MATH HELP please and THAnk you!!?

Question

rewrite the fractions with their LCD.

1. 4/3x - 6y , 1/ 5x - 10y

2, 3/c^2 - 6c , 5/c^2 + 6c

THANKS!!!!!!!!!!!!! :)

Answer 1

1. 4/3x - 6y , 1/ 5x - 10y
= 4/(3(x-2y)) , 1/(5(x-2y))
= 20/(15(x-2y)) , 3/(15(x-2y))
The LCD is 15(x-2y)

2, 3/c^2 - 6c , 5/c^2 + 6c
= 3/c^2 - 6c^3/c^2 , 5/c^2 +6c^3/c^2
=(3-6c^3)/c^2 , (5+6c^3)/c^2
The LCD is c^2

Answer 2

You need to choose a denominator that is common to all the terms in the expression, and then convert each term to a form using that denominator.

So, for example, if you have a/b + c/d, your LCD is b*d, and you need to multiply the first term by d/d = 1, and the second term by b/b = 1, not changing the terms but putting them on a common denominator: ad/bd + cb/db

a d c b
---- + ---- (the formatting is a bit messed up)
b d d b

Remember that a term like 6c has a denominator of 1.

Answer 3

1.

4/(3x - 6y) , 1/(5x - 10y)

Note that each of the denominators factors.

3x - 6y = 3(x - 2y)
5x - 10y = 5(x - 2y)

What is the LCD of 3(x - 2y) and 5(x - 2y)? The answer to that is (3)(5)(x - 2y). Therefore,

4/(3x - 6y) = 4/[3(x - 2y)] = 4(5)/[(3)(5)(x - 2y)] = 20 / (15(x - 2y))

1/(5x - 10y) = 3/[15(x - 2y)]

Answer 4

Somebody already did 1.

2. 3/[c(c-6)] & 5/[c(c+6)]

The LCD is then c(c-6)(c+6)

3(c+6)/c(c+6)(c-6) & 5(c-6)/(c(c+6)(c-6)

(3c+18)/(c^3 -36c²) & (5c-30)/(c^3 - 36c²)