<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.

# Adding and Subtracting Rational Expressions where One Denominator is the LCD

## Combine fractions including variables, one-step LCM

0%
Progress
Practice Adding and Subtracting Rational Expressions where One Denominator is the LCD
Progress
0%
Adding and Subtracting Rational Expressions where One Denominator is the LCD

The length of a garden plot is $\frac{6x^2-5}{2x^2 + 4x - 6}$ . The width of the plot is $\frac{2x-7}{x+3}$ . How much longer is the garden plot than it is wide?

### Guidance

Recall when two fractions do not have the same denominator. You have to multiply one or both fractions by a number to create equivalent fractions in order to combine them.

$\frac{1}{2} + \frac{3}{4}$

Here, 2 goes into 4 twice. So, we will multiply the first fraction by ${\color{red}\frac{2}{2}}$ to get a denominator of 4. Then, the two fractions can be added.

${\color{red}\frac{2}{2} \cdot} \frac{1}{2} + \frac{3}{4} = \frac{2}{4} + \frac{3}{4} = \frac{5}{4}$

Once the denominators are the same, the fractions can be combined. We will apply this idea to rational expressions in order to add or subtract ones without like denominators.

#### Example A

Subtract $\frac{3x-5}{2x+8} - \frac{x^2-6}{x+4}$ .

Solution: Factoring the denominator of the first fraction, we have $2(x+4)$ . The second fraction needs to be multiplied by ${\color{red}\frac{2}{2}}$ in order to make the denominators the same.

$\frac{3x-5}{2x+8} - \frac{x^2-6}{x+4} &= \frac{3x-5}{2(x+4)} - \frac{x^2-6}{x+4} \cdot {\color{red}\frac{2}{2}} \\&= \frac{3x-5}{2(x+4)} - \frac{2x^2-12}{2(x+4)}$

Now that the denominators are the same, subtract the second rational expression just like in the previous concept.

$&= \frac{3x-5-(2x^2-12)}{2(x+4)} \\&= \frac{3x-5-2x^2+12}{2(x+4)} \\&= \frac{-2x^2+3x+7}{2(x+4)}$

The numerator is not factorable, so we are done.

#### Example B

Add $\frac{2x-3}{x+5} + \frac{x^2+1}{x^2-2x-35}$

Solution: Factoring the second denominator, we have $x^2-2x-35=(x+5)(x-7)$ . So, we need to multiply the first fraction by ${\color{red}\frac{x-7}{x-7}}$ .

$\overbrace{{\color{red}\frac{(x-7)}{(x-7)}} \cdot \frac{(2x-3)}{(x+5)}}^{FOIL} + \frac{x^2+1}{(x-7)(x+5)} &= \frac{2x^2-17x+21}{(x-7)(x+5)} + \frac{x^2+1}{(x-7)(x+5)} \\&= \frac{3x^2-17x+22}{(x-7)(x+5)}$

#### Example C

Subtract $\frac{7x+2}{2x^2+18x+40} - \frac{6}{x+5}$ .

Solution: Factoring the first denominator, we have $2x^2+18x+40=2(x^2+9x+20)=2(x+4)(x+5)$ . This is the Lowest Common Denominator, or LCD. The second fraction needs the 2 and the $(x+4)$ .

$\frac{7x+2}{2x^2+18x+40} - \frac{6-x}{x+5} &= \frac{7x+2}{2(x+5)(x+4)} - \frac{6-x}{x+5}{\color{red}\cdot \frac{2(x+4)}{2(x+4)}} \\&= \frac{7x+2}{2(x+5)(x+4)} - \frac{2(6-x)(x+4)}{2(x+5)(x+4)} \\&= \frac{7x+2}{2(x+5)(x+4)} - \frac{48+4x-2x^2}{2(x+5)(x+4)} \\&= \frac{7x+2-(48+4x-2x^2)}{2(x+5)(x+4)} \\&= \frac{7x+2-48-4x+2x^2}{2(x+5)(x+4)} \\&= \frac{2x^2+3x-46}{2(x+5)(x+4)}$

Intro Problem Revisit We need to subtract the width from the length.

$\frac{6x^2-5}{2x^2 + 4x - 6} - \frac{2x-7}{x+3}$

Factoring the first denominator, we have $2x^2+4x-6=(2x-2)(x+3)$ . So, we need to multiply the second fraction by ${\color{red}\frac{2x-2}{2x-2}}$ .

$\frac{6x^2-5}{2x^2 + 4x - 6} -\overbrace{{\color{red}\frac{(2x-2)}{(2x-2)}} \cdot \frac{(2x-7)}{(x+3)}}^{FOIL} &= \frac{4x^2-18x+14}{2x^2+4x-6} \\\frac{6x^2-5}{2x^2 + 4x - 6} - \frac{4x^2-18x+14}{2x^2+4x-6}\\\frac{6x^2-5-(4x^2-18x+14)}{2x^2+4x-6}\\\frac{6x^2-5-4x^2+18x-14}{2x^2+4x-6}\\\frac{2x^2+18x-19}{2x^2+4x-6}$

Therefore, the garden plot is $\frac{2x^2+18x-19}{2x^2+4x-6}$ longer than it is wide.

### Guided Practice

Perform the indicated operation.

1. $\frac{2}{x+1} - \frac{x}{3x+3}$

2. $\frac{x-10}{x^2+4x-24} + \frac{x+3}{x+6}$

3. $\frac{3x^2-5}{3x^2-12} + \frac{x+8}{3x+6}$

#### Answers

1. The LCD is $3x+3$ or $3(x+1)$ . Multiply the first fraction by $\frac{3}{3}$ .

$\frac{2}{x+1} - \frac{x}{3x+3} &= \frac{3}{3} \cdot \frac{2}{x+1} - \frac{x}{3(x+1)} \\&= \frac{6}{3(x+1)} - \frac{x}{3(x+1)} \\&= \frac{6-x}{3(x+1)}$

2. Here, the LCD $x^2+4x-24$ or $(x+6)(x-4)$ . Multiply the second fraction by $\frac{x-4}{x-4}$ .

$\frac{x-10}{x^2+4x-24} + \frac{x+3}{x+6} &= \frac{x-10}{(x+6)(x-4)} + \frac{x+3}{x+6} \cdot \frac{x-4}{x-4} \\&= \frac{x-10}{(x+6)(x-4)} + \frac{x^2-x-12}{(x+6)(x-4)} \\&= \frac{x-10+x^2-x-12}{(x+6)(x-4)} \\&= \frac{x^2-22}{(x+6)(x-4)}$

3. The LCD is $3x^2-12=3(x-2)(x+2)$ . The second fraction’s denominator factors to be $3x+6=3(x+2)$ , so it needs to be multiplied by $\frac{x-2}{x-2}$ .

$\frac{3x^2-5}{3x^2-12} + \frac{x+8}{3x+6} &= \frac{3x^2-5}{3(x-2)(x+2)} + \frac{x+8}{3(x+2)} \cdot \frac{x-2}{x-2} \\&= \frac{3x^2-5}{3(x-2)(x+2)} + \frac{x^2+6x-16}{3(x-2)(x+2)} \\&= \frac{3x^2-5+x^2+6x-16}{3(x-2)(x+2)} \\&= \frac{4x^2+6x-21}{3(x-2)(x+2)}$

### Practice

Find the LCD.

1. $x, \ 6x$
2. $x, \ x+1$
3. $x+2, \ x-4$
4. $x, \ x-1, \ x^2 - 1$

Perform the indicated operations.

1. $\frac{3}{x} - \frac{5}{4x}$
2. $\frac{x+2}{x+3} + \frac{x-1}{x^2+3x}$
3. $\frac{x}{x-7} - \frac{2x+7}{3x-21}$
4. $\frac{x^2+3x-10}{x^2-4} - \frac{x}{x+2}$
5. $\frac{5x+14}{2x^2-7x-15} - \frac{3}{x-5}$
6. $\frac{x-3}{3x^2+x-10} + \frac{3}{x+2}$
7. $\frac{x+1}{6x+2} + \frac{x^2-7x}{12x^2-14x-6}$
8. $\frac{-3x^2-10x+15}{10x^2-x-3} + \frac{x+4}{2x+1}$
9. $\frac{8}{2x-5} - \frac{x+5}{2x^2+x-15}$
10. $\frac{2}{x+2} + \frac{3x+16}{x^2-x-6} - \frac{2}{x-3}$
11. $\frac{6x^2+4x+8}{x^3+3x^2-x-3} + \frac{x-4}{x^2-1} - \frac{3x}{x^2+2x-3}$

### Explore More

Sign in to explore more, including practice questions and solutions for Adding and Subtracting Rational Expressions where One Denominator is the LCD.

Please wait...
Please wait...