Making fractions similar

Algebra: Adding and subtracting fractions

Making fractions similar

We can make two fractions

\[
\dfrac{\orange a}{\blue b} \quad \text{and} \quad \dfrac{\purple c}{\green d}
\]

similar by multiplying the denominators with each other:

\[
\dfrac{\orange a}{\blue b} = \dfrac{\orange a \green d}{\blue b \green d} \quad \text{and} \quad \dfrac{\purple c}{\green d} = \dfrac{\blue b \purple c}{\blue b \green d}
\]

Note that both fractions now have the same denominator \(\blue b \green d\) and are therefore similar.

Example

Make #\tfrac{\orange 2}{\blue x}# and #\tfrac{\purple 3}{\green y}# similar:

Brackets When making fractions similar it's important to place brackets around the denominator if the denominator consists of more than one term.

On the right hand side we see how #\tfrac{\orange{2}}{\blue{x}}# and #\tfrac{\orange{\purple 3}}{\green{y+1}}# can be made similar.

Example

\[\begin{array}{rcll}\dfrac{\orange{2}}{\blue{x}}&=& \dfrac{\orange{2} \green{(y+1)}}{\blue{x} \green{(y+1)}}\end{array}\]and\[ \begin{array}{rcll} \dfrac{\orange{\purple 3}}{\green{y+1}}&=& \dfrac{\orange{\purple 3} \blue{x}}{\blue{x} \green{(y+1)}} \end{array}\]

Sometimes there are common factors, and we do not have to multiply both denominators by each other to find a new denominator.

Then you are able to find a new denominator by multiplying by the missing factors in the denominator.

Example

Make #\tfrac{\orange{2}}{\blue{x y}}# and #\tfrac{\purple{3}}{\green{y z}}# similar:

\[\dfrac{\orange{2}}{\blue{x y}}= \dfrac{\orange{2} \green{z}}{\blue{x y} \green{z}} \qqquad \dfrac{\purple{3}}{\green{y z}}= \dfrac{\purple{3} \blue x}{{\blue x \green{y z}}} \]

IntegerWe have now seen how to make two fractions similar. But we can also make a fraction equal to an integer. We can do this by dividing the integer by #1# and next applying the same procedure as when making two fractions similar.

Make #\orange x# and #\tfrac{\purple{3}}{\green{y}}# similar:

\[\begin{array}{rcll}\orange{x} = \dfrac{\orange{x}}{\blue{1}} = \dfrac{\orange{x} \green{y}}{\green {y}} \\ \end{array}\]

Make the following two fractions similar:
\[\frac{y}{y+6} \text{ and } \frac{-9\cdot y}{y+7} \]
Choose the smallest denominator possible.

#\frac{y}{y+6}=# #\frac{y\cdot \left(y+7\right)}{\left(y+6\right)\cdot \left(y+7\right)}# and #\frac{-9\cdot y}{y+7}=# #\frac{-9\cdot y\cdot \left(y+6\right)}{\left(y+6\right)\cdot \left(y+7\right)}#

After all, when making fractions similar we choose the new denominator as the multiplication of both denominators: #\left(y+6\right)\cdot \left(y+7\right)#.

For #\frac{y}{y+6}# we find this new denominator by multiplying numerator and denominator by a factor #y+7#.
This gives: \[\frac{y}{y+6}=\frac{y\cdot \left(y+7\right)}{\left(y+6\right)\cdot \left(y+7\right)}\]

For #\frac{-9\cdot y}{y+7}# we find this new denominator by multiplying numerator and denominator by a factor #y+6#.
This gives: \[\frac{-9\cdot y}{y+7}=\frac{-9\cdot y\cdot \left(y+6\right)}{\left(y+6\right)\cdot \left(y+7\right)}\]

New example