Question 66395
Multiply:
{{{((-3a^2b)/35a^5)*((14a^3b^2)/-9b^4))}}} Ok, let's take it one step at-a-time.
First look at the numbers.
Do you see that you can cancel a -3 in the numerator of the first fraction and the -9 in the denominator of the second fraction, leaving a 3 in the denominator of the second fraction?
{{{((a^2b)/(35a^5))*((14a^3b^2)/(3b^4))}}} Now do you see that the 14 and the 35 are both divisible by 7? Then let's divide them by 7.
{{{(a^2b/5a^5)*(2a^3b^2/3b^4)}}} Now we can look at the variables, a and b.
Remember the rules for dividing factors with exponents?
If the base is the same (in this case, the bases are a and b), you subtract the exponents. Let's do the a's first and see what we get!
{{{(a^(2-2)b/5a^(5-2))*(2a^3b^2/3b^4)}}} Simplifying this a bit and remembering that {{{a^0 = 1}}}
{{{(b/5a^3)*(2a^3b^2/3b^4)}}} Now we can do the same thing with the {{{a^3}}}s
{{{(b/5)*(2b^2/3b^4)}}} Now do the b's
{{{(b/5)*(2/3b^2)}}} Finally, cancel the b's and simplify.
{{{(1/5)*(2/3b)}}} = {{{2/15b}}}...and there it is!