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2x+3=11

take 3 from both sides getting 2x=8
divide both sides by 2 gettin x=4

Btw, grats on Master rank

Can i have a harder one now?

good job

that's called only a letter equation? :D

Nope but i take whats coming lol

Let's do a slightly trickier one: (1/3)x+2=3

Ill take (1/3)x+2=3
1x+6=9, right sofar?

That's right thus far!

x+6=9
x=3

That's correct.

Very good! Let's do this one now: (2/3)x+(5/2)=(5/4)
(I love fractions.)

Please put that in equation form

\[\frac{2}{3}x+\frac{5}{2}=\frac{5}{4}\]

Hmm

Im supposed to "equalize the fractions" right?

Wait!

NONONONO

no clues :(

(2/3)x+(5/2)=(5/4)----->(4/6)x+(10/4)=(10/8)?

No that cant be it..

wait, just say yes or no, not why.

You rewrote the equation. :P

(2/3)x+(5/2)=(5/4)
(2/3)x=(5/4)-(5/2)
Right so far?

Yes

Now multiply both sides by 3 to get (3/3)x=(5/4)-(5/2)*3?

If so i have the answer

You have the right idea, but you're overlooking a few steps. :)

(3/3)x=(5/4)*3-(5/2)*3??

Let's first focus on trying to simplify this:\[\frac{5}{4}-\frac{5}{2}.\]Can you do that?

5-5
----
4-2?

:( i thought i knew this stuff...

Whats the logic in that?

So i need to find the greates common (something i cant remember)?

You can do that OR you could use the fact that\[\frac{a}{b}\pm\frac{c}{d}=\frac{ad\pm bc}{bd}.\]

How about if they have similar (the ones we have over the base)(whatever they call them)

The numerator? Do you mean something like\[\frac{3}{5}+\frac{3}{6}?\]

yeah

Oh wait, never mind i think i see why that wont work

It doesn't matter: they have different bases and you can't add them.

(2/3)x=(5/4)-(5/2) 5*2-4*5/2*5?

Im not sure i know what you mean by manipulate

You mean

make them higher so they match or something?

Oh you did mean like that

(5/2)
(5/4)-------->2.5/2?

Aw man.. im really worthless at this..

At this rate ill never learn quantum physics..

Well how? i cant go upward.

(5/2)---->(10/4)!!!!
(5/4)

Now i can have (15)/(4)

So i wasnt right?

(2/3)x=(5/4)-(10/4)
(2/3)x=(10-5)/4?

That's very close! But you switched the signs:\[\frac{5}{4}-\frac{10}{4}=\frac{5-10}{4}.\]

Oh right! so its (2/3)x=(-5)/4

That's correct! Now, let's solve for x. :)

Can we multiply both sides by 3 now?

Yes we can. :)

(3/3)x=(-5)/4*3

x=(-5)/4*3

Is it right? :D

Wait, should i divide both sides by 2/3?

x=(-5)/4*(2/3)

Please say its right..

Oh right, i knew it had something to do with multiplying with the denominator!

Now, let's do a cool little trick: what happens if I multiply\[\frac{2}{3}x\]by\[\frac{3}{2}?\]:)

5/5?

Close!\[\frac{3}{2}\cdot\frac{2}{3}x=\frac{6}{6}x=x.\]

so essentially, i was right?

Now, can you apply this to solve\[\frac{2}{3}x=-\frac{5}{4}?\]

Yes, you were right, but I don't think you understood the whole process, which now you do. :)

Both sides multiplied by 3/2?

That's correct.

x=11/4

If thats not right, im gonna go give up on physics and go cry in a corner..

-5/4*6/4 so 1/4?

-30/4?

Look at our final expression:\[-\frac{5}{4}\cdot\frac{3}{2}=-\frac{5\cdot3}{4\cdot2}=?\]

Wait, do you know what this means?

What exactly?

That im a complete and utter failure and that i will never become a physicist.

susanne its really sweet of u to help sumone like this with this devotion

n yehh that kid is damn focused on the job for sure!

and you will never become a physicist

Ok, im done sulking. Wanna do some math?

Fumble you lagrangeson.

those word came out of your mouth, fumble yourself

You suck. At least across tries to help me.

your saying i havent helped you?

You just said i wasnt gonna become a physicist! Mark my words, I WILL!

prove it then you little squirt

Screw you, ill get 2 phds

we'll see ...

Damn right well see!

across, 15/6?