Two or more fractions with the same denominator are said to have a common denominator. The following rules for adding and subtracting fractions with a common denominator can be derived from the basic algebraic properties of real numbers and the definition of a quotient.

Rules for Addition Subtraction of Fractions with Common Denominator

If $p$, $q$ and $r$ are real numbers, $q \ne 0$, then $\frac{p}{q} + \frac{r}{q} = \frac{{p + r}}{q}$   and   $\frac{p}{q} – \frac{r}{q} = \frac{{p – r}}{q}$

For instance, $\frac{3}{7} + \frac{2}{7} = \frac{{3 + 2}}{7} = \frac{5}{7}$   and   $\frac{3}{7} – \frac{2}{7} = \frac{{3 – 2}}{7} = \frac{1}{7}$

Again, the same rules apply to adding and subtracting fractional expressions.

Example:

Perform each operation

(a) $\frac{{5x}}{{2x – 1}} + \frac{{3x}}{{2x – 1}}$
(b) $\frac{{5x}}{{{{\left( {3x – 2} \right)}^2}}} – \frac{{3x}}{{{{\left( {3x – 2} \right)}^2}}}$

Solution:

(a) $\frac{{5x}}{{2x – 1}} + \frac{{3x}}{{2x – 1}} = \frac{{5x + 3x}}{{2x – 1}} = \frac{{8x}}{{2x – 1}}$
(b) $\frac{{5x}}{{{{\left( {3x – 2} \right)}^2}}} – \frac{{3x}}{{{{\left( {3x – 2} \right)}^2}}} = \frac{{5x – 3x}}{{{{\left( {3x – 2} \right)}^2}}} = \frac{{2x}}{{{{\left( {3x – 2} \right)}^2}}}$

To add or subtract fractions that do not have a common denominator, you must rewrite the fractions so they do have denominators. To do this, multiply the numerator and denominator of each fraction by an appropriate number. For instance,

$\frac{2}{3} + \frac{4}{5} = \frac{{2 \cdot 5}}{{3 \cdot 5}} + \frac{{3 \cdot 4}}{{3 \cdot 5}} = \frac{{10}}{{15}} + \frac{{12}}{{15}} = \frac{{10 + 12}}{{15}} = \frac{{22}}{{15}}$

More generally, if $q \ne s$, you can always add $p/q$ and $r/s$ as follows:
$\frac{p}{q} + \frac{r}{s} = \frac{{p \cdot s}}{{q \cdot s}} + \frac{{q \cdot r}}{{q \cdot s}} = \frac{{ps + qr}}{{qs}}$

Example:

Add the expression $\frac{{3x}}{{4x – 1}}$ and $\frac{{2x}}{{3x – 5}}$

Solution:

$\frac{{3x}}{{4x – 1}} + \frac{{2x}}{{3x – 5}} = \frac{{3x\left( {3x – 5} \right)}}{{\left( {4x – 1} \right)\left( {3x – 5} \right)}} + \frac{{\left( {4x – 1} \right)2x}}{{\left( {4x – 1} \right)\left( {3x – 5} \right)}}$
$= \frac{{3x\left( {3x – 5} \right) + \left( {4x – 1} \right)2x}}{{\left( {4x – 1} \right)\left( {3x – 5} \right)}} = \frac{{9{x^2} – 15x + 8{x^2} – 2x}}{{\left( {4x – 1} \right)\left( {3x – 5} \right)}}$
$= \frac{{17{x^2} – 17x}}{{\left( {4x – 1} \right)\left( {3x – 5} \right)}} = \frac{{17x\left( {x – 1} \right)}}{{\left( {4x – 1} \right)\left( {3x – 5} \right)}}$