Answer :
We start with the original system of equations:
[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
[tex]$$5x^2 + 6y^2 = 50$$[/tex]
by 7, we obtain
[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 and then by -1.
First, multiplying
[tex]$$7x^2 + 2y^2 = 10$$[/tex]
by 5 gives:
[tex]$$
5(7x^2 + 2y^2) = 5 \cdot 10,
$$[/tex]
which results in
[tex]$$
35x^2 + 10y^2 = 50.
$$[/tex]
Then, multiplying the entire equation by [tex]$-1$[/tex], we get:
[tex]$$
-35x^2 - 10y^2 = -50.
$$[/tex]
Step 3. Write the new system.
The equivalent system is now:
[tex]$$
\begin{cases}
35x^2 + 42y^2 = 350, \\
-35x^2 - 10y^2 = -50.
\end{cases}
$$[/tex]
This system corresponds to option 4.
[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
[tex]$$5x^2 + 6y^2 = 50$$[/tex]
by 7, we obtain
[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 and then by -1.
First, multiplying
[tex]$$7x^2 + 2y^2 = 10$$[/tex]
by 5 gives:
[tex]$$
5(7x^2 + 2y^2) = 5 \cdot 10,
$$[/tex]
which results in
[tex]$$
35x^2 + 10y^2 = 50.
$$[/tex]
Then, multiplying the entire equation by [tex]$-1$[/tex], we get:
[tex]$$
-35x^2 - 10y^2 = -50.
$$[/tex]
Step 3. Write the new system.
The equivalent system is now:
[tex]$$
\begin{cases}
35x^2 + 42y^2 = 350, \\
-35x^2 - 10y^2 = -50.
\end{cases}
$$[/tex]
This system corresponds to option 4.
