I am to simplify the expression $\frac{3x^2y^{-3}z}{12x^{-1}yz^3}$
I got $\frac{3x^2z}{12y^2z^3}$
Submitting my answer to the quiz and getting it wrong, I can see the correct answer is actually $\frac{x^3}{4y^4z^2}$
I was unsure how to get started or approach, but here's how I arrived at where I did:
$\frac{3x^2y^{-3}z}{12x^{-1}yz^3}$
(Get rid of the negative exponents by using power rule $x^{-n}$ = $\frac{1}{x^n}$)
$\frac{\frac{3x^2z}{y^3}}{\frac{12yz^3}{x}}$
(complex fraction, multiple with swapped numerator and denominator)
$\frac{3x^2z}{y^3}*\frac{12yz^3}{x}$
$\frac{3x^3z}{12y^2z^3}$
Where did I go wrong and how can I arrive at $\frac{x^3}{4y^4z^2}$ ? Seeking granular, spoon fed steps if possible?
Best Answer
Looking at the coefficients, we see that both $3$ and $12$ are divisible by $3$. So dividing both the top and bottom by $3$ gives us $1$ and $4$, so we have: $$ \frac{x^2(\color{red}{y^{-3}})z}{4(\color{blue}{x^{-1}})yz^3} $$
Next, let's get rid of negative exponents. Basically, we can make them positive and flip them to the other side: $$ \frac{(\color{blue}{x^{1}})x^2z}{4(\color{red}{y^{3}})yz^3} $$ Combine the exponents for like bases by adding together the exponents: $$ \frac{x^3z}{4y^4z^3} $$ Finally, for $z$, divide both the top and the bottom by $z$ to obtain: $$ \frac{x^3}{4y^4z^2} $$