Mark44 said:
Specific to the problem as stated, I'd argue that isolating ##x## , ie: "what is being solved, here" trumps improperizing a mixed fraction.
No algebra textbook worth its salt would give a worked example like what you wrote; i.e., of carrying a mixed number into a subsequent equation.
Is this your preference in table condiments ?
"Solve ##14=2\frac 1 3x##"
Okay... ##14=14## .
Solved.
Yes, the goal is to solve the equation, but keeping a mixed number in its original form as you did in your 2nd and 3rd equations is extra work and can be confusing to the reader.
For example, in one of your equations you have
$$\frac {14}{2\frac 1 3}$$
The denominator could easily be interpreted to mean 2 times ##\frac 1 3## rather than 2 plus ##\frac 1 3##.
My handwriting isn't so bad that ##2\frac1 3## looks like ##2*\frac 1 3##, nor ##2~~\frac 1 3##
if the problem were ##2\frac{1}{3}=2\frac{1}{3}x## would you feel the need to write ##\frac{7}{3}=\frac{7}{3}x## as an intermediary line ?
First off, I doubt that any algebra textbook would even have a problem like this. Second, since the number on the left side is the same as the coefficient of x, I would immediately write ##x = 1##. I wouldn't bother writing an intermediate equation with the two sides switched.
Everybody that reads this thread - including the original poster, by now - is capable of doing ##14=2\frac 1 3x## in their head and arriving at a solution for ##x##, without breaking too many fingers. What's your point ?
Do you feel the need to convert ##\pi## to ##3.14159...## as a first step in solving an equation ?
No, I wouldn't do this, not to mention that rewriting ##\pi## as a very rough approximation has nothing to do with what we're discussing here, namely, converting mixed numbers to improper fractions.
I know how to do that already, thanks. What I thought we were discussing was the priority of same.
My point for the two examples is that (1) there are occasions where you don't feel the need to do an improperization, and (2) an example of not converting strictly to numbers, right off the bat.
Look, here's my take... (polished up from the original post)
##\text{Solve}~(for~x)~:~~~14=2\frac 1 3 x##
$$\begin{align}
\text{using}~symmetric~property~of~equality~~~~~~~~2\frac 1 3 x&=14 \nonumber\\
\text{followed by}~arithmetic~reductions~~~~~~~~~~~~~x&=\frac{14}{2\frac{1}{3}}=\frac{14}{\frac{7}{3}}=\frac{42}{7}=6 \nonumber \\
\end{align}$$
##\text{Solution :}~~~~x=6##
It strikes me as being quite clear. The only question (in my mind) would be maybe doing the reduction in a vertical convention.
