Subtracting Fractions with Unlike Denominators

It is a bit harder to subtract fractions with unlike denominators than with like denominators.

We saw that it is very simple to subtract fractions with like denominators.

But how to transform fractions that have different denominators into fractions that have same denominators? In fact, very easy. We use equivalence of fractions for this.

Two Ways to Subtract Fractions with Unlike Denominators:

Suppose we have fractions $\frac{{a}}{{b}}$ and $\frac{{c}}{{d}}$ . Multiply both numerator and denominator of the first fraction by denominator ${d}$ of the second fraction: $\frac{{a}}{{b}}=\frac{{{a}{d}}}{{{b}{d}}}$ . Multiply both numerator and denominator of the second fraction by the denominator ${b}$ of the first fraction: $\frac{{c}}{{d}}=\frac{{{b}{c}}}{{{b}{d}}}$ . Now fractions have same common denominator ${b}{d}$ . Subtract them and perform reducing if possible. ${\color{red}{{\frac{{a}}{{b}}-\frac{{c}}{{d}}=\frac{{{a}{d}-{b}{c}}}{{{b}{d}}}}}}$ .
Suppose we have fractions $\frac{{a}}{{b}}$ and $\frac{{c}}{{d}}$ . Find least common multiple of denominators ${b}$ and ${d}$ : ${L}{C}{M}{\left({b},{d}\right)}$ . This will be common denominator. Find equivalent fractions, perform subtraction and reduce if possible.

Difference between first and second way is that second way usually have simpler calculations and we need to reduce result more seldom than when using first way.

Example 1. Find $\frac{{3}}{{4}}-\frac{{5}}{{7}}$ .

Find equivalent fractions.

$\frac{{3}}{{4}}=\frac{{{3}\cdot{\color{green}{{{7}}}}}}{{{4}\cdot{\color{green}{{{7}}}}}}=\frac{{21}}{{28}}$ .

$\frac{{5}}{{7}}=\frac{{{5}\cdot{\color{red}{{{4}}}}}}{{{7}\cdot{\color{red}{{{4}}}}}}=\frac{{20}}{{28}}$ .

Now, subtract fractions $\frac{{21}}{{28}}-\frac{{20}}{{28}}=\frac{{1}}{{28}}$ .

Reduce if possible: $\frac{{1}}{{28}}$ is irreducible.

Answer: $\frac{{1}}{{28}}$ .

Next example.

Example 2. Find $\frac{{5}}{{12}}-\frac{{7}}{{18}}$ .

Find equivalent fractions.

$\frac{{5}}{{12}}=\frac{{{5}\cdot{\color{green}{{{18}}}}}}{{{12}\cdot{\color{green}{{{18}}}}}}=\frac{{90}}{{216}}$ .

$\frac{{7}}{{18}}=\frac{{{7}\cdot{\color{red}{{{12}}}}}}{{{18}\cdot{\color{red}{{{12}}}}}}=\frac{{84}}{{216}}$ .

Now, subtract fractions $\frac{{90}}{{216}}-\frac{{84}}{{216}}=\frac{{6}}{{216}}$ .

Reduce if possible: $\frac{{6}}{{216}}=\frac{{1}}{{36}}$ .

Answer: $\frac{{1}}{{36}}$ .

Now, let's try to do above example using second way.

Example 3. Find $\frac{{5}}{{12}}-\frac{{7}}{{18}}$ .

Find least common multiple of denominators: ${L}{C}{M}{\left({12},{18}\right)}={36}$ .

Find equivalent fractions.

We need to multiply numerator and denominator of the first fraction by $\frac{{36}}{{12}}={3}$ to get 36 in denominator: $\frac{{5}}{{12}}=\frac{{{5}\cdot{\color{green}{{{3}}}}}}{{{12}\cdot{\color{green}{{{3}}}}}}=\frac{{15}}{{36}}$ .

We need to multiply numerator and denominator of the second fraction by $\frac{{36}}{{18}}={2}$ to get 36 in denominator: $\frac{{7}}{{18}}=\frac{{{7}\cdot{\color{red}{{{2}}}}}}{{{18}\cdot{\color{red}{{{2}}}}}}=\frac{{14}}{{36}}$ .

Now, subtract fractions $\frac{{15}}{{36}}-\frac{{14}}{{36}}=\frac{{1}}{{36}}$ .

Reduce if possible: $\frac{{1}}{{36}}$ is irreducible.

Answer: $\frac{{1}}{{36}}$ .

Note, that using second way we obtained answer without reducing fraction and calculations were simpler.

Example 4. Find $-\frac{{19}}{{8}}-\frac{{13}}{{16}}$ .

Find least common multiple of denominators: ${L}{C}{M}{\left({8},{16}\right)}={16}$ .

Find equivalent fractions.

We need to multiply numerator and denominator of the first fraction by $\frac{{16}}{{8}}={2}$ to get 16 in denominator: $-\frac{{19}}{{8}}=-\frac{{{19}\cdot{\color{green}{{{2}}}}}}{{{8}\cdot{\color{green}{{{2}}}}}}=-\frac{{38}}{{16}}$ .

Second fraction already has required denominator, so we don't need to find equivalent fraction.

Now, subtract fractions $-\frac{{38}}{{16}}-\frac{{13}}{{16}}=\frac{{-{38}-{13}}}{{16}}=-\frac{{51}}{{16}}$ .

Reduce if possible: $-\frac{{51}}{{16}}$ is irreducible.

Answer: $-\frac{{51}}{{16}}$ .

Next example.

Example 5. Find $\frac{{13}}{{6}}-\frac{{1}}{{2}}$ .

Find least common multiple of denominators: ${L}{C}{M}{\left({6},{2}\right)}={6}$ .

Find equivalent fractions:

First fraction already has required denominator so we don't need to find equivalent fraction.

We need to multiply second fraction by $\frac{{6}}{{2}}={3}$ to get 6 in denominator: $\frac{{1}}{{2}}=\frac{{{1}\cdot{\color{red}{{{3}}}}}}{{{2}\cdot{\color{red}{{{3}}}}}}=\frac{{3}}{{6}}$ .

Now, add fractions $\frac{{13}}{{6}}-\frac{{3}}{{6}}=\frac{{10}}{{6}}$ .

Reduce if possible: $\frac{{10}}{{6}}=\frac{{5}}{{3}}$ .

Answer: $\frac{{5}}{{3}}$ .

Now, it is time to do a couple of exercises.

Exercise 1. Find $\frac{{5}}{{3}}-\frac{{1}}{{4}}$ .

Answer: $\frac{{17}}{{12}}$ .

Next exercise.

Exercise 2. Find $\frac{{7}}{{24}}-\frac{{17}}{{18}}$ . using both ways and tell what way was easier.

Answer: $-\frac{{47}}{{72}}$ .

Next exercise.

Exercise 3. Find $-\frac{{18}}{{7}}-\frac{{5}}{{21}}$ .

Answer: $-\frac{{59}}{{21}}$ .

Next exercise.

Exercise 4. Find $\frac{{2}}{{10}}-{\left(-\frac{{9}}{{3}}\right)}$ .

Answer: $\frac{{16}}{{5}}$ .

Next exercise.

Exercise 5. Find ${1}-\frac{{2}}{{5}}$ .

Answer: $\frac{{3}}{{5}}$ . Hint: ${1}=\frac{{5}}{{5}}$ .