We have to simplify the fraction: 2/(2x*(6x+4)) and write it as a sum of simplified fractions.
=> 1/x*(6x + 4)
This can be written as (A / x) + (B / (6x + 4))
=> [A(6x + 4) + Bx]/ x*(6x + 4) = 1/x*(6x + 4)
=> 6Ax + 4A + Bx = 1
=> 4A = 1 and 6A + B = 0
=> A = 1/4 and B = -6*(1/4) = -3/2
The fraction 2/2x(6x+4) can be written as (1/4x) - 3/[2*(6x + 4)]
We'll have to get 2 irreducible fractions because of the 2 factors from denominator.2/2x(6x+4) = 1/x(6x+4)
The final ratio 1/x(6x+4) is the result of addition or subtraction of 2 elementary fractions, as it follows:
1/x(6x+4) = A/x + B/(6x+4) (1)
We'll multiply by x(6x+4) both sides:
1 = A(6x+4) + Bx
We'll remove the brackets:
1 = 6Ax + 4A + Bx
We'll factorize by x to the right side:
1 = x(6A+B) + 4A
Comparing both sides, we'll get:
6A+B = 0
4A = 1 => A = 1/4
6/4 + B = 0
B = -3/2
We'll substitute A and B into the expression (1) and we'll get the algebraic sum of 2 elementary fractions:
2/2x(6x+4) = 1/x(6x+4) = 1/4x - 3/2(6x + 4)
2/2x(6x+4) = 1/2x(3x+2)
Let 1/2x((3x+2) = A/2x+B/(3x+2)
=> 1 = A(3x+2)+B(2x) (1)
Put = x= 0 in (1)
=> 1 = A*2. So A = 1/2.
Put x=-2/3 in (1)
=> 1 = 0+B*2* (-2/3) . S B = -3/4.
=> = 2/2x(6x+4) = 1/2x(3x+2) = 1/4x - 3/4(3x+2).