How do you find the quotient #12/(3x^2)div6/x#?

1 Answer
Feb 12, 2017

See the entire solution process below:

Explanation:

First, we can rewrite this expression as:

#(12/(3x^2))/(6/x)#

Then, we can use this rule for dividing fractions:

#(color(red)(a)/color(blue)(b))/(color(green)(c)/color(purple)(d)) = (color(red)(a) xx color(purple)(d))/(color(blue)(b) xx color(green)(c))#

Substituting the terms from the problem gives:

#(color(red)(12)/color(blue)(3x^2))/(color(green)(6)/color(purple)(x)) = (color(red)(12) xx color(purple)(x))/(color(blue)(3x^2) xx color(green)(6)) = (12x)/(18x^2)#

Next, let's simplify the constants:

#(12x)/(18x^2) = ((6 xx 2)x)/((6 xx 3)x^2) = ((color(red)(cancel(color(black)(6))) xx 2)x)/((color(red)(cancel(color(black)(6))) xx 3)x^2) = (2x)/(3x^2)#

Now, we can use these rules for exponents to complete the solution:

#a^color(red)(1) = a# or #a = a^color(red)(1)# and #x^color(red)(a)/x^color(blue)(b) = 1/x^(color(blue)(b)-color(red)(a))#

#(2x)/(3x^2) = (2x^color(red)(1))/(3x^color(blue)(2)) = 2/(3x^(color(blue)(2)-color(red)(1))) = 2/(3x^1) = 2/(3x)#