how do I complete (x-3)/(4x-3y) mutliplied by (16x^2-9y^2)/ 3x^2-5x-12)?

Answer 1

First find factors of the quadratic
16x^2-9y^2 = (4x-3y) (4x+3y)
3x^2-5x-12 =(3x +4) (x-3)

so
(x-3)/(4x-3y) mutliplied by (16x^2-9y^2)/ (3x^2-5x-12)?

becomes

(x-3) (4x-3y) (4x+3y)
-----------------------------------------------------------
(4x-3y) (3x +4) (x-3)

or
(4x+3y)/ (3x+4)

Happy New Year!

Answer 2

Howdy! :) Season's Greetings & Happy New Year! I sure hope all is well. :) As for your question:

The key is to realize that the expressions in the second fraction can actually be factored. Once you notice that and do factor them, you will see that some of the factors will cancel out the first fraction.

In the second fraction, the numerator is: 16X^2 -- 9Y^2. This expression is a difference of squares. So it factors out into: (4X + 3Y)(4X -- 3Y).

Meanwhile, in the second fraction, the denominator is:
3X^2 -- 5X -- 12. This factors into: (3X + 4)(X -- 3).

So multiplying the two fractions becomes a simple operation:

(X -- 3) * (16X^2 -- 9Y^2)
------------- ----------------------- =
(4X -- 3Y) (3X^2 -- 5X -- 12)

(X -- 3) * (4X -- 3Y)(4X + 3Y)
------------- ----------------------------- =
(4X -- 3Y) (X -- 3)(3X + 4)

(4X + 3Y)
--------------
(3X + 4)

I hope that helps and clearly explains the issue(s) for you. :) I tried my best just the same. :) Good luck, take very good care, and have a great day. :)

Cheers! :)

Answer 3

(x -3) / (4x-3y) * (16x^2 - 9y^2) / (3x^2 - 5x - 12) =
put all the numerators on top and all the denominators on bottom
(x -3) (16x^2 - 9y^2) / (4x -3y)(3x^2 - 5x - 12) =
factor
(x-3)(4x -3y)(4x +3y) / (4x -3y)(3x +4)(x -3) =
(x-3) cancels top and bottom, (4x - 3y) also cancels
(4x + 3y) / (3x +4)
final answer

to be technical, you need to add, where x is not 3, since you cancelled (x -3) from the divisor. When x = 3, the divisor is zero.

Answer 4

i like the first answer:by math
but this one is by mathematics
Apply difference of 2 squares for the numerator of the second fraction and factorisation for the denominator ; you will end up getting [(4x+3y)(4x-3y)]/[(3x+4)(x-3)].
Multiplying by the first fraction u would get (x-3) cancelling out (x-3) and the denominator (4x-3y) cancelling out the numerator (4x-3y).The resulting answer will then be
(4x-3y)/(3x+4)

In case of anything u can contact me on [e-mail]

Answer 5

[(x - 3)/4x - 3y)][(16x^2 - 9y^2)/(3x^2 - 5x - 12)]

Factor where possible to see if anything cancels

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

(4x + 3y)/(3x + 4)

Answer 6

(x--3)/(4x--3y) * (16x^2 -- 9y^2)/ (3x^2 -- 5x -- 12)
= (x--3)*(4x+3y)(4x--3y) / (4x--3y)(x--3)(3x+4)
= (4x+3y) / (3x+4)

Answer 7

(x - 3) x (16x^2 - 9y^2) = 16x^3 - 9xy^2 - 48x^2 + 27y^2
(4x - 3y) x (3x^2 - 5x - 12) = 12x^3 - 20x^2 - 48x - 9yx^2 + 15xy + 36y

(16x^3 - 9xy^2 - 48x^2 + 27y^2) / (12x^3 - 20x^2 - 48x - 9yx^2 + 15xy + 36y)

Answer 8

by math