8x/(3x-6) . x2-4/(2x+4)
the anwer should be4/3x
2007-01-10 03:53:59 · 6 answers · asked by mondragon_sv86 3 in Science & Mathematics ➔ Mathematics
8x/(3x-6) . x2-4/(2x+4)
= (8x (x^2 -4)) / ((3x - 6)(2x+4)
=(8x (x+2)(x-2)) / (3(x-2)2(x+2))
= 8x / 6
= 4x/3
2007-01-10 04:00:10 · answer #1 · answered by mandeep 3 · 2⤊ 0⤋
I assume you have >>> [8x/3x - 6] * [x^2 - 4/2x + 4] correct?
First: factor any numerator and denominator...
[8x/3(x -2)] * [(x + 2)(x - 2)/2(x + 2)]
Sec: cross cancel "like" terms at this point (cancel out an expression that has been repeated twice) ...
(8x/3) * (1/2)
Third: multiply the numerators ...
[8x]/(2)(3)
8x/6
*Simplify the fraction into lowest terms, if possible ...
4x/3
2007-01-10 12:17:55 · answer #2 · answered by ♪♥Annie♥♪ 6 · 0⤊ 0⤋
8x/(3x-6) . (x2-4/(2x+4)
I assume the dot means multiply. If so:
8x/3(x-2) times [(x-2)(x+2)]/ 2(x+2)
The (x+2) and (x-2) factors cancel out leaving:
8x/3 times 1/2 = 4x/3
This differs from your stated answer, but this answer is correct.
2007-01-10 12:11:35 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
What is the equation as stated 8x/(3x-6) . x2-4/(2x+4) it looks ambiguous (what is meant by the ".", is "x2" times 2, or 2x)
Is it 8x/(3x-6) = 2x-4/(2x+4) ?
If so, simplify the equation by multiplying both sides of the equation by (3x-6) and (2x+4), that will get rid of the fractions and make it easier to solve
2007-01-10 12:07:58 · answer #4 · answered by srrl_ferroequinologist 3 · 0⤊ 0⤋
Your expression is:
[8x / (3x-6)] * [(x²-4) / (2x+4)] =
8x(x²-4) / (3x-6)(2x+4) =
Factoring in both terms of the division:
8x(x+2)(x-2) / 3*2 (x-2)(x+2) =
We cancel (x+2)(x-2) in both terms of division we get:
8x / 6 = 4x / 3
That's it!
Good luck!
2007-01-10 12:04:27 · answer #5 · answered by CHESSLARUS 7 · 2⤊ 0⤋
8x/3(x-2) . (x+2)(x-2)/2(x+2)
cancel out x-2 and x+2
so
8x/3 . 1/2
so 4x/3
2007-01-10 12:04:35 · answer #6 · answered by prat_apr89 1 · 1⤊ 0⤋