Question

(12x^3y^2/9xy)(18x^5y^3/4x^2y^2) simplify

Answer 1

\[\frac{ 12x^3 y^2 }{ 9xy } \times \frac{ 18x^5 y^3 }{ 4x^2 y^2 }\] First lets work with the fraction on the left. We can divide by the x and y, which is equivalent to dropping the exponent by one. So, x^3 / x is x^2. Also, 12/9 simplifies to 4 over 3, so:
\[\frac{ 4x^2 y }{ 3 } \times \frac{ 18x^5 y^3 }{ 4x^2 y^2 }\] Now lets work with the right fraction. We can do the same thing here. When we divide, we subtract the exponents. And 18/4 simplifies to 9/2, so:
\[\frac{ 4x^2 y }{ 3 } \times \frac{ 9x^3 y }{ 2 }\] We can cancel out the 3 on the left leaving a 3 on the top on the right, and cancel the twos:
\[\frac{ 2x^2 y }{ 1 } \times \frac{ 3x^3 y }{ 1 } = 6x^5 y^2\]

Answer 2

well cancel some common factors 1st

the index law to use is for division is

\[\frac{x^a}{x^b} = x^{a -b}\]

or subtract the powers

for multiplication use

\[x^a \times x^b = x^{a +b}\]

add the powers.

so looking at the 1st brackets you have

\[\frac{4}{3} x^{3 -1}y^{2 -1} = \frac{4}{3} x^2y\]

do the same for the 2nd set of brackets... then multiply your answers

Answer 3

or just wait for someone to post the answer