You've Unlocked a New GRE Question!

Text Solution:

Let’s begin with the first equation. Since here we have only one denominator, with a value of 3, we can eliminate the fraction by multiplying the entire equation by 3.

$\begin{array}{l}\Rightarrow 3\left(x-5=\frac{2}{3}\left(y-5\right)\right)\\ \Rightarrow 3x-3(5)=\frac{\cancel{3}(2)}{\cancel{3}}\left(y-5\right)\\ \Rightarrow 3x-15=2\left(y-5\right)\\ \Rightarrow 3x-15=2y-10\\ \Rightarrow 3x=2y+5\end{array}$

Equation one is simplified; let’s move to equation two. Since here we also have only one denominator, with a value of 6, we can multiply the entire equation by 6.

$\begin{array}{l}\Rightarrow 6\left(x+10=\frac{5}{6}\left(y+10\right)\right)\\ \Rightarrow 6x+6(10)=\frac{\cancel{6}(5)}{\cancel{6}}\left(y+10\right)\\ \Rightarrow 6x+60=5\left(y+10\right)\\ \Rightarrow 6x+60=5y+50\\ \Rightarrow 6x=5y-10\end{array}$

Now we have two equations without fractions and can use the combination method to solve for the variables. Before we add the equations, they must be manipulated so that one variable will cancel out in the process. To do this, multiply equation one by $–$2:

$-\text{2}\left(\text{3x}=\text{2y}+\text{5}\right)\to -\text{6x}=-\text{4y}-\text{1}0$

Adding this new form of equation one to our simplified version of equation two yields the following:

$\begin{array}{l}-6x=-4y-10\\ \underset{¯}{+\left(6x=5y-10\right)}\\ 0=y-20\\ y=20\end{array}$

Finally, substitute 20 for y in one of the equations in order to solve for the value of x. Let’s use the simplified version of equation two.

$\begin{array}{l}\Rightarrow 6x=5y-10\\ \Rightarrow 6x=5(20)-10\\ \Rightarrow 6x=100-10\\ \Rightarrow 6x=90\\ \Rightarrow x=15\end{array}$

Therefore, xy = 15(20) = 300.