Chapter 11.4, Problem 19MCQ

To determine

Add or subtract $\frac{4}{t-3}-\frac{1}{t-2}$

Answer to Problem 19MCQ

  $\frac{4t-6}{\left(t-2\right)\left(t-3\right)}$

Explanation of Solution

Given:

  $\frac{4}{t-3}-\frac{1}{t-2}$ ,

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

  $\frac{4}{t-3}-\frac{1}{t-2}$

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

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

  $\begin{array}{c}\frac{4}{t-3}-\frac{1}{t-2}=\frac{4}{t-3}\cdot \frac{\left(t-2\right)}{\left(t-2\right)}-\frac{1}{t-2}\cdot \frac{\left(t-3\right)}{\left(t-3\right)}\\ =\frac{4t-8}{\left(t-2\right)\left(t-3\right)}-\frac{1-3}{\left(t-2\right)\left(t-3\right)}\\ =\frac{4t-8-1+3}{\left(t-2\right)\left(t-3\right)}\\ =\frac{4t-6}{\left(t-2\right)\left(t-3\right)}\end{array}$

Thus, the given expression simplifies to $\frac{4t-6}{\left(t-2\right)\left(t-3\right)}$ .