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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 xx2+xy+y(x+y)2

Solution:

We observe that the denominators of two fractions are

   x2 + 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 xx2+xy and by x(x + y) (x + y) ÷ (x + y) (x + y) = x in case of y(x+y)2

Therefore, xx2+xy+y(x+y)2

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

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

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

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

= x2+xy+xyx(x+y)(x+y)

= x2+2xyx(x+y)(x+y)

= x(x+2y)x(x+y)(x+y)

= x(x+2y)x(x+y)2


2. Find the difference of mm2+mnnmn

Solution:

Here we observe that the denominators of two fractions are

   m2 + 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 mm2+mn and by m(m + n) (m – n) ÷ m - n = m(m + n) in case of nmn

Therefore, mm2+mnnmn

= mm(m+n)nmn

= m(mn)m(m+n)(mn)nm(m+n)(mn)m(m+n)

= m(mn)m(m+n)(mn)mn(m+n)m(m+n)(mn)

= m(mn)mn(m+n)m(m+n)(mn)

= m2mnm2nmn2m(m+n)(mn)

= m2m2nmnmn2m(m2n2)

 

3. Simplify the algebraic fractions: 1xy1x+y2yx2y2

Solution:

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

  (x – y)                   (x + y)                and                      x2 - y2     

= (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 1xy, by (x + y) (x – y) ÷ (x + y) = (x – y) in case of 1x+y and by (x + y) (x – y) ÷ (x + y) (x – y) = 1 in case of 2yx2y2

Therefore, 1xy1x+y2yx2y2

= 1xy1x+y2y(x+y)(xy)

= 1(x+y)(xy)(x+y)1(xy)(x+y)(xy)2y1(x+y)(xy)1

= (x+y)(x+y)(xy)(xy)(x+y)(xy)2y(x+y)(xy)

= (x+y)(xy)2y(x+y)(xy)

= x+yx+y2y(x+y)(xy)

= 0(x+y)(xy)

= 0






8th Grade Math Practice

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