## Need help with understanding equations

For the discussion of math. Duh.

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Kenshi
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Joined: Fri Jul 11, 2014 6:25 pm UTC

### Need help with understanding equations

So... i am really embarrassed to admit this, but i have never been able to understand even simple equations with an unknown, and i feel that its time to understand it.

My problem is that in an equation such as: 6/3+2+5*x+4 = 38, for example, i don't have any idea what number to "move" first, second, third and so on. should i subtract 4 first or multiply/divide with a number first in the example, i simply don't know.

Every YouTube tutorial i have seen simply says: first we subtract this number from both sides, and secondly we add this number to both sides, then we multiply by this on both sides. but they never explain the order, what kind of numbers to do first or why they do that particular numbers first, second, third and so on.

i know that if you multiply, divide, add, subtract from one side you have to do that to the other side as well, and i of course also know the order of pemdas.

So dear xkcd'ians help me understand. You may of course use any example you want to, you don't need to use the example that i provided.

BTW: this is not homework.

if you have any questions or need me to clarify i will gladly do so. English is not my native language.

sgfw
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### Re: Need help with understanding equations

Pretty much reverse PEMDAS. But that's not the way I think about it. I think about it more like this:

Numbers that are being added or subtracted are pretty far away from each other, so they aren't really that "attached", and can be moved first, unless they are being held in by parenthesis. After you moved all the stuff that's added or subtracted to the other side, you go on to the next step: multiplication/division. These numbers are closer to each other than numbers being added (often separated by only a line of dot), and so they are more attached to each other than addition/subtraction. Therefore, you can only get to them after you got rid of all of the addition and subtraction around them. Then exponents are the most "attached" numbers, so they go last- except for parenthesis. You can only get rid of parenthesis after everything around them is gone.

This may have just made things more confusing for you, but it's how my brain works.

Qaanol
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### Re: Need help with understanding equations

The top-level concept here is that you want to isolate the unknown. That is, since your goal is to find out what x equals, therefore you want to end up with something of the form “x = …”. So you are trying to get x by itself on one side.

With that end-goal in mind, can you now see how to go about answering your own question?

One way to think about it is “peeling back layers” around the unknown. Another way is more pragmatic, “If it doesn’t have an x in it, get it to the other side!”

In your example, you could start by moving all the terms without x to the right, or you could start by dividing through by 5, or you could start by dividing 6/3, or you could start by adding together the terms without an x on the left. It actually doesn’t matter which of those you do first. Of course, if you start by dividing by 5 then you’ll end up with a whole bunch of fractions, which makes for a more tedious calculation, so from a viewpoint of simplicity, it might be easier to hold off on the division until later. But you don’t have to.

About the only time where it really matters what order you do things, is when x part of a term raised to a power. Since exponentiation does not distribute over other operations, you really need to isolate the power term first before you take its root.
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skullturf
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### Re: Need help with understanding equations

sgfw wrote:Numbers that are being added or subtracted are pretty far away from each other, so they aren't really that "attached", and can be moved first, unless they are being held in by parenthesis. After you moved all the stuff that's added or subtracted to the other side, you go on to the next step: multiplication/division.

That's very similar to how my brain works, and very close to what I was going to say.

Of course, that doesn't necessarily mean it's the best explanation for everyone. Different things will "click" with different people. But in case it helps, I'm going to expand on the above a little.

When I see something like

6/3+2+5*x+4

I think of the multiplications or divisions as stronger "bonds" holding things together tightly, so I envision the above as a sum of four "things"

Code: Select all

`6/3     +     2     +     5*x     +     4`

so it's "really" just a sum of four terms. And with sums, order doesn't matter, so it's up to you which of the four terms you want to "move" first. (Bearing in mind that the ultimate goal is to get x by itself.)

ConMan
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### Re: Need help with understanding equations

Let me chip in with another analogy (because you can't have too many of those!) An equation is like a set of balanced scales. There are weights on both sides, but they match each other so that the scales don't tip. And that also means that the only way to keep them in balance is to never change the weight on one side without making an equivalent change to the weight on the other (or if we do something to the weights on one side that doesn't change how much they weigh, only how they're represented).

For example, let's start with a system that has no unknowns, but that shows you a few of the things you can do: 1 + 1 = 2. So on the right side of the scales, there's a single weight labeled "2", and on the left side there are two weights each labeled "1". So what kind of things can we do to these weights without upsetting the balance? I'll put the answers in spoilers, just to make sure you're following along:

Can we add another "1" box to each side?
Spoiler:
Yes we can - it gives us 1 + 1 + 1 = 2 + 1, and because we've done the same on both sides the scales will stay balanced.

Can we replace one of the "1" boxes with two boxes each labeled "1/2"?
Spoiler:
Yes we can - two boxes with weights of 1/2 each will weigh the same as a single box with weight of 1, so doing this doesn't change the weight on that side.

Can we take a weight off each side?
Spoiler:
No we can't - the weights are different, so we'd be doing something different to the weights on each side. On the left we'd be removing a "1", and on the right we'd remove the "2", leaving us with 1 = (0) which is clearly not going to balance.

Can we replace every weight with a balloon that pulls "up" as much as the original weight pulled "down"?
Spoiler:
Yes we can, as long as we do it to every weight. This is equivalent to multiplying everything by -1, giving (-1) + (-1) = (-2).

Can we remove half of each weight?
Spoiler:
Yes we can - we'd get 1/2 + 1/2 = 1.

Are we good up to here? My next post will start looking at how you actually solve equations with this.
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Cleverbeans
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### Re: Need help with understanding equations

Keep everything integers as long as you can. Group constants together first. At the end, when you get x=? put it back into the original equation to check your work. If it's wrong look for your mistake starting at the bottom and working back.
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ConMan
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### Re: Need help with understanding equations

Ok, so hopefully you're feeling comfortable playing around with the weights on the scales. I'm also going to add one little trick that I left off - multiplication is just fancy addition. So if you see 2 x 1, or even 2(1), that's just a little stack of two "1" weights sitting on top of each other. And if you're cutting all the weights in half, you can instead just take half of that stack.

So now, we're going to introduce a box labelled "x". How much does it weigh? We don't know. All we know is that when we put it on the scales with a bunch of other boxes, we can make it balance. So, for example, x+3 = 10 means that if we put the "x" box* on the left side of the scales with a "3" box, and a "10" box on the right, then everything balances. How, then, do we measure the weight of the "x"?

Well, we can take the "3" off the left, but to keep things balanced we also have to take a weight of 3 off the right. We can replace the "10" with a "3" and a "7" (because we know 10=3+7), and take that "3" away, leaving us with "x" on the left and "7" on the right. Or in other words, x=7.

In general, when we're trying to solve an equation like 5x-9=3x+1, our broad aims are to (1) get all the unknown stuff on one side, and (2) get all the known stuff on the other side, then (3) adjust so that we have a nice simple x=(something) form. So let's take a look at that equation.

What, in the scales analogy, does "5x-9=3x+1" mean?
Spoiler:
On the left side of our scale, there are five boxes labeled "x" stacked on top of one another, and a balloon labeled "-9" pulling up. On the right side, there are three more "x" boxes stacked up, and one "1" box. All the "x" boxes weigh the same, and the scales are in balance.

Can we remove an "x" box from each side and keep the scales balanced?
Spoiler:
Yes we can - even though we don't know how much "x" weighs, we do know that all of the "x" boxes are the same weight, so removing one from each side doesn't upset the balance.

If a balloon and a weight on the same side of the scale have the same (but opposite sign) number on them, can we remove them together?
Spoiler:
Yes we can - a "1" weight and a "-1" balloon combine to make no net weight, so if we remove both of them we aren't changing the balance. Equivalently, we're saying that 1+(-1)=(0).

How can we move all the "x" stuff to the left side of the balance?
Spoiler:
Take away 3 "x" boxes from each side. Then there are two left on the left side, and none on the right. Algebraically, we start from 5x-9=3x+1, subtract 3x from both sides, and get 2x-9=1.

From there, how can we move all the non-"x" stuff to the right side of the balance?
Spoiler:
Put a "9" weight on both sides. Then we have 2x-9+9=1+9. The "9" weight and the "-9" balloon on the left are cancelling each other out, so we can take both of them away, leaving us with 2x+(0)=1+9, or just 2x=1+9. And the "1" and "9" weights can be combined into a single "10" weight, so 2x=10.

Finally, how can we simplify down to a single "x" box on the left?
Spoiler:
At this point, we have 2x=10, so two "x" boxes on the left and a single "10" on the right. Clearly we can't just subtract an x from (or put a "-x" balloon on) both sides, because then we'd be bringing "x" back onto the right. Instead, let's halve the weight on both sides - on the left, that means going from 2x to x (so the same effect as subtracting an x), and on the right we cut our "10" box in half, leaving a weight of 5. So we're left with x=5, meaning that the "x" box weighs the same as a "5" box.