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

# 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$$\frac{1}{2}$$

## 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$$\frac{1}{2}$$