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If p = \(\frac{1}{2}\) and \(\frac{1}{p - 1} = \frac{2}{p + x}\), find the value...


Question

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

Options

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

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

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

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

The correct answer is B.

Explanation:

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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Dicussion (1)

  • 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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