Chapter 11.4, Problem 4P

To determine

Difference $\frac{c}{3c-1}-\frac{4}{c-2}$

Answer to Problem 4P

  $\frac{c^2-14c+4}{\left(3c-1\right)\left(c-2\right)}$

Explanation of Solution

Given:

  $\frac{c}{3c-1}-\frac{4}{c-2}$ ,

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

  $\frac{c}{3c-1}-\frac{4}{c-2}$

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

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

  $\begin{array}{c}\frac{c}{3c-1}-\frac{4}{c-2}=\frac{c}{\left(3c-1\right)}\cdot \frac{\left(c-2\right)}{\left(c-2\right)}-\frac{4}{\left(c-2\right)}\cdot \frac{\left(3c-1\right)}{\left(3c-1\right)}\\ =\frac{c\cdot c+c\cdot \left(-2\right)}{\left(3c-1\right)\left(c-2\right)}-\frac{4\cdot 3c+4\cdot \left(-1\right)}{\left(3c-1\right)\left(c-2\right)}\\ =\frac{c^2-2c}{\left(3c-1\right)\left(c-2\right)}-\frac{12c-4}{\left(3c-1\right)\left(c-2\right)}\\ =\frac{c^2-2c-12c+4}{\left(3c-1\right)\left(c-2\right)}\\ =\frac{c^2-14c+4}{\left(3c-1\right)\left(c-2\right)}\end{array}$

Thus, the given expression simplifies to $\frac{c^2-14c+4}{\left(3c-1\right)\left(c-2\right)}$ .