If y is not equal to 4, x is not equal to 0...

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If y is not equal to 4, x is not equal to 0, and
$$\frac{y^2-16}{3x}=\frac{y-4}{6}$$
then in terms of x, y equals:

$$\left(A\right)\ \frac{\left(x+8\right)}{2}$$

$$\left(B\right)\ \frac{\left(x-8\right)}{2}$$

$$\left(C\right)\ -\frac{3x}{2}$$

$$\left(D\right)\ \frac{\left(-3x+8\right)}{2}$$

$$\left(E\right)\ \frac{\left(3x-8\right)}{2}$$

The OA is B.

Can any expert help me with this PS question please? I don't have it clear. Thanks.

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by [email protected] » Fri Oct 27, 2017 9:55 am
Hi LUANDATO,

Many Test Takers would approach this question algebraically (which is fine); however, it's designed in such a way that you can also get to the solution by TESTing VALUES (although we are not allowed to use X=4 or Y=0 in our work).

Let's TEST Y = 5.... that makes the equation...

9/3X = 1/6
3X = 54
X = 18

Thus, we're looking for an answer that equals 5 when X=18. There's only one answer that matches...

Final Answer: B

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Rich
Contact Rich at [email protected]
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by Scott@TargetTestPrep » Tue Nov 19, 2019 6:16 pm
BTGmoderatorLU wrote:If y is not equal to 4, x is not equal to 0, and
$$\frac{y^2-16}{3x}=\frac{y-4}{6}$$
then in terms of x, y equals:

$$\left(A\right)\ \frac{\left(x+8\right)}{2}$$

$$\left(B\right)\ \frac{\left(x-8\right)}{2}$$

$$\left(C\right)\ -\frac{3x}{2}$$

$$\left(D\right)\ \frac{\left(-3x+8\right)}{2}$$

$$\left(E\right)\ \frac{\left(3x-8\right)}{2}$$

The OA is B.

Can any expert help me with this PS question please? I don't have it clear. Thanks.
Cross multiplying, we have:

6(y + 4)(y - 4) = 3x(y - 4)

6(y + 4) = 3x

6y + 24 = 3x

6y = 3x - 24

y = (3x - 24)/6

y = (x - 8)/2

Answer: B

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