 # You've Unlocked a New GMAT® Question!

Sep 18, 2023

### Question

easy

If $\frac{2x}{3}-2y=\frac{5}{2}$ and $\frac{x}{5}-2y=3$, what is the value of y?

$-\frac{15}{4}$

$-\frac{45}{28}$

0

$\frac{45}{52}$

$\frac{165}{26}$

PRESENTED BY : Scott Woodbury-Stewart Founder and Expert GMAT Instructor Solution:

While this is a system of two equations, we can apply the same strategy of clearing the fractions from each equation using the LCM. The first equation has denominators of 3 and 2. The multiples of 3 are 3, 6, …. The multiples of 2 are 2, 4, 6, …. Thus, we see that the LCM of 3 and 2 is 6. We can multiply the entire first equation by 6 and simplify:

$\begin{array}{l}⇒6\left(\frac{2x}{3}-2y=\frac{5}{2}\right)\\ ⇒\frac{\stackrel{2}{\overline{)6}}\left(2x\right)}{\overline{)3}}-6\left(2y\right)=\frac{\stackrel{3}{\overline{)6}}\left(5\right)}{\overline{)2}}\\ ⇒2\left(2x\right)-6\left(2y\right)=3\left(5\right)\\ ⇒4x-12y=15\end{array}$

For the second equation, we see that the LCM of the denominators is the sole denominator, which is 5. Thus, we multiply the entire equation by 5 and simplify:

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

To determine the value of y, we can solve the system of equations by using the substitution method. In order to do so, let’s isolate x in the second equation by adding 10y to both sides:

$⇒x=10y+15$

We can now substitute 10y + 15 for x in the first equation:

$\begin{array}{l}⇒4x-12y=15\\ ⇒4\left(10y+15\right)-12y=15\\ ⇒40y+60-12y=15\\ ⇒28y=-45\\ ⇒y=-\frac{45}{28}\end{array}$

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