View Full Version : Simplying an x and y fraction with a negative exponent

GhostRider2552

06-22-2009, 01:08 AM

Hey guys, I really need your help here, I don't know what section of grade 10 math this is but here's the fraction:

((9x^3 y^-1)/(4x^-2 y^5))^-2

In words: Nine x cubed, y to the negative 1, divided by four x to the negative two, y to the five, all of that to the negative two.

If you could simplify this and show me the steps that would really help, thanks guys!

All it says is to "Simplify" it.

red and white kop!

06-22-2009, 03:20 AM

develop the denominator by putting all terms to power -2. then you can make things easier by thanging from denominator to numerator or vice-versa all those negative exponent terms and use the exponent division rules to cut out the redundant values. then you get 144y^9 / x

Subhotosh Khan

06-22-2009, 07:35 AM

Hey guys, I really need your help here, I don't know what section of grade 10 math this is but here's the fraction:

((9x^3 y^-1)/(4x^-2 y^5))^-2

In words: Nine x cubed, y to the negative 1, divided by four x to the negative two, y to the five, all of that to the negative two.

If you could simplify this and show me the steps that would really help, thanks guys!

All it says is to "Simplify" it.

[\frac {9x^3y^{-1}}{4x^{-2}y^5}]^{-2}

= \, [\frac {4x^{-2}y^5}{9x^3y^{-1}}]^{2}

= \, \frac {16}{81}[x^{(-2 - 3)}y^{(5 + 1)}]^{2}

Now continue....

soroban

06-22-2009, 08:13 AM

Hello, GhostRider2552!

Just follow your basic rules . . . carefully.

. . \begin{array}{cccc}\text{Mult'n Rule:} & a^m\cdot a^n &=& a^{m+n} \\ \\[-3mm] \text{Division Rule:} & \dfrac{a^m}{a^n} &=& a^{m-n} \\ \\[-2mm] \text{Power Rule:} & (a^m)^n &=& a^{mn} \end{array} \qquad\begin{array}{cccc}\text{Product Rule:} & (ab)^n &=& a^nb^n \\ \\[-2mm] \text{Quotient Rule:} & \left(\dfrac{a}{b}\right)^n &=& \dfrac{a^n}{b^n} \end{array}

\text{Simplify: }\;\left(\frac{9x^3 y^{-1}}{4x^{-2} y^5}\right)^{-2}

\begin{array}{cccc}\text{We have:}& \dfrac{(9x^3y^{-1})^{-2}} {(4x^{-2}y^5)^{-2}} & \text{Quotient Rule} \\ \\ = & \dfrac{9^{-2}(x^3)^{-2}(y^{-1})^{-2}} {4^{-2}(x^{-2})^{-2}(y^5)^{-2}} & \text{Product Rule} \\ \\ = & \dfrac{4^2x^{-6}y^2}{9^2x^4y^{-10}} & \text{Power Rule} \\ \\ = & \dfrac{16y^{12}}{81x^{10}} & \text{Division Rule} \end{array}

Denis

06-22-2009, 08:37 AM

[(9x^3 y^-1) / (4x^-2 y^5)]^-2

Just another way; start by eliminating the "outside power -2":

(9^-2 x^-6 y^2) / (4^-2 x^4 y^-10)

Move the negative powers:

(4^2 y^12) / (9^2 x^10)

(16 y^12) / (81 x^10)

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