Decopose in simply fractions. 2/2x(6x+4)

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to simplify the fraction: 2/(2x*(6x+4)) and write it as a sum of simplified fractions.

2/(2x*(6x+4))

=> 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)]

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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)

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

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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).

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