which system is equivalent to $\begin{cases}3x^{2}-4y^{2}=25\\-6x^{2}-2y^{2}=11end{cases}$…

which system is equivalent to $\begin{cases}3x^{2}-4y^{2}=25\\-6x^{2}-2y^{2}=11end{cases}$? $\begin{cases}3x^{2}-4y^{2}=25\\12x^{2}+4y^{2}=22end{cases}$ $\begin{cases}3x^{2}-4y^{2}=25\\-12x^{2}+4y^{2}=22end{cases}$ $\begin{cases}6x^{2}-8y^{2}=25\\-6x^{2}-2y^{2}=11end{cases}$ $\begin{cases}6x^{2}-8y^{2}=50\\-6x^{2}-2y^{2}=11end{cases}$

which system is equivalent to $\begin{cases}3x^{2}-4y^{2}=25\\-6x^{2}-2y^{2}=11end{cases}$? $\begin{cases}3x^{2}-4y^{2}=25\\12x^{2}+4y^{2}=22end{cases}$ $\begin{cases}3x^{2}-4y^{2}=25\\-12x^{2}+4y^{2}=22end{cases}$ $\begin{cases}6x^{2}-8y^{2}=25\\-6x^{2}-2y^{2}=11end{cases}$ $\begin{cases}6x^{2}-8y^{2}=50\\-6x^{2}-2y^{2}=11end{cases}$

Answer

Explanation:

Step1: Multiply first - equation

To find an equivalent system, we can multiply one of the equations in the original system by a non - zero constant. Multiply the first equation $3x^{2}-4y^{2}=25$ by 2. $2(3x^{2}-4y^{2})=2\times25$ $6x^{2}-8y^{2}=50$ The second equation in the original system is $-6x^{2}-2y^{2}=11$, and it remains the same.

Answer:

$\begin{cases}6x^{2}-8y^{2}=50\-6x^{2}-2y^{2}=11\end{cases}$