Chapter 11.4, Problem 3LC

To determine

Add or subtract $\frac{4}{6b^2}+\frac{5}{8b^3}$

Answer to Problem 3LC

  $\frac{16b+15}{24b^3}$

Explanation of Solution

Given:

  $\frac{4}{6b^2}+\frac{5}{8b^3}$ ,

Concept used:

The LCD (Least Common Denominator) is the smallest number that is divisible by all the denominators.

Calculation:

The LCD (Least Common Denominator) is the smallest number that is divisible by all the denominators.

The given expression can be written as

  $\frac{4}{6b^2}+\frac{5}{8b^3}$

Here observe that the denominators are not same, so first we will make both denominators equal to LCD. The LCD of $6b^2\text{and}8b^3$ is $24b^3$ .

So in order to subtract the fractions, we make both the denominators equal to LCD by multiplying the first fraction by $4b$ up and down and second fraction by $3$ . And then when denominators are equal, we directly subtract the numerators and put over the common denominator. So, we get

  $\begin{array}{c}\frac{4}{6b^2}+\frac{5}{8b^3}=\frac{4}{6b^2}\cdot \frac{4b}{4b}+\frac{5}{8b^3}\cdot \frac{3}{3}\\ =\frac{16b}{24b^3}+\frac{15}{24b^3}\\ =\frac{16b+15}{24b^3}\end{array}$

Thus, the given expression simplifies to $\frac{16b+15}{24b^3}$ .