Fractions with Monomial Denominators

Monomial expressions are algebraic expressions with only one term e.g., y, 2x, b, 4m³, etc. 

A monomial denominator is a denominator with only one term. These monomial denominators may include coefficients and variables that are multiplied together. e.g.

\( \frac{3}{7x}, \frac{4y}{3}, \frac{3x}{5xy} \scriptsize \: etc \)

Example 5.1.1: 

Simplify these fractions 

a. \(\frac{2a}{5} \: + \: \frac{3a}{5} \)

b. \(\frac{6}{a} \: + \: \frac{5}{a} \)

c. \(\frac{3x}{8} \: – \: \frac{2x}{6} \)

d. \(\frac{8}{2x} \: – \: \scriptsize 4 \)

e. \(\frac{2}{y} \: – \: \frac{3}{2y} \: – \: \frac{3}{4y}\)

Solution

a. \( \underset{Same\:denominator}{\frac{2a}{5} \: + \: \frac{3a}{5}} \\ = \frac{2a \: + \: 3a}{5} \\ = \frac{5a}{5}\\ = \scriptsize a\)

To add fractions with the same denominator, simply add the numerators then copy the common denominator. Always reduce your final answer to its lowest term.

b. \( \underset{Same\:denominator}{\frac{6}{a} \: + \: \frac{5}{a}} \\ = \frac{6 \: + \: 5}{a} \\ = \frac{11}{a}\)

c. \( \underset{Different\:denominators}{\frac{3x}{8} \: – \: \frac{2x}{6}}\)

The Denominators are 8 and 6 which are different.

The first step is to find the LCM in order to make the denominators of the fractions the same. To solve the equation, the denominators of both fractions must be the same.

1st Method

So we are going to create two new fractions that are equivalent to the fractions in the question

2	8	6
2	4	3
2	2	3
3	1	3
	1	1

L.C.M of 8 and 6 = 2 × 2 × 2 × 3

L.C.M of 8 and 6 = 24

For the first fraction, we can multiply the top and bottom by 3 to get an equivalent fraction with a denominator of 24:

\(\frac{3x}{8} = \frac{3x \: \times \: 3}{8 \: \times \: 3} = \frac{9x}{24} \)

For the second fraction, we can multiply the top and bottom by 4 to get an equivalent fraction with a denominator of 24:

\( \frac{2x}{6} = \frac{2x \: \times \: 4}{6 \: \times \: 4} = \frac{8x}{24} \)

\(\frac{3x}{8} \: – \: \frac{2x}{6} = \frac{9x}{24} \: – \: \frac{8x}{24}\)

Now we can do the subtraction by subtracting the top numbers of the new equivalent fractions:

\( \frac{9x}{24} \: – \: \frac{8x}{24} \)

= \( \frac{9x \: – \: 8x}{24} \)

= \( \frac{x}{24}\)

2nd Method (The most common method)

\( \underset{Different\:denominators}{\frac{3x}{8} \: – \: \frac{2x}{6}}\)

L.C.M = 24

1st Step: We divide the L.C.M. by the first denominator and then multiply the answer with the first numerator.

\( \frac{3x}{8} \) (Denominator = 8, Numerator = 3x)

\( \scriptsize 24 \: \div \: 8 = 3\)

\( \scriptsize 3 \: \times \: 3x = 3(3x)\)

2nd Step: We divide the L.C.M. by the second denominator and then multiply the answer with the second numerator.

\( \frac{2x}{6} \) (Denominator = 6, Numerator = 2x)

\( \scriptsize 24 \: \div \: 6 = 4\)

\( \scriptsize 4 \: \times \: 2x = 4(2x)\)

3rd Step: We then subtract the two results we have gotten above the L.C.M

∴ \( \frac{3x}{8} \: – \: \frac{2x}{6} \)

= \( \frac{3(3x) \: – \: 4(2x)}{24} \)

= \( \frac{9x \: – \: 8x}{24} \)

= \( \frac{x}{24} \)

d. \(\frac{8}{2x} \: – \: \scriptsize 4 \)

= \(\frac{8}{2x} \: – \: \frac{4}{1} \)

Denoinators are 2x and 1

2	2x	1
x	x	1
	1	1

LCM of 2x and 1 = \( \scriptsize 2 \: \times \: x = 2x \)

1st Step: We divide the L.C.M. by the first denominator and then multiply the answer with the first numerator.

\( \frac{8}{2x} \) (Denominator = 2x, Numerator = 8)

\( \scriptsize 2x \: \div \: 2x = 1\)

\( \scriptsize 1 \: \times \: 8 = 1(8)\)

2nd Step: We divide the L.C.M. by the second denominator and then multiply the answer with the second numerator.

\( \frac{4}{1} \) (Denominator = 1, Numerator = 4)

\( \scriptsize 2x \: \div \: 1 = 2x\)

\( \scriptsize 2x \: \times \: 4 = 2x(4)\)

3rd Step: We then subtract the two results we have gotten above the L.C.M

∴ \(\frac{8}{2x} \: – \: \frac{4}{1} \)

= \( \frac {1(8) \: – \: 2x(4)}{2x} \)

= \( \frac{8 \: – \: 8x}{2x} \)

= \( \frac{8(1 \: – \: x)}{2x} \)

= \( \frac{\not{8}^4(1 \: – \: x)}{\not{2}x} \)

= \( \frac{4(1 \: – \: x)}{x} \)

= \( \frac{4 \: – \: 4x}{x} \)

e. \(\frac{2}{y} \: – \: \frac{3}{2y} \: – \: \frac{3}{4y}\)

Denoinators are y, 2y and 4y

y	y	2y	4y
2	1	2	4
2	1	1	2
	1	1	1

LCM of y, 2y and 2y = \( \scriptsize y \: \times \: 2 \: \times \: 2 = 4y\)

L.C.M = 4y

1st Step: We divide the L.C.M. by the first denominator and then multiply the answer with the first numerator.

\( \frac{2}{y} \) (Denominator = y, Numerator = 2)

\( \scriptsize 4y \: \div \: y = 4\)

\( \scriptsize 4 \: \times \: 2 = 4(2)\)

2nd Step: We divide the L.C.M. by the second denominator and then multiply the answer with the second numerator.

\( \frac{3}{2y} \) (Denominator = 2y, Numerator = 3)

\( \scriptsize 4y \: \div \: 2y = 2\)

\( \scriptsize 2 \: \times \: 3 = 2(3)\)

3rd Step: We divide the L.C.M. by the second denominator and then multiply the answer with the second numerator.

\( \frac{3}{4y} \) (Denominator = 4y, Numerator = 3)

\( \scriptsize 4y \: \div \: 4y = 1\)

\( \scriptsize 1 \: \times \: 3 = 1(3)\)

4th Step: We then subtract the results we have gotten above the L.C.M

∴ \(\frac{2}{y} \: – \: \frac{3}{2y} \: – \: \frac{3}{4y}\)

= \( \frac {4(2) \: – \: 2(3)\: – \: 1(3)}{4y} \)

= \( \frac{8 \: – \: 6 \: -\: 3}{4y} \)

= \( \frac{8 \: – \:9}{4y} \)

= \( \frac{-1}{4y} \)

Evaluation Questions: