Math question, please help!!!?

Question

If you can, please give me step by step. Those who know ofcourse. Points to the right answer!

1. (7a^3b/3ab^5)/(21a^4b^2/14a^3b) = ?

and

2. (4y/3x)+(15x^2/2y) = ?

Answer 1

(7 a^3 b) ...... (21 a^4 b^2)
________ / __________ is the question, it can be

(3 a b^5) ...... (14 a^3 b)

written as,

(7 a^3 b) ..... (14 a^3 b)
________ X __________

(3 a b^5) ...... (21 a^4 b^2)

simplifying

7 a^2 ......... 2
______ X ________

3 b ^4 .... ... 3 a b

so answer would be

14 a
-------
9 b^5


second question

take least common multiple or LCm of the denominators.

which is 6xy

so divide the LCM separately with the denominators and multiply the respective results with the numerator.

so we have
(4y * 2y) + (15 x^2 * 3 x)
----------------------------------
6xy
which is equal to

8 y^2 + 45 x^3
-------------------
6xy

Answer 2

I think you might be missing something, a * between some of your numbers and letters in #1?? Maybe?? Or I'm confused at the way it's typed.

For problem #2;
1) find a common denominator (3x2y)
2) multiply (4y/3x) by (2y/2y) = (8y^2/(3x2y))
3) multiply (15x^2/2y) by (3x/3x) = (45x^3/(3x2y))
4) now add the two fractions together =(45x^3+8y^2)/(3x2y)
That's it, unless I misunderstood the type of answer you're looking for.

Answer 3

1. the answer is 14a/9b^5
use cross multiplication and simplify by cancelling the powers using indices

Answer 4

Using the operation "log" (logarithm) will make it easier to solve.

Answer 5

sorry idk