which system is equivalent to $\begin{cases}5x^{2}+6y^{2}=50\\7x^{2}+2y^{2}=10end{cases}$…

which system is equivalent to $\begin{cases}5x^{2}+6y^{2}=50\\7x^{2}+2y^{2}=10end{cases}$? $\begin{cases}5x^{2}+6y^{2}=50\\-21x^{2}-6y^{2}=10end{cases}$ $\begin{cases}5x^{2}+6y^{2}=50\\-21x^{2}-6y^{2}=30end{cases}$ $\begin{cases}35x^{2}+42y^{2}=250\\-35x^{2}-10y^{2}=-50end{cases}$ $\begin{cases}35x^{2}+42y^{2}=350\\-35x^{2}-10y^{2}=-50end{cases}$

which system is equivalent to $\begin{cases}5x^{2}+6y^{2}=50\\7x^{2}+2y^{2}=10end{cases}$? $\begin{cases}5x^{2}+6y^{2}=50\\-21x^{2}-6y^{2}=10end{cases}$ $\begin{cases}5x^{2}+6y^{2}=50\\-21x^{2}-6y^{2}=30end{cases}$ $\begin{cases}35x^{2}+42y^{2}=250\\-35x^{2}-10y^{2}=-50end{cases}$ $\begin{cases}35x^{2}+42y^{2}=350\\-35x^{2}-10y^{2}=-50end{cases}$

Answer

Explanation:

Step1: Multiply the second - equation by a constant

We have the system $\begin{cases}5x^{2}+6y^{2}=50\7x^{2}+2y^{2}=10\end{cases}$. Multiply the second equation $7x^{2}+2y^{2}=10$ by $- 5$. [ \begin{align*} -5\times(7x^{2}+2y^{2})&=-5\times10\ -35x^{2}-10y^{2}&=-50 \end{align*} ]

Step2: Multiply the first - equation by a constant

Multiply the first equation $5x^{2}+6y^{2}=50$ by $7$. [ \begin{align*} 7\times(5x^{2}+6y^{2})&=7\times50\ 35x^{2}+42y^{2}&=350 \end{align*} ]

Answer:

$\begin{cases}35x^{2}+42y^{2}=350\-35x^{2}-10y^{2}=-50\end{cases}$