Question

PLEASE HELP!!!!!

If 12xy^3 + 4x^2y^5 over 4xy^2 is completely simplified to 3x^ay^b + 1x^cy^d, where a, b, c, and d represent integer exponents, what is the value of a?

Answer 1

If we want to calculate 12xy^3/4xy^2, we divide the constants (12/4) and arrive 3, but then we simply subtract the power of the variable in the denominator from the power of the variable in the numerator. Here for x, that is 1 - 1, leaving us with 0. Thus the value of a is 0, and the x^0 term would just become a 1, allowing us to essentially remove it since it simply is multiplying by 1.

Answer 2

The first thing to do in this case is to perform the division and then compare with the simplified expression to determine the values of a, b, c and d.
We have then:
(12xy ^ 3 + 4x ^ 2y ^ 5) / (4xy ^ 2) =
3x (1-1) and ^ (3-2) + x ^ (2-1) and ^ (5-2) =
3x (0) and ^ (1) + x ^ (1) and ^ (3)
Then, comparing with the simplified expression:
3x ^ ay ^ b + 1x ^ cy ^ d
The value of a is
a = 0
answer
the value of a is 0