How can you solve (6x^-2y^7)/(3x^-7y^-3)6x−2y73x−7y−3?

2 Answers
smendyka
Mar 28, 2018

See a solution process below:

Explanation:

First, rewrite the expression as:

6/3(x^-2/x^-7)(y^7/y^-3) =>63(x−2x−7)(y7y−3)⇒

2(x^-2/x^-7)(y^7/y^-3)2(x−2x−7)(y7y−3)

Now, use this rule of exponents to simplify the expression:

x^color(red)(a)/x^color(blue)(b) = x^(color(red)(a)-color(blue)(b))xaxb=xa−b

2(x^color(red)(-2)/x^color(blue)(-7))(y^color(red)(7)/y^color(blue)(-3)) =>2(x−2x−7)(y7y−3)⇒

2x^(color(red)(-2)-color(blue)(-7))y^(color(red)(7)-color(blue)(-3)) =>2x−2−−7y7−−3⇒

2x^(color(red)(-2)+color(blue)(7))y^(color(red)(7)+color(blue)(3)) =>2x−2+7y7+3⇒

2x^5y^102x5y10

Roopali
Mar 28, 2018

2 x^5 y^102x5y10

Explanation:

any thing with a negative power will switch its place up and down. so, (6 x^7y^7 y^3)/(3x^2)6x7y7y33x2
now as on the numerator we have y^7 and y^3 adds up = y^10 (a^b * a^c =a^(c+b))y7andy3addsup=y10(ab⋅ac=ac+b)
(6x^7y^10)/(3x^2)6x7y103x2
we can separatex^7 = x^2 and x^5x7=x2andx5
now solve it: (cancel(6)color(red)(2)cancel(x^2)x^5 y^10)/ (cancel(3)cancel(x^2))

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