Middleschool mather here.
When solving for p
in the proportion equation below our teacher tells us to handle the variable in the denominator first -- as if order of rationalization matters here:
$$\frac{8}9 = \frac{12}p$$
School Solution
First multiply both sides by $\frac{p}1$: $$\frac{8}9 * \frac{p}1 = 12$$
Then multiply both sides by $\frac{9}8$: $$p = \frac{12}1 * \frac{9}8$$
Simplify: $$p = \frac{27}2$$
I see how starting here simplifies things beautifully. But I'm interested why I can't start somewhere else. Of course when I don't rationalize the variatic denominator first I end up with a different ( and wrong? ) answer:
First divide both sides by $\frac{1}{12}$: $$\frac{8}9\div\frac{1}{12} = \frac{1}p$$
Rationalize division as multiplication on left side: $$\frac{8}9*\frac{12}{1} = \frac{1}p$$
Which becomes: $$\frac{96}{9} = \frac{1}p$$
Multiply both sides by $\frac{p}{1}$: $$\frac{96}{9} * \frac{p}1 = 1$$
Multiply both sides by $\frac{9}{96}$: $$p = \frac{9}{96}$$
Uh oh, I'm in trouble here!!
Two Questions:
Is there a rule or property that governs the order for solving this that I'm not able to see? Is my math teacher teaching us rules but not explaining them?
I'm bad at math. Is my math in the second solution just wrong? Can my attempt be solved first by dividing both sides by $\frac{1}{12}$?
Thanks for all your patience and help
Answer
You multiplied the LHS by $1/12$ and the RHS by $12$ -- those need to be the same for both sides. In your case, you would have gotten
$$\frac89 \cdot \frac{1}{12}=\frac{1}{p}\\
\frac{8}{108}=\frac{1}{p}\\
\frac{2}{27}=\frac{1}{p}$$
where the last line is simplification of $\frac{8}{108}$
No comments:
Post a Comment