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## Q. 1.6.2

Equations Solve each equation.

$(a) \frac{3x + 2}{x – 2} + \frac{1}{x} =\frac{ -2}{x²- 2x} (b) \frac{-4x}{x – 1 }+ \frac{ 4}{x + 1 }= \frac{-8}{x² – 1}$

## Verified Solution

(a)        $\frac{3x + 2}{x – 2} + \frac{1}{x} =\frac{ -2}{x²- 2x}$

$\frac{ 3x + 2}{x – 2} + \frac{1}{ x} =\frac{ -2 }{x(x – 2)}$       Factor the last denominator.

$x(x – 2)\left(\frac{3x + 2}{x – 2}\right) + x(x – 2)\left(\frac{1 }{x }\right) = x(x – 2) \left(\frac{-2 }{x(x – 2)}\right)$   Multiply by x(x – 2), x ≠ 0, 2.

x(3x + 2) +(x – 2)= -2       Divide out common factors.

3x² + 2x + x – 2 = -2         Distributive property

3x²+ 3x = 0                        Standard form

3x(x + 1) = 0                     Factor.

$\fbox{Set each factor equal to 0.}$→3x = 0   or   x + 1 = 0       Zero-factor property

x = 0   or   x = -1             Proposed solutions

Because of the restriction x ≠ 0, the only valid proposed solution is -1. Check -1 in the original equation. The solution set is { -1}.

(b)     $\frac{-4x}{x – 1 }+ \frac{ 4}{x + 1 }= \frac{-8}{x² – 1}$

$\frac{-4x}{x – 1 }+ \frac{ 4}{x + 1 }= \frac{-8}{(x – 1)(x+1)}$     Factor.

The restrictions on x are x ≠ ±1. Multiply by the LCD, (x + 1)(x – 1).

$(x + 1)(x – 1)\left( \frac{-4x}{x – 1 }\right) + (x + 1)(x – 1) \left( \frac{4}{ x + 1}\right) = (x + 1)(x – 1) \left( \frac{-8 }{(x + 1)(x – 1)} \right)$

-4x(x + 1) + 4(x – 1) = -8   Divide out common factors.

-4x²- 4x + 4x – 4 = -8        Distributive property

-4x²+ 4 = 0                         Standard form

x² – 1 = 0                             Divide by -4.

(x + 1)(x – 1) = 0                  Factor.

x + 1 = 0    or    x – 1 = 0      Zero-factor property

x = -1          or    x = 1            Proposed solutions

Neither proposed solution is valid, so the solution set is ∅.