Chapter 6.1, Problem 44WE

To determine

To calculate: The simplified form of the expression $\frac{8a^2+6ab-5b^2}{16a^2-25b^2}$ and state restrictions on variable.

Answer to Problem 44WE

The simplified value of the expression $\frac{8a^2+6ab-5b^2}{16a^2-25b^2}$ is $\frac{2a-b}{4a-5b}$ and restriction is $a\ne \pm \frac{5b}{4}$.

Explanation of Solution

Given information:

The expression $\frac{8a^2+6ab-5b^2}{16a^2-25b^2}$.

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 $\frac{8a^2+6ab-5b^2}{16a^2-25b^2}$.

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 numerator by middle term splitting method and denominator by using the formula $X^2-Y^2=\left(X+Y\right)\left(X-Y\right)$ and then strike out common terms from the expression and simplify it as,

  $\begin{array}{c}\frac{8a^2+6ab-5b^2}{16a^2-25b^2}=\frac{8a^2-4ab+10ab-5b^2}{{\left(4a\right)}^2-{\left(5b\right)}^2}\\ =\frac{4a\left(2a-b\right)+5b\left(2a-b\right)}{\left(4a+5b\right)\left(4a-5b\right)}\\ =\frac{\left(4a+5b\right)\left(2a-b\right)}{\left(4a+5b\right)\left(4a-5b\right)}\\ =\frac{2a-b}{4a-5b}\end{array}$

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

So, denominator will be zero when,

  $\begin{array}{c}16a^2-25b^2=0\\ 16a^2=25b^2\\ a^2=\frac{25b^2}{16}\\ a=\pm \frac{5b}{4}\end{array}$

For denominator not to be zero, $a\ne \pm \frac{5b}{4}$ .

Thus, the simplified value of the expression $\frac{8a^2+6ab-5b^2}{16a^2-25b^2}$ is $\frac{2a-b}{4a-5b}$ and restriction is $a\ne \pm \frac{5b}{4}$.