Chapter 11.4, Problem 42PPE

To determine

Add or subtract $\frac{t}{2t-3}+11$

Answer to Problem 42PPE

  $\frac{23t-33}{\left(2t-3\right)}$

Explanation of Solution

Given:

  $\frac{t}{2t-3}+11$ ,

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

  $\frac{t}{\left(2t-3\right)}-\frac{11}{1}$

Here observe that the denominators are not same, so first we will make both denominators equal to LCD. The LCD of $1\text{and}\left(2t-3\right)$ is $\left(2t-3\right)$ .

So in order to subtract the fractions, we make both the denominators equal to LCD by multiplying the first fraction by $1$ up and down and second fraction by $\left(2t-3\right)$ . And then when denominators are equal, we directly subtract the numerators and put over the common denominator. So, we get

  $\begin{array}{c}\frac{t}{\left(2t-3\right)}-\frac{11}{1}=\frac{t}{\left(2t-3\right)}\cdot \frac{1}{1}+\frac{11}{1}\cdot \frac{\left(2t-3\right)}{\left(2t-3\right)}\\ =\frac{t}{\left(2t-3\right)}+\frac{11\cdot 2t+11\cdot \left(-3\right)}{\left(2t-3\right)}\\ =\frac{t+22t-33}{\left(2t-3\right)}\\ =\frac{23t-33}{\left(2t-3\right)}\end{array}$

Thus, the given expression simplifies to $\frac{23t-33}{\left(2t-3\right)}$ .