## Add and Subtract Fractions with Different Denominators

Read this text for more examples and guidance. It gives a good review of how to find common denominators. Pay attention to the "How To" section at the beginning for an overview of the process. Do Examples 4.67 – 4.72 and check your answers.

#### EXAMPLE 4.67

##### Solution
 $\dfrac{1}{2}+\dfrac{1}{3}$ Find the LCD of 2, 3. \begin{aligned}2 &=2 \\3 &=3 \\\hline \mathrm{LCD} &=2 \cdot 3 \\\mathrm{LCD} &=6\end{aligned} $\dfrac{1 \cdot 3}{2 \cdot 3}+\dfrac{1 \cdot 2}{3 \cdot 2}$ Change into equivalent fractions with the LCD 6. $\dfrac{1 \cdot 3}{2 \cdot 3}+\dfrac{1 \cdot 2}{3 \cdot 2}$ Simplify the numerators and denominators. $\dfrac{3}{6}+\dfrac{2}{6}$ Add. $\dfrac{5}{6}$

Remember, always check to see if the answer can be simplified. Since $5$ and $6$ have no common factors, the fraction $\frac{5}{6}$ cannot be reduced.

#### EXAMPLE 4.68

##### Solution

 $\dfrac{1}{2}-\left(-\dfrac{1}{4}\right)$ Find the LCD of 2 and 4. \begin{aligned}2 &=2 \\4 &=2 \cdot 2 \\\hline \mathrm{LCD} &=2 \cdot 2 \\\mathrm{LCD} &=4\end{aligned} Rewrite as equivalent fractions using the LCD 4. $\dfrac{1 \cdot 2}{2 \cdot 2}-\left(-\dfrac{1}{4}\right)$ Simplify the first fraction. $\dfrac{2}{4}-\left(-\dfrac{1}{4}\right)$ Subtract. $\dfrac{2-(-1)}{4}$ Simplify. $\dfrac{3}{4}$

One of the fractions already had the least common denominator, so we only had to convert the other fraction.

#### EXAMPLE 4.69

##### Solution

 $\dfrac{7}{12}+\dfrac{5}{18}$ Find the LCD of 12 and 18. \begin{aligned}12 &=2 \cdot 2 \cdot 3 \\18 &=2 \cdot 3 \cdot 3 \\\hline \mathrm{LCD} &=2 \cdot 2 \cdot 3 \cdot 3 \\\mathrm{LCD} &=36\end{aligned} Rewrite as equivalent fractions with the LCD. $\dfrac{7 \cdot 3}{12 \cdot 3}+\dfrac{5 \cdot 2}{18 \cdot 2}$ Simplify the numerators and denominators. $\dfrac{21}{36}+\dfrac{10}{36}$ Add. $\dfrac{31}{36}$

Because $31$ is a prime number, it has no factors in common with $36$. The answer is simplified.

#### EXAMPLE 4.70

##### Solution
 $\dfrac{7}{15}-\dfrac{19}{24}$ Find the LCD. \begin{aligned}&15=\\&24=2 \cdot 2 \cdot 2 \cdot 3\\\hline &\begin{aligned}&\mathrm{LCD}=2 \cdot 2 \\&\mathrm{LCD}=120\end{aligned}\end{aligned} $15$ is 'missing' three factors of $2$$24$ is 'missing' a factor of $5$ Rewrite as equivalent fractions with the LCD. $\dfrac{7 \cdot 8}{15 \cdot 8}-\dfrac{19 \cdot 5}{24 \cdot 5}$ Simplify each numerator and denominator. $\dfrac{56}{120}-\dfrac{95}{120}$ Subtract. $-\dfrac{39}{120}$ Rewrite showing the common factor of 3. $-\dfrac{13 \cdot 3}{40 \cdot 3}$ Remove the common factor to simplify. $-\dfrac{13}{40}$

#### EXAMPLE 4.71

##### Solution
 $-\dfrac{11}{30}+\dfrac{23}{42}$ Find the LCD. \begin{aligned} 30 &=2 \cdot 3 \cdot 5 \\ 42 &=2 \cdot 3 \cdot 7 \\ \hline \mathrm{LCD} &=2 \cdot 3 \cdot 5 \cdot 7 \\ \mathrm{LCD} &=210 \end{aligned} Rewrite as equivalent fractions with the LCD. $-\dfrac{11 \cdot 7}{30 \cdot 7}+\dfrac{23 \cdot 5}{42 \cdot 5}$ Simplify each numerator and denominator. $-\dfrac{77}{210}+\dfrac{115}{210}$ Add. $\dfrac{38}{210}$ Rewrite showing the common factor of 2. $\dfrac{19 \cdot 2}{105 \cdot 2}$ Remove the common factor to simplify. $\dfrac{19}{105}$

#### EXAMPLE 4.72

##### Solution

The fractions have different denominators.

 $\dfrac{3}{5}+\dfrac{x}{8}$ Find the LCD. \begin{aligned}5 &=\\8 &=2 \cdot 2 \cdot 2 \\\hline \mathrm{LCD} &=2 \cdot 2 \cdot 2 \cdot 5 \\\mathrm{LCD} &=40\end{aligned} Rewrite as equivalent fractions with the LCD. $\dfrac{3 \cdot 8}{5 \cdot 8}+\dfrac{x \cdot 5}{8 \cdot 5}$ Simplify the numerators and denominators. $\dfrac{24}{40}+\dfrac{5 x}{40}$ Add. $\dfrac{24+5 x}{40}$

We cannot add $24$ and $5x$ since they are not like terms, so we cannot simplify the expression any further.