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.

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.