How do you simplify #(28x^-2)/(7y^-3)# and write it using only positive exponents?

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Answer 1 · 2017-08-04T04:06:11

See a solution process below:

First, simplify the constants:

#((7 xx 4)x^-2)/(7y^-3) =>#

#((color(red)(cancel(color(black)(7))) xx 4)x^-2)/(color(red)(cancel(color(black)(7)))y^-3) =>#

#(4x^-2)/y^-3#

Next, use this rule of exponents to eliminate the negative exponent for #y#:

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

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

#4x^-2y^color(red)(3)#

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

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

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

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

#(4y^3)/x^color(red)(2)#