How do you simplify #((4a^3 b)/(a^2 b^3)) ((3b^2)/(2a^2 b^4))#?

2 Answers
Jun 11, 2018

#\frac{6}{ab^4}#

Explanation:

First of all, you can simplify each fraction on its own: to make things easier to see, you can rewrite the first fraction as a multiplication of three fractions, each involving only one variable, or the numeric part:

#\frac{4a^3b}{a^2b^3} = 4 * \frac{a^3}{a^2} * \frac{b}{b^3}#

Now you can simplify exponents:

#4 * \frac{a^cancel(3)}{cancel(a^2)} * \frac{cancel(b)}{b^{cancel(3) 2}} = 4*a*\frac{1}{b^2} = \frac{4a}{b^2}#

We can do the same for the second fraction:

#\frac{3b^2}{2a^2b^4} = \frac{3}{2} * \frac{1}{a^2} * \frac{b^2}{b^4}#

We can simplify exponents on the #b# part:

#\frac{3}{2} * \frac{1}{a^2} * \frac{b^cancel(2)}{b^{cancel(4)2}} = \frac{3}{2a^2b^2}#

So, the multiplication between the two fraction has become

#\frac{4a}{b^2}*\frac{3}{2a^2b^2}#

As a final step, we can cross simplify factor, which means that we can simplify the same factors if they appear at the numerator of one fraction and the denominator of the other. In our case, we have #4a# at the numerator of the first fraction, and #2a^2# at the numerator of the second, which simplify into

#\frac{cancel(4)^2cancel(a)}{b^2}*\frac{3}{cancel(2)a^cancel(2)b^2} = \frac{2*3}{a*b^2*b^2} = \frac{6}{ab^4}#

Jun 11, 2018

#=6/(ab^4)#

Explanation:

#(4a^3b)/(a^2b^3) xx (3b^2)/(2a^2b^4)" "larr# multiplying fractions

#=(cancel4^2a^3b)/(a^2b^3) xx (3b^2)/(cancel2a^2b^4)" "larr# cancel the numbers

#=(6a^3b^3)/(a^4b^7)" "larr# multiply to get one fraction

#=6/(ab^4)" "larr# subtract the indices of like bases