Just a simple order of operations question

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May 30th 2012, 05:22 AM
SplashDamage

Just a simple order of operations question

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.
May 30th 2012, 05:42 AM
skeeter

Re: Just a simple order of operations question

Quote:

Originally Posted by SplashDamage

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.

what you are really doing by "moving the 12 over" is multiplying by 12 ... you need to multiply every term by 12 to keep the equation balanced.

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

$12 \cdot 4 + (a-1) = 12 \cdot 24$

now you can finish it
May 30th 2012, 05:45 AM
emakarov

Re: Just a simple order of operations question

When you move things to the other side of an equation, you in fact apply some function to both sides of the equation. It is important that the same function is applied to both sides. The function is chosen in order to make one side of the equation simpler.

Suppose the equation is L1(L2(x)) = R(x) and suppose we know that F(L1(x)) = x for all x, i.e., F and L1 are inverse functions. Then we can apply F to both sides of the equation to get F(L1(L2(x))) = F(R(x)). Together with F(L1(x)) = x this gives L2(x) = F(R(x)), i.e., we simplified the left-hand side.

In your example 4 + ((a - 1) / 12) = 24, we have L1(x) = 4 + x, L2(x) = (a - 1) / 12 and R(x) = 24. This way, L1(L2(x)) = 4 + ((a - 1) / 12). The inverse function to L1(x) = 4 + x is F(x) = -4 + x. Indeed, F(L1(x)) = -4 + (4 + x) = x. Therefore, applying F(x) to both sides of the equation, we have -4 + (4 + ((a - 1) / 12)) = -4 + 24, or (a - 1) / 12 = 20.

As noted in the previous post, the function F(x) applied to both sides of an equation does not have to cancel the top-level function L1(x). We could apply F(x) = x * 12 to both sides of the original equation. Then the equation becomes

F(4 + ((a - 1) / 12)) = F(24)

(4 + ((a - 1) / 12)) * 12 = 24 * 12

4 * 12 + ((a -1) / 12) = 24 * 12

48 + (a - 1) = 24 * 12
May 30th 2012, 05:49 AM
Reckoner

Re: Just a simple order of operations question

Quote:

Originally Posted by SplashDamage

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?

Multiplying both sides by 12 is a valid operation. We can multiply, divide, add, or subtract a constant from any equation as long as we do the same thing to both sides. But we must be careful to perform the operation on the entire left- and right-hand sides. So, if we start by multiplying by 12, we'll have to distribute:

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

$\Rightarrow12\cdot\left(4+\frac{a-1}{12}\right)=12\cdot24$

$\Rightarrow12\cdot4+12\cdot\frac{a-1}{12}\right)=288$

$\Rightarrow48+(a-1)=288$

and we could continue from there. The reason you would probably begin by subtracting the four is so that you only have to multiply the 12 by two terms instead of three:

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

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

and, because addition is commutative and associative, this becomes

$(4-4)+\frac{a-1}{12}=20$

$\Rightarrow\frac{a-1}{12}=20$ .

Edit: I was a little slow. The previous posts give good explanations.
May 30th 2012, 05:53 AM
HallsofIvy

Re: Just a simple order of operations question

Since you title this "order of operations", you apparently understand that if you were given a value of a, and asked to evaluate the left side, you would (1)subtract 1 from a, (2) divide that result by 12, (3) add 4.

In order to solve for x, you have to "undo" that which means you must do the opposite of each step in the opposite order.

Think of it this way- when you get dressed in the morning you put on your socks first and only then your shoes (obviously) and in the evening you have to first take off your shoes then your socks. You can't very well take of your socks before you take off your shoes!

So you must (1) subtract 4, (2) multiply by 12, (3) add 1 to go from (a-1)/12+ 4 to a. Doing the same thing on the right side gives (1)24- 4= 20, (2)12(20)= 240, (3)240+ 1= 241= a.
May 30th 2012, 05:59 AM
SplashDamage

Re: Just a simple order of operations question

I finally get it. Thank you guys so much. For some reason I just assumed that you could remove the 12 without changing the rest, but in actuallity, you need to keep both sides of the equation =, and my only multiplying 12 from part of the lefthand side, you screw it up. Thanks guys.

One more thing. I wanted to know if it was possible to simplify this fraction further. I tried it, and I do not think it is possible. I checked by trying to extract the highest common factor of the numerator and the denomonator, but I don't think there is one.

Also is x^2 a common factor?

17x^4 / 4yx^2
May 30th 2012, 06:07 AM
Reckoner

Re: Just a simple order of operations question

Quote:

Originally Posted by SplashDamage

One more thing. I wanted to know if it was possible to simplify this fraction further. I tried it, and I do not think it is possible. I checked by trying to extract the highest common factor of the numerator and the denomonator, but I don't think there is one.

Also is x^2 a common factor?

17x^4 / 4yx^2

Yes, $x^2$ is a common factor:

$\frac{17x^4}{4x^2y} = \frac{17x^2x^2}{4x^2y} = \frac{17x^2}{4y},\quad x,y\neq0$ .
May 30th 2012, 06:25 AM
SplashDamage

Re: Just a simple order of operations question

Oh I never thought of breaking up the x^4 like that.

Thanks man. And here's another one (sorry guys, I have just been trying to revise how to work with fractions because i am at it)

(16x + 4x^3) / (x^2 + 2x - 8)

I tried to extract common factors which made it

(4x(4 + x^2)) / ((x - 2)(x + 4))

I do not know where to go from here. Should I have tried to cancel out the 2x from the bottom half first?
May 30th 2012, 06:29 AM
SplashDamage

Re: Just a simple order of operations question

I tried this other approach.

(16x + 4x^3) / (x^2 + 2x - 8)

split the 4x^3 from the top half to make

(16x + 4x + x^2) / (x^2 + 2x - 8)

take the x^2 from both sides

(16x + 4x) / (2x - 8)

add the x's on the top, and then take the 2x from the top

18x / -8

is this correct?
May 30th 2012, 06:34 AM
SplashDamage

Re: Just a simple order of operations question

Ugh no. You can't split 4x^3 up into 4x and x^2 can you.

Wait what if I put a multiplication sign inbetween?

EDIT: Nope that doesnt work
May 30th 2012, 06:52 AM
Reckoner

Re: Just a simple order of operations question

Quote:

Originally Posted by SplashDamage

Oh I never thought of breaking up the x^4 like that.

Thanks man. And here's another one (sorry guys, I have just been trying to revise how to work with fractions because i am at it)

(16x + 4x^3) / (x^2 + 2x - 8)

I tried to extract common factors which made it

(4x(4 + x^2)) / ((x - 2)(x + 4))

You have factored correctly, and completely. Now we cancel out any common factors. But, it appears there are no common factors:

$\frac{16x + 4x^3}{x^2 + 2x - 8}$

$=\frac{4x\left(4 + x^2\right)}{(x-2)(x+4)}$

$=\frac{4\cdot x\cdot\left(4 + x^2\right)}{(x-2)\cdot(x+4)}$

Since there is nothing that can be canceled, there is no further simplification to be done.

Quote:

I do not know where to go from here. Should I have tried to cancel out the 2x from the bottom half first?

The $2x$ is not a factor of the denominator, it is merely a part of the larger factor $x^2+2x-8$ . You can only cancel out entire factors, you cannot cancel out individual terms within factors.

Quote:

Originally Posted by SplashDamage

I tried this other approach.
...

Quote:

Originally Posted by SplashDamage

Ugh no. You can't split 4x^3 up into 4x and x^2 can you.

There's no addition here, so we cannot split these up into two terms. We can split them up into two factors, $4x^3=(4x)\left(x^2\right)$ , but this isn't really helpful because the $4x^3$ is not a factor of the numerator, it is only a term.
May 30th 2012, 07:08 AM
SplashDamage

Re: Just a simple order of operations question

So once I got to the point where I couldn't cancel anything out, what do I do?

Does that mean that the original form was the most simplified?

And is that like a be all/end all for simplifying these fractions? If there is nothing to cancel after getting the highest common factor then it is always the most simplified?
May 30th 2012, 07:18 AM
Reckoner

Re: Just a simple order of operations question

Quote:

Originally Posted by SplashDamage

And is that like a be all/end all for simplifying these fractions? If there is nothing to cancel after getting the highest common factor then it is always the most simplified?

Yes. In general, to reduce a fraction, completely factor the numerator and denominator, eliminate any factors that are common to both, and what remains is the reduced form. So yes, the original expression could not be reduced.

You may be able to leave the fraction in its factored form as your final answer, though. It's really personal preference as to which form is "simpler". Just make sure there are no common factors left.
May 30th 2012, 07:30 AM
SplashDamage

Re: Just a simple order of operations question

In this example (which is actually from a previous test that i failed hard)

http://desmond.imageshack.us/Himg820...ng&res=landing

I am not sure what to do.

In the top half, with the bracketed group that is ^2, is it supposed to be split up into two groups like one would usually do with something like (x + 5)^2?

And similarly, if you have a group that has a ^3 (just like the bottom half), what do you do with that?
May 30th 2012, 07:34 AM
SplashDamage

Re: Just a simple order of operations question

I mean, if you had something like (x + 3) (y + 4) (x + y)... How do you multiply them all together? Like you would with just two.

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