multiply. (4x-12)/(x^2+6x) times 5x/(3-x)?

Question

please help! Im trying to figure these out but Im lost!

1. multiply. (4x-12)/(x^2+6x) times 5x/(3-x)

2. Divide. 3/x^7 divided by 6/x^3

3. Divide. (2x-6)/21 divided by (5x-15)/12

Answer 1

1. OK, multiplication of fractions. You multiply the numerators and denominators separately, right? Right, so we can go ahead and combine it so it looks like this instead...

(4x-12)(5x) / (x²+6x)(3-x)

Now, Since we want to get this in simplest terms, the easiest way to do it is to reduce each item before multiplying it out. 4x-12 can be factored, as can (x² + 6x), giving us instead...
(4)(x-3)(5x) / (x)(x+6)(3-x)
(20x)(x-3) / (x)(x+6)(3-x)
We can divide 20x by x to leave 20, giving us...
20(x-3) / (x+6)(3-x)
Here's another simple trick... (x-n) * -1 = (n-x)... so...
20(-1)(3-x) / (x+6)(3-x)
-20 / (x+6)

So long as x ≠ (does not equal) -6, 3, or 0.

2. Divide. Division of fractions is the same as multiplying by the reciprocal, so...
[ 3 / x^7 ] ÷ [ 6 / x³ ] =
[ 3 / x^7 ] × [ x³ / 6 ]
Now, since order of multiplication doesn't mattrer, we can rearrange that to say...
[ x³ / x^7 ] × [ 3 / 6]
Now, since dividing exponential expressions with the same base is done by subtracting the exponent, and 3-7 = -4, we have
[ 1 / x^4 ] × [ 1 / 2 ] =
1 / 2(x^4)

3. Again, division is multiplying by the reciprocal, so...
[ (2x-6) / 21 ] ÷ [ (5x-15) / 12 ] =
[ (2)(x-3) / (3)(7) ] × [ (3)(4) / (5)(x-3) ]
[ (2)(3)(4)(x-3) ] × [ (3)(5)(7)(x-3) ]
Cancel out like terms, and we have
(2)(4) / (5)(7)
8/35
So long as x ≠ (does not equal) 5

Answer 2

Since you know basic math, factoring, principle addition of simple fractions.

No need for further explanation.

If you just follow carefully and understand the steps below, you will know how its done:


((4x - 12) / (x^2 + 6x))(5x/(3 - x))

((4x - 12) / (x^2 + 6x))(5x/(-x + 3))

(4((4x/4) - (12/4)) / (x^2 + 6x))(5x/(-x + 3))

(4(x - 3)) / (x^2 + 6x))(5x/(-x + 3))

(4(x - 3)) / x(x^2/x + 6x/x))(5x/(-x + 3))

(4(x - 3)) / x(x + 6))(5x/(-x + 3))

(4(x - 3))(5x) / (x(x + 6))(-x + 3)

(4(x - 3)(5x)) / ((x)(x + 6)(-x + 3))

(4)(5)(x)(x - 3) / (x)(x + 6)(-x + 3))

CANCEL SIMILAR TERMS, NEGATE AS NECESSARY:

(4)(5)(x)(x - 3) / (x)(x + 6)(-x + 3))

(4)(5)(x)(-1)(-x + 3) / (x)(x + 6)(-x + 3))

(4)(5)(-1) / (x + 6)

-20 / (x + 6)




(3/x^7) / (6/x^3)

(3/x^7) (x^3/6)

3x^3 / 6x^7

3x^3 / (2)(3x^7)

1 / (2)(x^(7-3))

1 / (2)(x^4)

1 / 2x^4




((2x - 6)/21) / ((5x - 15)/12)

((2(2x/2 - 6/2))/21) / ((5x - 15)/12)

((2(x - 3))/21) / ((5x - 15)/12)

((2(x - 3))/21) / (5(5x/5 - 15/5)/12)

((2(x - 3))/21) / (5(x - 3)/12)

((2(x - 3))/21) (12/5(x - 3))

((2(x - 3))(12) / (21)5(x - 3))

(2)(x - 3)(12) / (21)(5)(x - 3)

(2)(12)(x - 3) / (21)(5)(x - 3)

(2)(12) / (21)(5)

(2)(2^2)(3) / (21)(5)

(2)(2^2)(3) / (21)(5)

(2^3)(3) / (21)(5)

(2^3)(3) / (3)(7)(5)

(2^3) / (7)(5)

8/35

Answer 3

1. multiply.
(4x - 12) / (x^2 + 6x) times 5x / (3 - x) =
(4(x - 3) / (x(x + 6)))( 5x / (-1(x - 3)) =
4(5x)(x - 3) / - (x(x - 3)(x + 6)) =
- (4/(x + 6))(5x/x)((x - 3)/(x - 3)) =
- 4/(x + 6)

2. Divide.
(3/x^7) / (6/x^3) =
(3/x^7)(x^3/6) =
(3/6)(x^3/x^7) =
1/(2x^4)

3. Divide.
(2x - 6) / 21 divided by (5x - 15) / 12 =
((2x - 6) / 21) / ((5x - 15) / 12) =
((2x - 6) / 21) (12 / (5x - 15) ) =
2(12)(x - 3) / (21(5)(x - 3) =
2(3)(4)/(3(5)(7) = 8/35

Answer 4

1)Numerator(4x-12) times 5x or 4(x-3)5x
Denominator (x^2+6x)(3-x) or x(x+6)(3-x)
20x(x-3)/x(x+6)(3-x)
(3-x)=-(x-3) and cancel out the x in numerator and denominator
20(x-3)/-(x+3)(x+6)=-20/x+6

2) 3/x^7 divided by 6/x^3 Invert the second term and multiply
3x^3/6x^7=1/2x^4

3) (2x-6)/21 divided by (5x-15)/12 Invert the second term and multiply
(2x-6)12/21(5x-15)=12(2)(x-3)/21(5)(x-3)
cancel out the x-3 terms
24/21(5)=8/7(5)=8/35

Answer 5

Generally, before going into these problems "hammer and tongs", look for tricks.
In the first one, if you factor 4x-12, you wind up with 4(x-3), but you also can turn this into
-4 (3-x). But that's in the denominator in the other expression. So whammo!!, out go two term and you are left with a -4. Isn't math wonderful?

In multiplication of fractions, (A/B) * (C/D) = AC/BD
Then you usually need to "clean up" your result by factoring out like terms in the numerator and denominator.

In division of fractions, "flip" the divisor (the divided by term) and then proceed as if it was a multiplication. So in problem 2, you wind up with the multiplication: 3 /x^7 * x^3/6 which is
3 x^3 / 6 x^7. Here you can work on the 3/6 and the x^3/x^7 to wind up with 1/ (2x^4).

In problem 3, "flip" and then notice that (2x-6) and (5x-15) can both be factored in terms that both include (x-3), so that gets wiped out, and all you have left are plain old numbers.

Answer 6

(4x - 12)/(x^2 + 6x) * 5x / (3 - x)
factor the top and bottom
-4(3 - x) / x(x+6) * 5x / (3 - x)
The 3 - x cancel, x cancel
-4 / (x + 6) * 5
-20/(x + 6)
----------------------------------------
3/x^7 / (6/x^3) division of fractions; multiply by the reciprocal of the latter term
3/x^7 * (x^3/6)
3/6 = 1/2; x^3/x^7 = 1/x^4
thus it is 1/2x^4
----------------------------------
(2x - 6)/21 / (5x - 15)/12
12*(2x - 6) / (21[5x - 15])
24(x - 3) / 105(x - 3)
24/105
=8/35

Feel free to e-mail me for clarification on any specific point. Hope I helped.

Answer 7

if the situation is 3x - (a million/4)x = 5x + (3/6)x an obtrusive answer is 0, because of the fact that x is in each and every term If the situation is 3x - a million/(4x) = 5x + 3(6x) -2x - 6/(24x) = 12/(24x) -2x = 18 / (24x) -x = 9 / (24x) -24x² = 9 x² = -9/24 x = 3i/2?6 Grandpa

Answer 8

ahh the 1st one is alot of work
second one is a complex fraction soo the answer is 3x^3/6x^7 if i recall correctly
and the third is 12(2x-6)/21(5x-15) just simplify