Math help (Elementary algebra)

[Yuri-G]

hm, I'm a math whiz, got all As in math, 5 on the AP Calc test (although that's not much to brag about), and I still couldn't do that in my head. easy as shit on paper, but not in my head. maybe I've just done too many drugs.

I also did it weird:

5/2 - x/3 = 3x
15/2 - x = 9x
15/2 = 10x
15 = 20x
15/20 = x
3/4 = x

[Kaylia]

hm, I'm a math whiz, got all As in math, 5 on the AP Calc test (although that's not much to brag about), and I still couldn't do that in my head. easy as shit on paper, but not in my head. maybe I've just done too many drugs.

I also did it weird:

5/2 - x/3 = 3x
15/2 - x = 9x
15/2 = 10x
15 = 20x
15/20 = x
3/4 = x

Not every brain work the same, but instead of multiplying the whole equation, try putting similar term (x) on the same denominator (/3 here). Instead of calculating all 3 term, you only have to change the 3x to 9x/3. After this, you can proceed with normal algebra algebra

[kuronosan]

So you cheated to get extra credit. Glad you learned something.

[Overburn]

Even though somehow you ended up with the correct answer I just wanted to point out you made a mistake in the middle. If you add something to one side you need to add exactly the same thing to the other to maintain the equality.

In the bold part on the right hand side of the equation it should be + 2x instead of + 2

The only reason I say this is because in many math classes I took your work in between was more important than the final answer and you would lose points for that.

Thanks I've gotten in trouble with that before. My old teacher assumed that if we did +2 under a thing with a variable, it would be a + to that variable. If we didn't want to add to a variable we would do (+number) to the end of the equation.

[zoobernut]

hm, I'm a math whiz, got all As in math, 5 on the AP Calc test (although that's not much to brag about), and I still couldn't do that in my head. easy as shit on paper, but not in my head. maybe I've just done too many drugs.

I also did it weird:

5/2 - x/3 = 3x
15/2 - x = 9x
15/2 = 10x
15 = 20x
15/20 = x
3/4 = x

I did it in my head but I also did it the exact same way as you did. Just seemed more simple to keep track of one fraction instead of multiple fractions.

[Xiona]

as to not make a second thread i have a problem which is embarrassing that i cant figure it out.

Example: If x - y = 9, then (x - y/3) - (y - x/3) =

(A) -4 (B) -3 (C) 0 (D) 12 (E) 27

(x - y/3) - (y - x/3) =
x - y/3 - y + x/3 =
4x/3 - 4y/3 =
4(x - y)/3 =
4(9)/3 =
12

The answer is (D).

my question is... where the hell does the 4 come from. ive looked it up in my old algebra book, googled etc and still cant seem to see what rule im missing as its been forever.

[xopher]

The real question is, when do you ever use this shit after school?

[Woozie]

as to not make a second thread i have a problem which is embarrassing that i cant figure it out.



my question is... where the hell does the 4 come from. ive looked it up in my old algebra book, googled etc and still cant seem to see what rule im missing as its been forever.

Example: If x - y = 9, then (x - y/3) - (y - x/3) =

(A) -4 (B) -3 (C) 0 (D) 12 (E) 27

(x - y/3) - (y - x/3) =
x - y/3 - y + x/3 =

To show you where the 4 comes from, I'm going to rearange the terms first and then simplify:

http://latex.codecogs.com/gif.latex?...y-\frac{y}{3}=

http://latex.codecogs.com/gif.latex?...}-\frac{y}{3}=

(can you see why 3x/3 is equal to x? the 3's cancel).

http://latex.codecogs.com/gif.latex?...frac{3y+y}{3}=

http://latex.codecogs.com/gif.latex?...}-\frac{4y}{3}

Two ways you can think about it:
1 + 1/3 = 4/3. If you had a pie plus one third, then you have four thirds (because the whole pie is just three thirds. Those three plus the extra one gives you four). As you know, when adding two of the same variables, you just add the coefficients. For example, 5x+9x=14x because 9+5=14. Or 6x+2x=8x because 8+2=6. Same thing here:
x+x/3 is the same thing as 1x+1/3. So what's 1+1/3? It's 4/3.

Or, if you remember adding fractions, you'll remember that they must be given common denominators before you can add them.

http://latex.codecogs.com/gif.latex?...3}=\frac{4}{3}

How did I know to go from 1 to 3/3 (and not 4/4 or 5/5 or something)? Because I needed the denominators to match so that I could add the fractions.

Sorry for being a little redundant in my explanations. This stuff is taught in so many different ways. I'm hoping one of my explanations will jog your memory.

[Xiona]

yea i get it now haha. its been at least 11 years but yea that was one of those things where its so simple you completely over look it. thanks a bunch

[Cream Soda]

K, how do i show this?

3(2z-5) = z-15

6z-15 = z-15
6z = z

z = 0

I got that it's 0 (helping my friend out) but i can't for the life of me show why it's 0

6z = z

divide both by z

6 = 1? >.>

what step am i missing out on to show it's 0? or are you supposed to just know?

[Zilong]

uhhh

6z = z
6z - z = 0
5z = 0
z = 0/5
z = 0

You shouldn't factor out variable from an equation...at least not when only 1 variable is involved

[Eliseos]

K, how do i show this?

3(2z-5) = z-15

6z-15 = z-15
6z = z

z = 0

I got that it's 0 (helping my friend out) but i can't for the life of me show why it's 0

6z = z

divide both by z

6 = 1? >.>

what step am i missing out on to show it's 0? or are you supposed to just know?

When you have 6z=z, just subtract the z on the left hand side and you get 0.

EDIT: Beat'd

[Nystul]

K, how do i show this?

3(2z-5) = z-15

6z-15 = z-15
6z = z

z = 0

I got that it's 0 (helping my friend out) but i can't for the life of me show why it's 0

6z = z

divide both by z

6 = 1? >.>

what step am i missing out on to show it's 0? or are you supposed to just know?

You can't divide by Z because you're supposed to solve for it.

6z=z

subtract z from both sides:
5z=0
divide both sides by 5

z=0

[Lornkeit]

6z = z

6z-z = 0

5z = 0

z = 0

I probably did everything wrong <_<

Edit: Nvm, lol!

[Eliseos]

This reminds me of http://www.colbertnation.com/the-col...6/math-is-hard

[Xno Kappa]

Divide by 0.

[Demosthenes11]

The real question is, when do you ever use this shit after school?

i love this question, because it's only ever asked by people that never took calculus

the first semester of calc AB it becomes painfully obvious why you had to take 12 years+ of math to get there. It's like reading - you can't read a dense college book before slowly reading all levels of books before it

[Cream Soda]

Thanks. At least I only feel half dumb. Still got the right answer!

[Woozie]

Well, the 6 = 1 step actually proves by contradiction that the answer is zero. So you did kinda prove it, but you just don't realize it. You could word a proof like this.

6z = z implies z=0. Proof:

Given this perfectly valid equation, suppose z != 0. Then dividing both sized by z is a legal operation because I can divide any number by any real non-zero number. But dividing both sides by z gives 6 =1, a contradiction. Thus, z must equal zero.


Of course, no teacher expects you to sit and prove this. Any time you get 6z = z, the answer is zero. If you get 5z = z, the answer is zero. If you get 658 z = z, then z = 0. Why? Because zero is the only number you can multiply by another number (that isn't 1) without changing its value. Multiply 1 by any number that isn't 1, and you get a new number. Multiply 17 by anything other than 1, and you get something that isn't 17. So yeah, any time you see an expression like 6z = z, you automatically conclude that the answer is zero without worrying about explaining to the teacher why this is the case.

Another way to think about your solution is to subtract z from both sides and get 5z = 0. If you multiply two numbers by zero, and your answer is zero, then one of the numbers were already zero. So 5*0 is zero, 0*800 is zero. a*b is zero if and only if either a is zero, or z is zero (or both). So 5*z is zero if and only if 5 = 0, or z = 0 (or both. But 5 clearly isn't zero, so z has to be zero).

When you divided both sizes of the equation by z and got 6=1, you got a nonsense answer. So why is it that a perfectly valid algebraic operation gives you nonsense? Because dividing both sides of an equation by a number is only valid if the number you divide by isn't zero. Since z was zero, technically your last step was illegal, which is why you got a non-sensible answer. When we do algebra, and we start dividing by x's and y's, we take for granted that the value of the variables aren't zero. Technically, when we divide by sides by a variable and say that our original expression implies our new expression, what we're really saying is "If z isn't zero, then our original equation implies our new equation". If our new equation is non-sense, then our assumption of non-zero is non sense.

Sorry for the long explanation of such a simple concept. I just want you to understand what's really going on here. The short answer is "Yes, you are supposed to just know this."

[Cream Soda]

Was actually a good read. Thanks.