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If x^2/9 – 4/y^2 = 12, what is the value of x?

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by GMATGuruNY » Sat Sep 01, 2018 2:40 am
BTGmoderatorDC wrote:If x^2/9 - 4/y^2 = 12, what is the value of x?

(1) x/3 + 2/y = 6

(2) x/3 - 2/y = 2
x²/9 - 4/y² = 12

(x/3)² - (2/y)² = 12

The resulting equation is the DIFFERENCE OF TWO SQUARES:
a² - b² = (a+b)(a-b).
Thus, the equation can be rephrased as follows:
(x/3 + 2/y)(x/3 - 2/y) = 12.

Statement 1:
Substituting x/3 + 2/y = 6 into (x/3 + 2/y)(x/3 - 2/y) = 12, we get:
(6)(x/3 - 2/y) = 12
x/3 - 2/y = 2.

Adding together x/3 + 2/y = 6 and x/3 - 2/y = 2, we get:
(x/3 + 2/y) + (x/3 - 2/y) = 6+2
2(x/3) = 8
x/3 = 4
x = 12.
SUFFICIENT.

Statement 2:
Substituting x/3 - 2/y = 2 into (x/3 + 2/y)(x/3 - 2/y) = 12, we get:
(x/3 + 2/y)(2) = 12
x/3 + 2/y = 6.

Adding together x/3 + 2/y = 6 and x/3 - 2/y = 2, we get:
(x/3 + 2/y) + (x/3 - 2/y) = 6+2
2(x/3) = 8
x/3 = 4
x = 12.
SUFFICIENT.

The correct answer is D.
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edited:

by deloitte247 » Sun Sep 02, 2018 9:40 am
$$\frac{x^2}{9}-\frac{4}{y^2}=12$$
factorizing this will give
$$\left(\frac{x}{3}+\frac{2}{y}\right)\cdot\left(\frac{x}{3}-\frac{2}{y}\right)=12$$
$$Let\ \left(\frac{x}{3}+\frac{2}{y}\right)=a,\ and\ \left(\frac{x}{3}-\frac{2}{y}\right)=b$$
$$So,\ a\cdot b=12$$

Statement 1;
$$\frac{x}{3}+\frac{2}{y}=6\ ---\left(eqn\ 1\right)$$
we have a*b=12 where
$$a=\frac{x}{3}+\frac{2}{y}$$,
From statement 1, we can deduce that a=6. Therefore, 6*b = 12
$$b=\frac{12}{6}=2$$

Remember that 6 = (x/3) - (2/y)
$$Remember\ that\ b=\frac{x}{3}-\frac{2}{y}\ \ \ \ \ \left(exactly\ as\ statement\ 2\right)$$
$$\frac{x}{3}-\frac{2}{y}=\ 2\ ----\ \left(eqn\ 2\right)$$
Add equation 1 and 2 together, we have
$$\frac{2x}{3}=8$$ $$x=\frac{24}{2}=12$$
STATEMENT 1 SUFFICIENT

Statement 2:
$$\frac{x}{3}-\frac{2}{y}=2---eqn\ 1$$
$$We\ have\ a\cdot b=12$$
$$where\ b=\frac{x}{3}-\frac{2}{y}$$
From statement 2, we can deduce that b=2
a*2=12
a=6

$$Remember\ that\ a=\frac{x}{3}+\frac{2}{y}\ \ \ \ \ \ \left(same\ as\ statement\ 1\right)$$
$$Therefore,\ 6=\frac{x}{3}+\frac{2}{y}$$
Going through the whole simultaneous equation using the elimination method, we will get x=12.
Hence, Statement 2 id also SUFFICIENT.

Therefore, option D is correct because STATEMENT 1 & 2 ALONE are SUFFICIENT.
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by fskilnik@GMATH » Mon Sep 03, 2018 4:43 am
BTGmoderatorDC wrote:If x^2/9 - 4/y^2 = 12, what is the value of x?

(1) x/3 + 2/y = 6

(2) x/3 - 2/y = 2
\[\left( {\frac{x}{3} - \frac{2}{y}} \right)\left( {\frac{x}{3} + \frac{2}{y}} \right) = {\left( {\frac{x}{3}} \right)^{\,2}} - {\left( {\frac{2}{y}} \right)^{\,2}} = 12\,\,\,\,\,\,\,\left( * \right)\]
\[? = x\]
\[\left( 1 \right)\, + \left( * \right)\,\,\,\,\,\left\{ \begin{gathered}
\frac{x}{3} + \frac{2}{y} = 6 \hfill \\
\frac{x}{3} - \frac{2}{y} = 2\,\,\,\,\left( { = \frac{{12}}{6}} \right) \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( + \right)} \,\,\,\,\,2\left( {\frac{x}{3}} \right) = 8\,\,\,\,\,\, \Rightarrow \,\,\,\,x\,\,\,{\text{unique}}\]
\[\left( 2 \right)\, + \left( * \right)\,\,\,\,\,\left\{ \begin{gathered}
\frac{x}{3} - \frac{2}{y} = 2 \hfill \\
\frac{x}{3} + \frac{2}{y} = 6\,\,\,\,\left( { = \frac{{12}}{2}} \right) \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( + \right)} \,\,\,\,\,2\left( {\frac{x}{3}} \right) = 8\,\,\,\,\,\, \Rightarrow \,\,\,\,x\,\,\,{\text{unique}}\]

The above follows the notations and rationale taught in the GMATH method.
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