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\(\frac{1}{2}\)

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\(\frac{1}{2}\)