# Addition and Subtraction of Fractions

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,

More generally, if $q \ne s$, you can always add $p/q$ and $r/s$ as follows:

Example:

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

Solution:

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