Factoring and solving system of equations

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Need to show all work on how to solve this; it is the top one divided by the second line

Mar 20th, 2015

OK, first let's split this into:

[ 9x^5y^-6 ] multiplied by [ 1 over 5(x^4 y^2)^-2 ]     (dividing by x is the same as multiplying by 1/x)

Simplify the denominator (right hand terms). That part is the same as:

(x^4 y^2)^2 over 5 - when you take the negative of an exponent it moves it between the numerator and denominator. E.g. x^-2 is the same as 1/x^2; or x^2 is the same as 1/x^-2.

Now we multiply out the brackets. When we multiply together two terms x^n and x^m we add their exponents; when we raise one to another power like (x^n)^m, we multiply the exponents to get x^nm. Do that here to each exponent:

(x^4 y^2)^2 over 5

= x^8 times y^4 over 5

Now, we have: 

9 times x^5 times y^-6 times x^8 times y^4 over 5

We can combine the x terms:

x^5 times x^8 = x^(5+8) = x^13

And the y terms:

y^-6 times y^4 = y^(4-6) = y^-2

To give:

9 times x^13 times y^-2 over 5

Simplified to:

(9x^13) / (5y^2)

Mar 20th, 2015

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