How do you simplify #(6r^2p^3)/(4rp^4)# and find any non permissible values?

1 Answer
Jun 29, 2017

Answer:

See a solution process below:

Explanation:

First, rewrite the expression as:

#(6/4)(r^2/r)(p^3/p^4) =>#

#((3 xx 2)/(2 xx 2))(r^2/r)(p^3/p^4) =>#

#((3 xx color(red)(cancel(color(black)(2))))/(2 xx color(red)(cancel(color(black)(2)))))(r^2/r)(p^3/p^4) =>#

#3/2(r^2/r)(p^3/p^4)#

Next, use these rules of exponents to simplify the #r# terms:

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

#3/2(r^color(red)(2)/r^color(blue)(1))(p^3/p^4) => #

#3/2(r^(color(red)(2)-color(blue)(1)))(p^3/p^4) => #

#3/2(r^1)(p^3/p^4) => #

#3/2(r)(p^3/p^4) => #

#(3r)/2(p^3/p^4)#

Now, use these rules of exponents to simplify the #p# terms:

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

#(3r)/2(p^color(red)(3)/p^color(blue)(4)) =>#

#(3r)/2(1/p^(color(blue)(4)-color(red)(3))) =>#

#(3r)/2(1/p^1) =>#

#(3r)/2(1/p) =>#

#(3r)/(2p)#

From the original expression, because we cannot divide by #0#:

#4rp^4 != 0#

Or

#r != 0# and #p != 0#