Just a simple order of operations question

[Sudharaka]

SplashDamage's question from Math Help Forum,

if I have 4 + ((a - 1) / 12) = 24

Why can I not move the 12 over to the other side first? Making it 4 + a - 1 = 24 * 12

Howecome I have to move the 4 over first? I never really knew why, possibly there is an order to moving things over the = sign?

Thanks for any help.

Hi SplashDamage,

There is no such thing as "moving" a term to the other side. A equal sign means that both quantities same quantitative amount. Therefore you can add, subtract, divide or multiply both sides of the equality by the same number and equality will still hold.

So you cannot "move" 12 over to the other sides, but you can multiply the two sides by 12. Similarly you cannot move 4 over to the other side, but you can subtract 4 from both sides.

Suppose if you want to find \(a\) in the equation, \(4 + \dfrac{a - 1}{12} = 24\).

First we can multiply both sides of the equation by 12,

\[12\left(4+\frac{a - 1}{12}\right)=12\times 24\]

\[\Rightarrow (12\times 4)+\left(12\times \frac{a-1}{12}\right)=288\]

\[\Rightarrow 48+a-1=288\]

\[\Rightarrow 47+a=288\]

Now we shall subtract 47 from both sides,

\[47+a-47=288-47\]

\[a=241\]

Another method is to subtract 4 first, starting from the original equation.

\[4 + \frac{a - 1}{12} = 24\]

\[\Rightarrow 4 + \frac{a - 1}{12}-4=24-4\]

\[\Rightarrow \frac{a - 1}{12}=20\]

Now we shall multiply both sides by 12,

\[\Rightarrow 12\times \dfrac{a - 1}{12}=20\times 12\]

\[\Rightarrow a - 1=240\]

Finally add 1 to both sides,

\[a=241\]

 

[Jameson]

In the US the phrase "move to the other side" is used a lot in classrooms and it's a potential problem because students must understand what is really going on to use this "shortcut". Normally when given [math]x+7=10[/math] and asked to solve for x many teachers use the method "move 7 to the other side and change the sign" which is another way of saying "subtract 7 from both sides of the equation keeping the equality a true statement" but students can quickly forget this. The way I remember it being taught was by first doing everything the "long way" then after mastering that we were taught the "switch signs by moving across the equal sign" method.

If this "trick" is incorrectly applied to multiplication for example then the student will have lots of confusion unfortunately.

 

[checkitagain]

SplashDamage's question from Math Help Forum,
ld.

Suppose if you want to find \(a\) in the equation, \(4 + \dfrac{a - 1}{12} = 24\).

First we can multiply both sides of the equation by 12,

\[12\left(4+\frac{a - 1}{12}\right)=12(24)\]

\[\Rightarrow 12(4) + 12 \bigg( \dfrac{a-1}{12} \bigg) = 288\]

\[\Rightarrow 48+a-1=288\]

\[\Rightarrow 47+a=288\]

Now we shall subtract 47 from both sides,

\[47+a-47=288-47\]

\[a=241\]Another method is to subtract 4 first, starting from the original equation.

\[4 + \frac{a - 1}{12} = 24\]

\[\Rightarrow 4 + \frac{a - 1}{12}-4=24-4\]

\[\Rightarrow \frac{a - 1}{12}=20\]

Now we shall multiply both sides by 12,

\[\Rightarrow 12\bigg( \dfrac{a - 1}{12}\bigg) = 12(20)\]

\[\Rightarrow a - 1 = 240\]

Finally add 1 to both sides,

\[a=241\]

1) Please do not use the "times" symbol in multiplication with algebraic expressions or equations.
Here, at the outset, you used a set of parentheses for multiplication. Continue with that use.

2) Please be consistent with the symbol used and the order it was used.
In the top line, you set the standard by having the parentheses placed
around the second quantity being multiplied. So, continue using the same symbols
for the operation and placing them in the same relative places for each product.

3) Please do not completely solve (read: do the work) for the student here, as per
the MHB rules.

 

[skeeter]

Please do not completely solve (read: do the work) for the student here, as per
the MHB rules.]

the OP (student) has no idea that this particular solution even exists ... has the urge for a "math-help fix" become so desperate that one must cross-post from another site?

 

[Jameson]

the OP (student) has no idea that this particular solution even exists ... has the urge for a "math-help fix" become so desperate that one must cross-post from another site?

This issue has been brought up before by dwsmith. You can read the past discussion and add anything http://www.mathhelpboards.com/showthread.php?936-problems-from-other-sites if you're interested. I believe what Sudharaka is doing is trying to help and promote MHB at the same time. It's unfortunate that we aren't more well-known but that will come with time and I expect the new school year to bring in many new members. Until then we're working on site improvements and in particular a big one which I hope will impress everyone :). This thread seems to be off topic and I'm not helping that at all so I'll close it.

As always you can PM me or make a thread in the feedback forums if you have any issues, thoughts, comments, questions or criticisms.