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If p = $\frac{1}{2}$ and \(\frac{1}{p - 1} = \frac{2}{p + ...

If p = $\frac{1}{2}$ and $\frac{1}{p - 1} = \frac{2}{p + x}$ , find the value of x

A.

-2 $\frac{1}{2}$

B.

-1 $\frac{1}{2}$

C.

1 $\frac{1}{2}$

D.

2 $\frac{1}{2}$

View answer & explanation

Correct answer is B

p = $\frac{1}{2}; \frac{1}{p - 1} = \frac{2}{p + x}$

$\frac{1}{\frac{1}{2} - 1} = \frac{2}{\frac{1}{2} + x}$

$\frac{1}{\frac{1 - 2}{2}} = \frac{2}{\frac{1 + 2x}{2}}$

$\frac{1}{-\frac{1}{2}} = \frac{2}{\frac{1 + 2x}{2}}$

-2 = $\frac{4}{1 + 2x} -2(1 + 2x) = 4$

1 + 2x = $\frac{4}{-2}$

1 + 2x = -2

2x = -2 - 1

2x = -3

x = - $\frac{3}{2}$

x = -1 $\frac{1}{2}$

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