# Sum and Difference of Algebraic Fractions

Learn step-by-step how to solve sum and difference of algebraic fractions with the help of few different types of examples.

1. Find the sum of $$\frac{x}{x^{2} + xy} + \frac{y}{(x + y)^{2}}$$

Solution:

We observe that the denominators of two fractions are

x$$^{2}$$ + xy                    and                    (x + y)$$^{2}$$

= x(x + y)                                          = (x + y) (x + y)

Therefore, L.C.M of the denominators = x(x + y) (x + y)

To make the two fractions having common denominator both the numerator and denominator of these are to be multiplied by x(x + y) (x + y) ÷ x(x + y) = (x + y) in case of $$\frac{x}{x^{2} + xy}$$ and by x(x + y) (x + y) ÷ (x + y) (x + y) = x in case of $$\frac{y}{(x + y)^{2}}$$

Therefore, $$\frac{x}{x^{2} + xy} + \frac{y}{(x + y)^{2}}$$

= $$\frac{x}{x(x + y)} + \frac{y}{(x + y)(x + y)}$$

= $$\frac{x \cdot (x + y)}{x(x + y) \cdot (x + y)} + \frac{y \cdot x}{(x + y)(x + y) \cdot x}$$

= $$\frac{x(x + y)}{x(x + y)(x + y)} + \frac{xy}{x(x + y)(x + y)}$$

= $$\frac{x(x + y) + xy}{x(x + y)(x + y)}$$

= $$\frac{x^{2} + xy + xy}{x(x + y)(x + y)}$$

= $$\frac{x^{2} + 2xy}{x(x + y)(x + y)}$$

= $$\frac{x(x + 2y)}{x(x + y)(x + y)}$$

= $$\frac{x(x + 2y)}{x(x + y)^{2}}$$

2. Find the difference of $$\frac{m}{m^{2} + mn} - \frac{n}{m - n}$$

Solution:

Here we observe that the denominators of two fractions are

m$$^{2}$$ + mn                    and                     m - n

= m(m + n)                                           = m - n

Therefore, L.C.M of the denominators = m(m + n) (m – n)

To make the two fractions having common denominator both the numerator and denominator of these are to be multiplied by m(m + n) (m – n) ÷ m(m + n) = (m - n) in case of $$\frac{m}{m^{2} + mn}$$ and by m(m + n) (m – n) ÷ m - n = m(m + n) in case of $$\frac{n}{m - n}$$

Therefore, $$\frac{m}{m^{2} + mn} - \frac{n}{m - n}$$

= $$\frac{m}{m(m + n)} - \frac{n}{m - n}$$

= $$\frac{m \cdot (m - n)}{m(m + n) \cdot (m - n)} - \frac{n \cdot m(m + n)}{(m - n) \cdot m(m + n)}$$

= $$\frac{m(m - n)}{m(m + n)(m - n)} - \frac{mn(m + n)}{m(m + n)(m - n)}$$

= $$\frac{m(m - n) - mn(m + n)}{m(m + n)(m - n)}$$

= $$\frac{m^{2} - mn - m^{2}n - mn^{2}}{m(m + n)(m - n)}$$

= $$\frac{m^{2} - m^{2}n - mn - mn^{2}}{m(m^{2} - n^{2})}$$

3. Simplify the algebraic fractions: $$\frac{1}{x - y} - \frac{1}{x + y} - \frac{2y}{x^{2} - y^{2}}$$

Solution:

Here we observe that the denominators of the given algebraic fractions are

(x – y)                   (x + y)                and                      x$$^{2}$$ - y$$^{2}$$

= (x – y)               = (x + y)                                         = (x + y) (x – y)

Therefore, L.C.M of the denominators = (x + y) (x – y)

To make the fractions having common denominator both the numerator and denominator of these are to be multiplied by (x + y) (x – y) ÷ (x – y) = (x + y) in case of $$\frac{1}{x - y}$$, by (x + y) (x – y) ÷ (x + y) = (x – y) in case of $$\frac{1}{x + y}$$ and by (x + y) (x – y) ÷ (x + y) (x – y) = 1 in case of $$\frac{2y}{x^{2} - y^{2}}$$

Therefore, $$\frac{1}{x - y} - \frac{1}{x + y} - \frac{2y}{x^{2} - y^{2}}$$

= $$\frac{1}{x - y} - \frac{1}{x + y} - \frac{2y}{(x + y)(x - y)}$$

= $$\frac{1 \cdot (x + y)}{(x - y) \cdot (x + y) } - \frac{1 \cdot (x - y)}{(x + y) \cdot (x - y)} - \frac{2y \cdot 1}{(x + y)(x - y) \cdot 1}$$

= $$\frac{(x + y)}{(x + y)(x - y)} - \frac{(x - y)}{(x + y)(x - y)} - \frac{2y}{(x + y)(x - y)}$$

= $$\frac{(x + y) - (x - y) - 2y}{(x + y)(x - y)}$$

= $$\frac{x + y - x + y - 2y}{(x + y)(x - y)}$$

= $$\frac{0}{(x + y)(x - y)}$$

= 0