i need help on dividing 3y-12 all over 2y+4 by 6y-24 all over 4y+8?

Question

i need to kow what steps i supose to take

Answer 1

you will have an equation that looks like: (3y-12)/(2y+4)/(6y-24)/(4y+8)

when you rearrange it you get: (3y-12)/(2y+4) x (4y+8)/(6y-24)

Cross multiply 3y-12 with 6y-24. the 3y-12 will cancel out on both ends and the 6y-24 will be left with 2.

Same process for 4y+8 and 2y+4 to get the 2y+4 to cancel out and leave you with 2.

Since the equation 3y-12 was reduced to 1 and 2y+4 was reduced to 1 you get 1/1=1

Same with the next equation: 2/2=1

Solution = 1.

Answer 2

[(3y-12) / (2y+4)] / [(6y-24) / (4y+8)]

= [(3y-12) / (2y+4)] x [ (4y+8) / (6y-24)]
replace the division sign by multiplication and inverse the fraction in denominator

= [3(y-4)/2(y+2)] x [4(y+2) / 6(y-4)]

=(3/2) x (4/6)

= (3/2) x (2/3)

= 1

Answer 3

If what you mean is (3y-12/2y+4)/(6y-24/4y+8), then just multiply the fraction on top by the reciprocal of the other.

Answer 4

in order to divide fractions, multiply the first fraction by the reciprocal of the second fraction

in this case:

(3y-12)/(2y+4) x (4y+8)/(6y-24)
...you do the rest

Answer 5

3y-12 / 2y +4 /6y-24 /4y+8 =
.
(3y-12 /2y+4) x(4y+8 /6y-24)=
.
(3y-12)/(2y+4) x2(2y+4)/2(3y-12)=
.
1/1 x1/1 =1

Answer 6

= 3.(y - 4).4.(y + 2) / (2(y + 4).6.(y - 4))
= (y + 2) / (y + 4)