Chapter 6.1, Problem 49WE

To determine

To calculate: The simplified form of the expression $\frac{4a^2-b^2-2a+b}{{\left(2a+b\right)}^2-1}$ and state restrictions on variable.

Answer to Problem 49WE

The simplified value of the expression $\frac{4a^2-b^2-2a+b}{{\left(2a+b\right)}^2-1}$ is $\frac{2a-b}{2a+b+1}$ and restriction is $b\ne \frac{\pm 1-b}{2}$.

Explanation of Solution

Given information:

The expression $\frac{4a^2-b^2-2a+b}{{\left(2a+b\right)}^2-1}$.

Formula used:

The simplest form of an algebraic fraction is when numerator and denominator have no common factor other than 1 and $-1$ . Simplification is done by factorizing numerator and denominator.

Calculation:

Consider the expression $\frac{4a^2-b^2-2a+b}{{\left(2a+b\right)}^2-1}$.

Recall that the simplest form of an algebraic fraction is when numerator and denominator have no common factor other than 1 and $-1$ . Simplification is done by factorizing numerator and denominator.

So, factorize bothnumerator and denominator by using the formula $X^2-Y^2=\left(X+Y\right)\left(X-Y\right)$ and then take out and strike common terms and simplify it as,

  $\begin{array}{c}\frac{4a^2-b^2-2a+b}{{\left(2a+b\right)}^2-1}=\frac{{\left(2a\right)}^2-{\left(b\right)}^2-\left(2a-b\right)}{\left(2a+b+1\right)\left(2a+b-1\right)}\\ =\frac{\left(2a+b\right)\left(2a-b\right)-\left(2a-b\right)}{\left(2a+b+1\right)\left(2a+b-1\right)}\\ =\frac{\left(2a-b\right)\left(2a+b-1\right)}{\left(2a+b+1\right)\left(2a+b-1\right)}\\ =\frac{\left(2a-b\right)}{\left(2a+b+1\right)}\end{array}$

To restrict the variable, exclude the value that makes the denominator zero.

So, denominator will be zero when,

  $\begin{array}{c}{\left(2a+b\right)}^2=1\\ 2a+b=\pm 1\\ 2b=\pm 1-b\\ b=\frac{\pm 1-b}{2}\end{array}$

For denominator not to be zero, $b\ne \frac{\pm 1-b}{2}$ .

Thus, the simplified value of the expression $\frac{4a^2-b^2-2a+b}{{\left(2a+b\right)}^2-1}$ is $\frac{2a-b}{2a+b+1}$ and restriction is $b\ne \frac{\pm 1-b}{2}$.