which system is equivalent to \\(\\begin{cases}5x^2 + 6y^2 = 50 \\\\ 7x^2 + 2y^2 = 10\\end{cases}\\)?\n\\(\\b…

Explanation:

Step1: Analyze the second equation

We have the original second equation (7x^{2}+2y^{2} = 10). To find an equivalent system, we can multiply this equation by a constant to eliminate (y^{2}) or (x^{2}) when combined with the first equation. Let's check the options by manipulating the second equation.

Step2: Multiply the second equation by - 3

Multiply (7x^{2}+2y^{2}=10) by (- 3). Using the distributive property (a(b + c)=ab+ac), we get (-3\times(7x^{2})+(-3)\times(2y^{2})=-3\times10), which simplifies to (-21x^{2}-6y^{2}=- 30)? Wait, no, wait. Wait, let's check the first option's second equation: (-21x^{2}-6y^{2}=10) (wrong, because (-3\times10=-30)). Second option: (-21x^{2}-6y^{2}=30) (wrong, (-3\times10 = - 30)). Wait, maybe we multiply the first equation by 7 and the second by 5? Let's try that.

Step3: Multiply first equation by 7 and second by 5

First equation (5x^{2}+6y^{2}=50) multiplied by 7: (7\times(5x^{2}+6y^{2})=7\times50), so (35x^{2}+42y^{2}=350). Second equation (7x^{2}+2y^{2}=10) multiplied by 5: (5\times(7x^{2}+2y^{2})=5\times10), so (35x^{2}+10y^{2}=50). Now, multiply the second equation (after multiplying by 5) by - 1: (-35x^{2}-10y^{2}=-50). So the system becomes (\begin{cases}35x^{2}+42y^{2}=350\-35x^{2}-10y^{2}=-50\end{cases}), which is the fourth option.

Wait, let's re - check the multiplication of the second equation by - 3. The original second equation is (7x^{2}+2y^{2}=10). Multiply both sides by - 3: (-21x^{2}-6y^{2}=-30). But none of the first two options have - 30. So let's check the third and fourth options. Third option: first equation multiplied by 7: (5x^{2}\times7 + 6y^{2}\times7=50\times7\Rightarrow35x^{2}+42y^{2}=350)? Wait, 50×7 = 350, yes. Second equation multiplied by 5: (7x^{2}\times5+2y^{2}\times5 = 10\times5\Rightarrow35x^{2}+10y^{2}=50). Then multiply the second equation (after multiplying by 5) by - 1: (-35x^{2}-10y^{2}=-50). So the system is (\begin{cases}35x^{2}+42y^{2}=350\-35x^{2}-10y^{2}=-50\end{cases}), which is the fourth option.

Answer:

(\begin{cases}35x^{2}+42y^{2}=350\-35x^{2}-10y^{2}=-50\end{cases}) (the fourth option: (\boldsymbol{\begin{cases}35x^{2}+42y^{2}=350\-35x^{2}-10y^{2}=-50\end{cases}}))