Solve Rational Equation Problems

In algebra solving rational equation is very simple,we have few rules to solve any type of equations. Whatever we do on  one side  of the equation, we must do to the other side also. If you have fractions, we can try to eliminate them by multiplying by the common denominator. If there are quadratics involved in our equations, we must get all the  terms to one side with zero on the other.The basic rational expression is in the form of fraction.where there is at least one variable in the denominator.

Solved Example Based on Rational Equation :

Ex 1:Solve `3/x+6=2/(4x)`

Sol:

Step 1: The given equation is

`3/x+6=2/(4x)`

subtracting both sides by `2/(4x)`

`3/x+6-(2/(4x))=(2/(4x))-(2/(4x))`

Step 2: Rearrange the equation.

`3/x-2/(4x)+6=0`

subtracting both sides by 6.

`3/x-2/(4x)+6-6=0-6`

`3/x-2/(4x)=-6`

Step 3: Find the l.c.d(least common denominator) x and 4x

L.C.D=4x

`(12-2)/(4x)=-6`

`10/(4x)=-6`

Step 4:Multiply both sides by 4x.

`10/(4x)xx4x=-6xx4x`

10=-24x

Step 5:Divide both sides by -24

`10/(-24)=(-24x)/(-24)`

`x=-5/12`

The solution is  [x=-5/12]

Example Based on Rational Equation:

Ex 2: Solve `x/(x-2)+1/(x-4)=2/(x^2-6x+8)`

Sol:

Step 1:first factor the `x^2-6x+8`

factors=(x-4)(x-2)

Step 2:convert common denominator to all

`(x/(x-2))((x-4)/(x-4))+(1/(x-4))((x-2)/(x-2))=2/((x-2)(x-4))`

`(x^2-4x)/((x-2)(x-4))+(x-2)/((x-2)(x-4))=2/((x-2)(x-4))`

`(x^(2)-4x)+(x-2)=2`

`x^(2)-4x+x-2=2`

`x^(2)-3x-4=0`

`(x-4)(x+1)=0`

`x=4 or x=-1`

If we submit the x=4 in the denominator,it is division by zero,so x=4 is not considerable.

x=-1   is the answer