Question

which system is equivalent to \\(\

$$\begin{cases}5x^2 + 6y^2 = 50 \\\\ 7x^2 + 2y^2 = 10\\end{cases}$$

\\)?
\\(\

$$\begin{cases}\\ \\ 5x^2 + 6y^2 = 50 \\\\ -21x^2 - 6y^2 = 10\\end{cases}$$

\\)
\\(\

$$\begin{cases}\\ \\ 5x^2 + 6y^2 = 50 \\\\ -21x^2 - 6y^2 = 30\\end{cases}$$

\\)
\\(\

$$\begin{cases}\\ 35x^2 + 42y^2 = 250 \\\\ -35x^2 - 10y^2 = -50\\end{cases}$$

\\)
\\(\

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

\\)

Explanation:

Step1: Analyze the second equation

We have the second equation \(7x^{2}+2y^{2}=10\). To make the coefficients of \(y^{2}\) match or to manipulate the system, we can multiply this equation by a constant. Let's check the options. For the fourth option, let's multiply the first original equation \(5x^{2}+6y^{2}=50\) by 7: \(7\times(5x^{2}+6y^{2})=7\times50\), which gives \(35x^{2}+42y^{2}=350\)? Wait, no, 507 is 350? Wait, no, 57 = 35, 67=42, 507 = 350? Wait, no, the first original equation is \(5x^{2}+6y^{2}=50\). If we multiply the second equation \(7x^{2}+2y^{2}=10\) by - 5: \(-5\times(7x^{2}+2y^{2})=-5\times10\), which is \(-35x^{2}-10y^{2}=-50\). Now multiply the first equation by 7: \(7\times(5x^{2}+6y^{2}) = 7\times50\), so \(35x^{2}+42y^{2}=350\)? Wait, no, 507 is 350? Wait, the fourth option has \(35x^{2}+42y^{2}=350\) and \(-35x^{2}-10y^{2}=-50\). Let's check the third option: multiply first equation by 7: 57=35, 67=42, 507=350? No, 507 is 350? Wait, no, 507 is 350? Wait, the first original equation is \(5x^{2}+6y^{2}=50\). If we multiply by 7: \(35x^{2}+42y^{2}=350\). The second equation \(7x^{2}+2y^{2}=10\) multiplied by - 5: \(-35x^{2}-10y^{2}=-50\). So the system becomes \(

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

\)? Wait, no, the fourth option is \(

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

\)? Wait, no, the fourth option in the problem is \(

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

\)? Wait, let's re - check.

Wait, let's take the original system \(

$$\begin{cases}5x^{2}+6y^{2}=50\\7x^{2}+2y^{2}=10\end{cases}$$

\)

Multiply the first equation by 7: \(7\times(5x^{2}+6y^{2})=7\times50\)

\(35x^{2}+42y^{2}=350\)

Multiply the second equation by - 5: \(-5\times(7x^{2}+2y^{2})=-5\times10\)

\(-35x^{2}-10y^{2}=-50\)

So the equivalent system is \(

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

\), which is the fourth option. Wait, but let's check the third option: third option has \(35x^{2}+42y^{2}=250\) (which is wrong, 50*7 = 350) and \(-35x^{2}-10y^{2}=-50\) (correct for the second equation multiplied by - 5). Wait, no, my mistake. Wait, let's check the fourth option:

Fourth option: \(

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

\)

First equation: 5x²+6y² = 50, multiply by 7: 35x² + 42y²=350 (correct, 50*7=350)

Second equation:7x² + 2y²=10, multiply by - 5: - 35x²-10y²=-50 (correct, 10*(-5)=-50)

So this system is equivalent. Let's check other options:

First option: second equation is - 21x² - 6y²=10. If we multiply original second equation by - 3: - 3(7x²+2y²)=-310→-21x² - 6y²=-30, not 10. So first option wrong.

Second option: second equation - 21x² - 6y²=30. Multiply original second equation by - 3: - 21x² - 6y²=-30, not 30. So wrong.

Third option: first equation 35x² + 42y²=250. Multiply original first equation by 7: 35x² + 42y²=350, not 250. So wrong.

Fourth option: first equation 35x² + 42y²=350 (507=350), second equation - 35x² - 10y²=-50 (10(-5)=-50). So correct.

Step2: Verify the operations

• Multiply the first equation \(5x^{2}+6y^{2}=50\) by 7:
• \(7\times5x^{2}+7\times6y^{2}=7\times50\)
• \(35x^{2}+42y^{2}=350\)
• Multiply the second equation \(7x^{2}+2y^{2}=10\) by - 5:
• \(-5\times7x^{2}+(-5)\times2y^{2}=-5\times10\)
• \(-35x^{2}-10y^{2}=-50\)

So the equivalent system is \(

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

\)

Answer:

\(\boldsymbol{

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

}\) (the fourth option in the list)