Chapter 6.6, Problem 23WE

To determine

Simplify: the given expression as fraction.

Answer to Problem 23WE

The simplified form of $\frac{b-1}{b}-\frac{3}{b-2}+1$ is $\frac{2-4b}{b^2-2b}$.

Explanation of Solution

Given information:

The expression $\frac{b-1}{b}-\frac{3}{b-2}+1$.

Formula used:

Least common denominator: It is the smallest multiple of denominator of set of fractions.

  $\frac{b-1}{b}-\frac{3}{b-2}+1$

Recall that Least common denominator is the smallest multiple of denominator of set of fractions.

  $=\frac{b-1}{b}-\frac{3}{b-2}+\frac{1}{1}$

Least common denominator for $\frac{b-1}{b},\frac{3}{b-2},\frac{1}{1}$ is $b\left(b-2\right)$ . So,

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

The simplified form of $\frac{b-1}{b}-\frac{3}{b-2}+1$ is $\frac{2-4b}{b^2-2b}$.