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For the following equation, I know the correct answer is $9$: $$ x / 3 + 2 = 5 $$ You subtract $2$ from each side, and the multiply each side by $3$...

But why do you subtract the $2$ first?

Doesn't the order of operations say I should do division before addition?

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4  
It does, as far as the division of $x$ is concerned. However, when solving for $x$ you undo all operations around it, which means that you enact them in reverse order. –  abiessu Sep 5 at 1:56
11  
The order of operations isn't referring to the order you perform operations, it's there to tell you what an expression written on paper actually means. So the expression that you've written means that $x$ has first been divided by $3$, and then $2$ has been added to it, and the result is $5$. You are free to perform further operations in any order you like, as long as you perform the same operation to both sides! –  James Machin Sep 5 at 1:58
    
Is there some reason why my answer was down-voted. In particular, @user138021, do you have some objection to it? –  Michael Hardy Sep 5 at 18:23
    
@MichaelHardy I did not down-vote your answer. Maybe it was down-voted because many of the other answers say that the order of operations does not matter as long as the equation remains balanced, while your answer implies that there is a specific order? But, again, I did not down-vote and obviously still have a lot to learn, so I am only speculating. –  user138021 Sep 6 at 19:51

5 Answers 5

up vote 22 down vote accepted

Any order of operations is fine, as long as you are consistent in applying the same operation to both sides of the equation in every step. If you want to multiply by $3$ first:

$$ \frac{x}{3} + 2 = 5 $$

Multiply both sides by $3$:

$$ x + 6 = 15 $$

Subtract $6$ from both sides:

$$ x = 15 - 6 $$

Just work out the solution:

$$ x = 9 $$

Look at equations as a scale balance: anything you do will keep the scale in balance as long as you do it to both sides. The trick is to come up with a sequence of operations that eventually leaves the unknown on one side, and only knowns (in this case, numbers) on the other.

As a bad example, when we have $x+6=15$ we could, if we wanted to, decide to add $100$ to both sides. The equation would still be perfectly valid, but we wouldn't be any closer to finding $x$!

Finding a good sequence of operations will be challenging at first, but after some practice, it will become second nature.

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PS: if you were taught to solve equations in terms of "a $+$ when moved across the $=$ becomes a $-$", etc., I would suggest that you forget all of that and think instead in terms of a scale balance. This was certainly a breakthrough moment for me.

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No, you're getting confused. The "order of operations" doesn't matter here, it's only for solving an expression (i.e. $4(5)-\left(\frac52\right)^3$). Here you just do something to two sides that are equal. You can choose whatever you want to do to both sides, but you have to make sure that you are doing it to both sides. You don't have to subtract 2 first, you can also multiply by 3 first:

$$3\left(\frac{x}3+2\right)=3(5) \\ x+6=15 \\ x=9$$

And you will get the same answer as subtracting first:

$$\left(\frac{x}3+2\right)-2=(5)-2\\\frac{x}3=3\\x=9$$

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1  
Well, you can do anything you want as long as it is reversible! (i.e. you cannot multiply by $0$, or square, else you're gonna get extra solutions...) –  Pedro Tamaroff Sep 5 at 2:34
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@PedroTamaroff you can square as long as both members have the same sign. –  Silviu Burcea Sep 5 at 12:06

Order of operations more so tells you how you should read the problem unambiguously. Here we have "$x$ is divided by $3$, THEN added with $2$ to achieve $5$." To figure out what $x$ was originally, we sort of backtrack.

What did the equation look like before we added $2$?

$$x/3=5-2$$

Then what was it before we divided by $3$?

$$x=(5-2)\times3$$

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As regards correctness of the solution, the order is irrelevant as long as the operations are legal. $$\frac x3+2=5.$$ First times $3$: $$x+3\cdot2=3\cdot5,\\x+6=15,$$ followed by minus $6$: $$x=15-6,\\x=9.$$ Or first minus $2$: $$\frac x3=5-2,\\\frac x3=3,$$ followed by times $3$: $$x=3\cdot3,\\x=9.$$ It's a matter of taste, except that in the first case you perform two multiplies and one subtract, vs. one subtract and one multiply in the second.

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What gets done last gets undone first. Since you do division first, you undo division last.

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2  
Despite its silly sounding. this is a perfectly valid answer. Sometimes to undo math, you just have to do all the parts backwards –  Asimov Sep 5 at 17:32
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Is there some reason for the down-votes here? Down-votes should be explained. –  Michael Hardy Sep 5 at 18:22
    
They should be. Maybe they just thought that it wasn't explained enough. –  Asimov Sep 5 at 20:55

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