# How to solve Algebraic fractions?

Here I’m back with the same topic Algebraic fractions….
don’t stressed…
I said with same topic but the advance level of the Algebraic fractions….
YES..!!

## Algebraic Fractions

#### Solution of algebraic fractions having two or more terms and having unknown variables and also those equations which are containing TRIGONOMETRIC functions.

Don’t worry,
let me tell you we’ll pick the topic trigonometry later on, I’m just explaining here the solutions of fractions based upon Trigonometry functions.

### Solutions of Equations based upon two variables :-

firstly let me tell you if we are looking forward to solve equations having two variables, for those Solutions we must have at least two equation with same variables
let’s​ proceed:-

 (original expression) =2x+3⋅x−1x−1−3xx−1⋅x+3x+3  =2x+3⋅x−1x−1−3xx−1⋅x+3x+3 rather we can solve this step with an easy method by multiplying denominators to the numerator and multiplying denominators to each other i.e. =2(x−1)−3x(x+3)(x+3)(x−1)  =2(x−1)−3x(x+3)(x+3)(x−1) (keep the denominator the same; add the numerators) =2x−2−3×2−9x(x+3)(x−1)  =2x−2−3×2−9x(x+3)(x−1) (multiply out the numerator) =−3×2−7x−2(x+3)(x−1)
likewise for Trigonometry functions :-

so, you can analyse how the fraction solution becomes easier without finding out the LCM of denominators.

see some more examples having exponents as well… but do remember we will apply BODMAS RULE first in the Solutions…

HOW TO APPLY BODMAS RULE

for brief explanation check this picture based solution

### what if the equation is given with two variables:-

for this kind of Solutions those two variables​ will work as unlike variables ( which cannot be added, subtracted, multiplied, or divided)
they will remain as they are.
You can have a quick look here under this kinds of solutions
Solve:-

x/2 + y/3

(3x + 2y)/ 6

one more check this out-

3xx−2+5xx2−4
Solution:
3xx−2+5xx2−4
= 3xx−2+5xx2−(2)2  f fg h
= 3xx−2+5x(x+2)(x−2) g
= 3x×(x+2)(x−2)(x+2)+5x(x+2)(x−2)
= 3x(x+2)−5x(x−2)(x+2)
= 3×2+6x−5x(x−2)(x+2)
= 3×2+x(x−2)(x+2)
= x(3x+1)(x−2)(x+2)

firstly in above solution the denominators made similar through Identity
a² – b² = (a+b)(a-b)

then after there is no need to proceed the Solutions like above explained, so it is concluded that if the denominators are similar Solutions can be proceed from numerators​ themselves with respect to their signs.

### What if there is three fraction terms are given:-

Don’t worry if there is any problem like this solution will be processed in the similar way here’s​ check out the solutions

1×2−3x+2+1×2−5x+6+1×2−4x+x² – 4x +3

Solution:
1×2−3x+2+1×2−5x+6+1×2−4x+3
= 1×2−2x−x+2+1×2−3x−2x+6+1×2−x−3x+3
= 1x(x−2)−1(x−2)+1x(x−3)−2(x−3)+1x(x−1)−3(x−1)
= 1(x−2)(x−1)+1(x−3)(x−2)+1(x−1)(x−3)
= 1×(x−3)(x−2)(x−1)(x−3)+1×(x−1)(x−3)(x−2)(x−1)+1×(x−2)(x−1)(x−3)(x−2)
= (x−3)(x−2)(x−1)(x−3)+(x−1)(x−3)(x−2)(x−1)+(x−2)(x−1)(x−3)(x−2)
= (x−3)+(x−1)+(x−2)(x−1)(x−2)(x−3)
= (3x−6)(x−1)(x−2)(x−3)
= 3(x−2)(x−1)(x−2)(x−3)
= 3(x−1)(x−3)

here is also made the denominators similar and processed similarly.

Hope you have grasp the concepts regarding solving algebraic fractions. If you have any doubt then please share that in comment box, I’ll be pleased to clear those doubts and also do subscribe for our new post direct to your Mail ID.
Thanks,
Happy learning…