How do you simplify #(5/(4x^2)) / (8/x^3)#?

Question

1804 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-03T04:55:09

#frac{frac{5}{4x^2}}{frac{8}{x^3}} = frac{5x}{32}#

Multiply the "lowest common multiple" to both the numerator and the denominator.

A nice common multiple of #4x^2# and #x^3# is #4x^3#.

So trying to multiply out

#frac{frac{5}{4x^2} * (4x^3)}{frac{8}{x^3} * (4x^3)} = frac{5x}{32}#

Answer 2 · 2016-09-03T06:54:11

#(5x)/32#

When you have a fraction over a fraction, you can make use of the following as the first step in simplifying.

#(color(red)(a)/color(blue)(b))/(color(blue)(c)/color(red)(d)) = (color(red)(axxd))/(color(blue)(bxxc))#

#(5/(4x^2))/(8/x^3) = (5xxx^3)/(4x^2xx8)" "larr" simplify"#

=#(5x)/32#