\[({3x \over y})({3x^2y^{-2} \over 12y^3})\]\[x^ax^b=x^{a+b}\]\[{x^a \over x^b}=x^{a-b}\]Memorize these
OpenStudy (anonymous):
okay
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OpenStudy (anonymous):
We'll do one part at a time \[3x \times 3x^2=9x^3\]any questions on this part
OpenStudy (anonymous):
no you can continue!
OpenStudy (anonymous):
I want you to do the multiplication of the denominator
OpenStudy (anonymous):
do you find you LCD?
OpenStudy (anonymous):
Multiplication of fractions you just simply multiply across
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OpenStudy (anonymous):
\[\frac{ a }{ b } \times \frac{ c }{ d }=\frac{ ac }{ bd }\]
OpenStudy (anonymous):
oh okay I thought you needed to find your common denominator first.
OpenStudy (anonymous):
So what is this\[y \times 12y^3\]
OpenStudy (anonymous):
12y^4
OpenStudy (anonymous):
yup now we have to simplify the y's in the numerator and the denominator.\[\frac{ y^{-2} }{ y^4 }=y^{-2-4}=y^{-6}=\frac{ 1 }{ y^6 }\]
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OpenStudy (anonymous):
Okay I'm understanding but we also have our 9/12= 3/4
OpenStudy (anonymous):
Now we are left with \[\frac{ 3x }{ y }*\frac{ 3x^2 }{ 12y^5 }=\frac{ 9x^3 }{12y^6 }\]Just simplify the constants and we get the answer the other guy posted
OpenStudy (anonymous):
oh yeah I forgot about our x^3 \[3x ^{3}/4y ^{6}\]