Bunuel wrote:

Official Solution:

Given that \(5x=125-3y+z\) and \(\sqrt{5x}-5-\sqrt{z-3y}=0\), then what is the value of \(\sqrt{\frac{45(z-3y)}{x}}\)?

A. 5

B. 10

C. 15

D. 20

E. cannot be determined

Rearranging both expressions we'll get: \(5x-(z-3y)=125\) and \(\sqrt{5x}-\sqrt{z-3y}=5\). Denote \(\sqrt{5x}\) as \(a\) and \(\sqrt{z-3y}\) as \(b\).

So we have that \(a^2-b^2=125\) and \(a-b=5\). Now, \(a^2-b^2=(a-b)(a+b)=125\) and as \(a-b=5\) then \((a-b)(a+b)=5*(a+b)=125\), which means that \(a+b=25\).

Thus we get two equations with two unknowns: \(a+b=25\) and \(a-b=5\). Solving for \(a\) gives \(a=15=\sqrt{5x}\) so, \(x=45\). Solving for \(b\) gives \(b=10=\sqrt{z-3y}\).

Finally, \(\sqrt{\frac{45(z-3y)}{x}}=\sqrt{\frac{45*10^2}{45}}=10\).

Answer: B

Will sqrt(10^2) not yield +/- 10 ?