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






8th Grade Math Practice

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