How do you simplify #\frac { 4x ^ { 2} + x ^ { 3} y ^ { - 3} } { x \cdot y ^ { 0} }#?

1 Answer
Sep 29, 2017

See a solution process below:

Explanation:

First, use this rule of exponents to simplify the denominator:

#a^color(red)(0) = 1#

#(4x^2 + x^3y^-3)/(x * y^color(red)(0)) => (4x^2 + x^3y^-3)/(x * 1) => (4x^2 + x^3y^-3)/x#

Next, rewrite the expression as the sum of two fractions:

#(4x^2 + x^3y^-3)/x => (4x^2)/x + (x^3y^-3)/x#

Use these two rules of exponents to simplify the #x# terms:

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

#(4x^color(red)(2))/x^color(blue)(1) + (x^color(red)(3)y^-3)/x^color(blue)(1) => 4x^(color(red)(2)-color(blue)(1)) + x^(color(red)(3)-color(blue)(1))y^-3 => 4x^1 + x^2y^-3 =>#

#4x + x^2y^-3#

Now, use this rule of exponents to eliminate the negative exponent:

#x^color(red)(a) = 1/x^color(red)(-a)#

#4x + x^2y^color(red)(-3) => 4x + x^2/y^color(red)(- -3) => 4x + x^2/y^3#