Addition and Subtraction of unlike fractions
Example 1: What is \(1\over 2\)+ \(1\over 8\)?
Recall that same sized or like fractions can simply be added by counting the number of parts in the numerators.
Look at the figures. Parts in both shapes are of different sizes.
So, we cannot simply add them, We will first have to convert them to same sized parts or like fractions
Since \({1\over 2}={4\over 8}\), . We can add \(4\over 8\) and \(1 \over 8\).
Addition and Subtraction of unlike fractions
Example 1: What is \(1\over 2\)+ \(1\over 8\)?
Recall that same sized or like fractions can simply be added by counting the number of parts in the numerators.
Look at the figures. Parts in both shapes are of different sizes.
So, we cannot simply add them, We will first have to convert them to same sized parts or like fractions
Since \({1\over 2}={4\over 8}\), . We can add \(4\over 8\) and \(1 \over 8\).
Addition and Subtraction of unlike fractions
Example 1: What is \(1\over 2\)+ \(1\over 8\)?
Recall that same sized or like fractions can simply be added by counting the number of parts in the numerators.
Look at the figures. Parts in both shapes are of different sizes.
So, we cannot simply add them, We will first have to convert them to same sized parts or like fractions
Since \({1\over 2}={4\over 8}\), . We can add \(4\over 8\) and \(1 \over 8\).
Addition and Subtraction of unlike fractions
Example 1: What is \(1\over 2\)+ \(1\over 8\)?
Recall that same sized or like fractions can simply be added by counting the number of parts in the numerators.
Look at the figures. Parts in both shapes are of different sizes.
So, we cannot simply add them, We will first have to convert them to same sized parts or like fractions
Since \({1\over 2}={4\over 8}\), . We can add \(4\over 8\) and \(1 \over 8\).
Addition and Subtraction of unlike fractions
Example 1: What is \(1\over 2\)+ \(1\over 8\)?
Recall that same sized or like fractions can simply be added by counting the number of parts in the numerators.
Look at the figures. Parts in both shapes are of different sizes.
So, we cannot simply add them, We will first have to convert them to same sized parts or like fractions
Since \({1\over 2}={4\over 8}\), . We can add \(4\over 8\) and \(1 \over 8\).
Addition and Subtraction of unlike fractions
Example 1: What is \(1\over 2\)+ \(1\over 8\)?
Recall that same sized or like fractions can simply be added by counting the number of parts in the numerators.
Look at the figures. Parts in both shapes are of different sizes.
So, we cannot simply add them, We will first have to convert them to same sized parts or like fractions
Since \({1\over 2}={4\over 8}\), . We can add \(4\over 8\) and \(1 \over 8\).
\(\frac{4}{8} + \frac{1}{8} = \frac{5}{8}\)
So, \(\frac{4}{8} + \frac{1}{8} = \frac{5}{8}\)
Let’s add \(4\over 9\) and \(1\over 6\) .
Since, the fractions are unlike fractions, we will convert them to like fractions.
Step 1: Let’s find LCM of 9 and 6
Step 2: 18 is the LCM, which will be the common denominator
Step 3: Let’s find the missing numbers
\(8 \over 18\) and \(3 \over 18\)are like fractions
\(\frac{8}{{18}} + \frac{3}{{18}} = \frac{{11}}{{18}}\)
So,\(\frac{4}{9} + \frac{1}{6} = \frac{{11}}{8}\)
Example 2: Let’s subtract \(1\over 2\) from \(4\over 5\) .
The shapes do not have equal sized parts. So, we cannot subtract \(1\over 2\) from \(4\over 5\) .
We will first convert them to shapes with equal sized parts or like fractions.
Let’s find equivalent fraction for both \(1\over 2\) and \(4\over 5\) such that they have a common denominator
10 is the LCM. So, 10 will be the common denominator
Let’s find the missing numbers.
Now, we can subtract \(5 \over 10\) from \(8\over 10\),just subtracting the numerators
\(\frac{8}{{10}} - \frac{5}{{10}} = \frac{3}{{10}}\)