [tex]\(x^2\)[/tex], we can multiply Equation 1 and Equation 2 by factors that will allow the terms involving [tex]\(x^2\)[/tex] to be combined easily:

- Multiply Equation 1 by 7:
[tex]\[
35x^2 + 42y^2 = 350
\][/tex]

- Multiply Equation 2 by 5, and then negate the result to help with combination:
[tex]\[
-35x^2 - 10y^2 = -50
\][/tex]

3. The transformed system of equations is:
[tex]\[
\begin{array}{l}
35x^2 + 42y^2 = 350 \\
-35x^2 - 10y^2 = -50
\end{array}
\][/tex]

This transformed system of equations corresponds to the fourth option:
[tex]\[
\left\{
\begin{aligned}
35x^2 + 42y^2 & = 350 \\
-35x^2 - 10y^2 & = -50
\end{aligned}
\right.
\][/tex]

Therefore, the equivalent system is:
[tex]\[
\left\{
\begin{aligned}
35x^2 + 42y^2 & = 350 \\
-35x^2 - 10y^2 & = -50
\end{aligned}
\right.
\][/tex]