multiplication of alegbraic fractions?

Question

simplify (6x^2) / (4x^2y-12xy) * (x^2+x-12) / 3x^2+12x)

show working please

Answer 1

I presume you mean

[(6x^2) / (4x^2y-12xy)] * [(x^2+x-12) / (3x^2+12x)]

There seems to be something wrong. Are you sure about those "y" terms?

Answer 2

To multiply fractions you can simply multiply the numerators and the denominators

(6x^2) / (4x^2y - 12xy) * (x^2 + x - 12) / 3x^2 + 12x)

(6x^2)(x^2 + x - 12)/(4x^2y - 12xy)(3x^2 + 12x)

(6x^2)(x + 4)(x - 3)/4xy(x - 3)3x(x + 12)

(6x^2)(x + 4)(x - 3)/12x^2y(x - 3)(x + 12)

(6x^2)(x + 4)/12x^2y(x + 12)

edit: I thought it could be factored down further, just couldn't see how. I'll let Paritosh Vasava take the credit for that!

Answer 3

(6x^2) / (4x^2y-12xy) * (x^2+x-12) / 3x^2+12x)
=(6x^2) / (4x^2y-12xy) * (x^2+x-12) / 3x^2+12x)
=(6x^2)*(x^2+x-12) / (4x^2y-12xy) *(3x^2+12x)
=(6x^2)*(x-3)*(x+4) / (x-3)*4xy *(x+4)*3x [Cancelling (x-3)]
=(6x^2)*(x+4) / 4xy *(x+4)*3x [Cancelling (x+4)]
=(6x^2) / 12*x^2 y [Cancelling 6*x^2]
=1/2y

I hope i answered your question.

All the best.

Answer 4

(6x²) / (4x²y-12xy) * (x²+x-12) / (3x²+12x)

To simplify this, you can pull the factors out of each side.

[2x*3x/2x*2y*(x-3)] * [(x-3)(x+4)/3x(x+4)]

The 2x on the top and bottom of the left cancels out.
The (x+4) on the top and bottom of the right cancels out.
The 3x on the top of the left, and bottom of the right, cancel out.
The (x-3) on the bottom of the left, and top of the right, cancel out.
[1/2y] * [1/1]
1/2y