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by GMATGuruNY » Mon Oct 23, 2017 4:22 pm
LUANDATO wrote:If x ≠0 and

$$\frac{\left(x+1\right)^2}{-1}-x\ =\ \frac{y}{x}$$

Then y = ?

A. 2x
B. 2x-1
C. 2x+1
D. 2x^2-1
E. 2x^2+1
Let x=-1.
Plugging x=-1 into the given equation, we get:

$$\frac{\left(-1+1\right)^2}{-1}-(-1)\ =\ \frac{y}{-1}$$

0 + 1 = -y
1 = -y
y = -1.

Since the question stem asks for the value of y, the target value is -1.
Now plug x=-1 into the answers to see which yields the target value of -1.
Only C works:
2x+1 = (2)(-1) + 1 = -2+1 = -1.

The correct answer is C.
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by Scott@TargetTestPrep » Sun Nov 17, 2019 7:06 pm
BTGmoderatorLU wrote:If x ≠0 and

$$\frac{\left(x+1\right)^2}{x}-x\ =\ \frac{y}{x}$$

Then y = ?

A. 2x
B. 2x-1
C. 2x+1
D. 2x^2-1
E. 2x^2+1

The OA is C.

Can any expert help me with this PS question please? I can't understand it. Thanks.
Multiplying the entire equation by x, we have:

(x + 1)^2 - x^2 = y

x^2 + 2x + 1 - x^2 = y

2x + 1 = y

Answer: C

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