Which system is equivalent to

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

A.

[tex]\[
\left\{
\begin{array}{r}
5x^2 + 6y^2 = 50 \\
-21x^2 - 6y^2 = 10
\end{array}
\right.
\][/tex]

B.

[tex]\[
\left\{
\begin{array}{r}
5x^2 + 6y^2 = 50 \\
-21x^2 - 6y^2 = 30
\end{array}
\right.
\][/tex]

C.

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

D.

[tex]\[
\left\{
\begin{array}{l}
35x^2 + 42y^2 = 350 \\
-35x^2 - 10y^2 = -50
\end{array}
\right.
\][/tex]

To find an equivalent system to the given set of equations:

1. Given System:
[tex]\[
\begin{cases}
5x^2 + 6y^2 = 50 \\
7x^2 + 2y^2 = 10
\end{cases}
\][/tex]

2. Finding an Equivalent System:
- We can create equivalent systems by performing algebraic operations such as multiplying the equations by constants.

3. Multiply First Equation by 7:
[tex]\[
7 \times (5x^2 + 6y^2) = 7 \times 50
\][/tex]
This results in:
[tex]\[
35x^2 + 42y^2 = 350
\][/tex]

4. Multiply Second Equation by 5:
[tex]\[
5 \times (7x^2 + 2y^2) = 5 \times 10
\][/tex]
This results in:
[tex]\[
35x^2 + 10y^2 = 50
\][/tex]

5. Subtract the Second Transformed Equation from the First:
[tex]\[
(35x^2 + 42y^2) - (35x^2 + 10y^2) = 350 - 50
\][/tex]
Simplifying gives:
[tex]\[
-35x^2 - 10y^2 = -50
\][/tex]

6. Equivalent System Found:
[tex]\[
\begin{cases}
35x^2 + 42y^2 = 350 \\
-35x^2 - 10y^2 = -50
\end{cases}
\][/tex]

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

This corresponds to the third choice in the options provided.