Chapter 1, Problem 58RE

To determine

To simplify $\frac{2}{x}+\frac{1}{x-2}+\frac{3}{{(x-2)}^2}$ by using appropriate operations.

Answer to Problem 58RE

The answer is $\frac{2x^2-3x+4}{x{(x-2)}^2}$ .

Explanation of Solution

Given information:

  $\frac{2}{x}+\frac{1}{x-2}+\frac{3}{{(x-2)}^2}$

Formula used:

Properties of fractions:- $\frac{A}{C}+\frac{B}{C}=\frac{A+B}{C}$

Difference of squares : $A^2-B^2=(A-B)(A+B)$

Calculations:

First Term $\frac{2}{x}$

Second Term $\frac{1}{x-2}$

Now,

  $=\frac{2}{x}+\frac{1}{x-2}$

writing the fraction in LCD (least common denominator) form we have,

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

  $\begin{array}{l}=\frac{2x-4+2}{x(x-2)}\\ =\frac{2x-2}{x(x-2)}\end{array}$

New term $\frac{2x-2}{x(x-2)}$ (i)

Third term $\frac{3}{{(x-2)}^2}$  (ii)

Solving equation (i) and (ii)

  $=\frac{2x-2}{x(x-2)}+\frac{3}{{(x-2)}^2}$

Taking $\frac{1}{(x-2)}$ common

  $=\frac{1}{(x-2)}\left\{\frac{2x-2}{x}+\frac{3}{(x-2)}\right\}$

writing the fraction in LCD (least common denominator) form we have,

  $\begin{array}{l}=\frac{1}{(x-2)}\left\{\frac{(2x-2)(x-2)}{x(x-2)}+\frac{3x}{x(x-2)}\right\}\\ =\frac{1}{(x-2)}\left\{\frac{2x^2-4x-2x+4}{x(x-2)}+\frac{3x}{x(x-2)}\right\}\\ =\frac{1}{(x-2)}\left\{\frac{2x^2-6x+4+3x}{x(x-2)}\right\}\\ =\frac{1}{(x-2)}\left\{\frac{2x^2-3x+4}{x(x-2)}\right\}\end{array}$

  $=\frac{2x^2-3x+4}{x{(x-2)}^2}$

After simplifying we get $\frac{2x^2-3x+4}{x{(x-2)}^2}$ .