How do you simplify #\frac { x ^ { 2} } { y ^ { - 3} } ( \frac { x ^ { 2} } { y } - \frac { 3z ^ { - 2} y } { x } )#?

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Answer 1 · 2017-10-21T01:31:29

#x^2/y^-3(x^2/y-(3z^-2y)/x)=x^4-(xy^4)/(9z^2)#

This may seem like a complicated problem due to all the variables and exponents. But, it's not.

Let's start with the original problem:

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

Let's first use the Distributive Property to simplify things a little bit:

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

We can then use some rules about multiplying together exponents with the same base as well as negative exponents to simplify a little bit more:

1) #a^b*a^c=a^(b+c)#

2) #a^b/c^b=(a-c)^b#

3) #a^-b=1/a^b#

#(x^2*x^2)/(y^-3*y^1)-(x^2/y^-3)(3z^-2y/x)#

Let's take care of the left side of the expression first:

#x^4/y^-2=x^4/(1/y^2)=x^4*y^2=x^4y^2#

Now let's work on the right side:

#(x^2/y^-3)(3z^-2y/x)=(x^2/(1/y^3))(1/((3z)^2)*y/x)=(x^2y^3)(y/(9xz^2))=(xy^3)(y/(9z^2))=(xy^4)/(9z^2)#

After all that, let's put them together, while not forgetting the #-# sign in the middle:

#x^4-(xy^4)/(9z^2)#

And, you're done!