Is there a better way to solve this linear equation with fractions?

[hackedagainanda]

Homework Statement

Solve:

$14 = 2\frac 1 {3}x$

Relevant Equations

None.

My method is converting 14 into $\frac {42} {3}, = \frac 7 3x$

Then $$ \frac {126} { 21}, = 6$$

 

[.Scott]

Multiplying both sides by 3: 42 = 7x
Dividing boths sides by 7: 6 = x

 

[hackedagainanda]

Thanks! I don't know why I didn't see that. I feel dumb.

 

[symbolipoint]

Simpler still using just multiplicative inverse and then simplify if necessary.

$$14 = 2\frac 1 {3}x$$

14=(7/3)x

(14)(3/7)=(7/3)x(3/7)

14(3/7)=x

6=x

 

[archaic]

Thanks! I don't know why I didn't see that. I feel dumb.

Why? You did exactly the same thing but didn't cancel the $3$ manually.

 

[zeffur7]

14=2+1/3x
14=7/3x
14*3/7=x
14/7*3=x
2*3=x
6=x

 

[TimeDoctor]

Another way is to do 14 = [( 3*2+1) /3] X
14= 7/3 X
14*3 = 7X
X= 42/7 = 6

 

[hmmm27]

$$ \begin{align}
14&=2\frac{1}{3}x \nonumber\\
2\frac{1}{3}x&=14 \nonumber\\
x&=\frac{14}{2\frac{1}{3}}=\frac{14}{\frac{7}{3}}=\frac{42}{7}=6 \nonumber
\end{align}$$

 

[Mark44]

$$ \begin{align}
14&=2\frac{1}{3}x \nonumber\\
2\frac{1}{3}x&=14 \nonumber\\
x&=\frac{14}{2\frac{1}{3}}=\frac{14}{\frac{7}{3}}=\frac{42}{7}=6 \nonumber
\end{align}$$

Algebra textbooks almost never present equations with mixed numbers such as $2~\frac 1 3$ that is shown here. Such a fraction should always be converted right away to an improper fraction, one in which the numerator is greater than the denominator.
The second line above should be $14 = \frac 7 3 x$, and the mixed number should not be carried along in subsequent equations.

 

[hmmm27]

Algebra textbooks almost never present equations with mixed numbers such as $2~\frac 1 3$ that is shown here. Such a fraction should always be converted right away to an improper fraction, one in which the numerator is greater than the denominator.
The second line above should be $14 = \frac 7 3 x$, and the mixed number should not be carried along in subsequent equations.

Specific to the problem as stated, I'd argue that isolating $x$, ie: "what is being solved, here" trumps improperizing a mixed fraction.

As far as the importance of improperizing a mixed fraction is concerned... 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 ? Do you feel the need to convert $\pi$ to $3.14159...$ as a first step in solving an equation ? (Bear in mind that both arguments aren't terribly related to the problem at hand... just trying to justify my interpretation)

 

[symbolipoint]

Algebra textbooks almost never present equations with mixed numbers such as $2~\frac 1 3$ that is shown here. Such a fraction should always be converted right away to an improper fraction, one in which the numerator is greater than the denominator.
The second line above should be $14 = \frac 7 3 x$, and the mixed number should not be carried along in subsequent equations.

That method is more a stylistic preference of choosing steps, but one can understand the clumsiness of so tardily converting the mixed number, so that converting it to improper fraction earlier is a neater way.

 

[hmmm27]

That method is more a stylistic preference of choosing steps, but one can understand the clumsiness of so tardily converting the mixed number, so that converting it to improper fraction earlier is a neater way.

Well, it started out as an objection to $6=x$ ; that's a reduction, not a solution.

 

[Mark44]

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. 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 $\frac 1 3$.

As far as the importance of improperizing a mixed fraction is concerned... 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.

Do you feel the need to convert $\pi$ to $3.14159...$ as a first step in solving an equation ? (Bear in mind that both arguments aren't terribly related to the problem at hand... just trying to justify my interpretation)

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.

 

[hmmm27]

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.

 

[Mark44]

Is this your preference in table condiments ?
"Solve $14=2\frac 1 3x$"
Okay... $14=14$ .
Solved.

No, you didn't solve the equation. What you did was to write an equation that is identically true.
Solving the equation would result in an equation with the variable isolated on one side, with its value on the other side.

My handwriting isn't so bad that $2\frac1 3$ looks like $2*\frac 1 3$, nor $2~~\frac 1 3$

This has nothing to do with your handwriting, good, bad, or indifferent.
Placing two expressions in juxtaposition implies that the operation is multiplication; e.g., $2x$ means 2 times x.
My point is that $2 \frac 1 3$ could be interpreted to mean 2 * 1/3 rather than 2 + 1/3. This is the reason that you almost never see mixed numbers in algebra textbooks.

Take it from me, someone who taught math for 20+ years, some at high school level, but most at college level, lots of people have difficulties with fractions. To keep mixed numbers in an equation is anything but "quite clear."

Why divide by $2\frac 1 3$, and drag this along for two additional steps? The OP was smart enough to convert the mixed number right away to an improper fraction.

A much more straightforward approach would be the following:
$2\frac 1 3 x = 14$
$\frac 7 3 x = 14$
$x = \frac 3 7 \cdot 14 = 6$ Multiply both sides by the multiplicative inverse of 7/3, and then simplify arithmetically.

 

[hmmm27]

No, you didn't solve the equation. What you did was to write an equation that is identically true.

No, what I didn't do was "solve for $x$", which phrase was omitted from the problem statement.

Solving the equation would result in an equation with the variable isolated on one side, with its value on the other side.

Yes (as far as I'm aware, convention requires variable on the left, equation on the right)

This has nothing to do with your handwriting, good, bad, or indifferent.
Placing two expressions in juxtaposition implies that the operation is multiplication; e.g., $2x$ means 2 times x.
My point is that $2 \frac 1 3$ could be interpreted to mean 2 * 1/3 rather than 2 + 1/3. This is the reason that you almost never see mixed numbers in algebra textbooks.

Take it from me, someone who taught math for 20+ years, some at high school level, but most at college level, lots of people have difficulties with fractions. To keep mixed numbers in an equation is anything but "quite clear."

Fair enough ; I haven't got the in situ experience that you have : being a part of the problem statement seems to validate usage : though, notwithstanding your objections, I don't see any problem with bouncing it around monolithically.

A few years teaching/applying a very specific and rigorous CS methodology/standard a few decades ago scarred me for life.

Why divide by $2\frac 1 3$, and drag this along for two additional steps? The OP was smart enough to convert the mixed number right away to an improper fraction.

A much more straightforward approach would be the following:
$2\frac 1 3 x = 14$
$\frac 7 3 x = 14$
$x = \frac 3 7 \cdot 14 = 6$ Multiply both sides by the multiplicative inverse of 7/3, and then simplify arithmetically.

To show every step, your last line should read $x=\frac 3 7\cdot14=\frac{42}{7}=6$ and now we have the same number of algebraic steps.

I can sortof see the point of :

Solve : $14=2\frac 1 3x$
$$ \begin{align}
14&=\frac 7 3x \nonumber\\
\frac {42}{7}&=x \nonumber\\
6&=x \nonumber\\
\end{align}
$$
Solution : $x=6$

but then I wouldn't have gotten to practice my LaTeX as much.

 

[Mark44]

No, what I didn't do was "solve for x", which phrase was omitted from the problem statement.

A reasonable person would understand that "Solve:" in the first line, together with an equation with a single variable, would mean "Solve for x:".

To show every step, your last line should read $x=\frac 3 7\cdot14=\frac{42}{7}=6$ and now we have the same number of algebraic steps.

But unless the teacher is very pedantic, showing every arithmetic step is not usually required in problems of this nature. A clever student would realize that $\frac 3 7 \cdot 14$ was the same as (i.e., equal to) $3 \cdot \frac {14} 7$ and mentally do the arithmetic to get 6.

but then I wouldn't have gotten to practice my LaTeX as much.

Well, then, that's certainly a plus!

 

[hmmm27]

A reasonable person would understand that "Solve:" in the first line, together with an equation with a single variable, would mean "Solve for x:".

A reasonable person would also assume that an expression - in this case a mixed fraction - used in a bare numerical problem statement (ie: no backstory like "the carpenter measured the beam as $3\frac 1 2$ inches", which could show context) would be reasonable fodder to use in the solution in toto (in this case until a computation is required).

But unless the teacher is very pedantic, showing every arithmetic step is not usually required in problems of this nature. A clever student would realize that $\frac 3 7 \cdot 14$ was the same as (i.e., equal to) $3 \cdot \frac {14} 7$ and mentally do the arithmetic to get 6.

Yes ; like I stated earlier, most people could do the whole thing in their head (unless they make it too complicated like the OP did, originally)

Well, then, that's certainly a plus!

Being p*ssed at having to decipher the half-assed ascii chickenscratches that most of the posts proferred was my reason for joining the thread in the first place. LaTeX is not that difficult a tool to use once you get the hang of it (which I'm still working on).

 

[symbolipoint]

@hackedagainanda was most likely satisfied at post #3.

 

[hmmm27]

Well he just mentally zigged when he should have zagged - not a big issue. Probably frantically looking for the Unsubscribe button.