High School

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

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

Answer :

Sure! Let's determine which system of equations is equivalent to the given system:

Original system:
1. [tex]\( 5x^2 + 6y^2 = 50 \)[/tex]
2. [tex]\( 7x^2 + 2y^2 = 10 \)[/tex]

To find an equivalent system, we'll use a common technique of creating a new system by transforming these equations. Here's how you can do it:

1. Multiply the first equation by 7:

Multiply each term in the equation [tex]\( 5x^2 + 6y^2 = 50 \)[/tex] by 7:
[tex]\[
7 \times (5x^2) + 7 \times (6y^2) = 7 \times 50
\][/tex]
This results in:
[tex]\[
35x^2 + 42y^2 = 350
\][/tex]

2. Multiply the second equation by 5:

Multiply each term in the equation [tex]\( 7x^2 + 2y^2 = 10 \)[/tex] by 5:
[tex]\[
5 \times (7x^2) + 5 \times (2y^2) = 5 \times 10
\][/tex]
This gives us:
[tex]\[
35x^2 + 10y^2 = 50
\][/tex]

3. Form the equivalent system and adjustment:

Now subtract the second result from the first result of step 1:
[tex]\[
(35x^2 + 42y^2) - (35x^2 + 10y^2) = 350 - 50
\][/tex]
[tex]\[
35x^2 + 42y^2 - 35x^2 - 10y^2 = 350 - 50
\][/tex]
[tex]\[
32y^2 = 300
\][/tex]
Solving this adjustment creates the appearance of the second equation as:
[tex]\[
-35x^2 - 10y^2 = -50
\][/tex]

So, the equivalent system of equations is:
1. [tex]\( 35x^2 + 42y^2 = 350 \)[/tex]
2. [tex]\( -35x^2 - 10y^2 = -50 \)[/tex]

Thus, the correct choice from the given options is:
[tex]\[
\left\{\begin{array}{r}35x^2+42y^2=350 \\ -35x^2-10y^2=-50\end{array}\right.
\][/tex]