Can someone give me a only letter equation to solve?

- anonymous

Can someone give me a only letter equation to solve?

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- across

2x+3=11

- anonymous

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

- anonymous

Btw, grats on Master rank

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## More answers

- anonymous

Can i have a harder one now?

- anonymous

good job

- anonymous

that's called only a letter equation? :D

- anonymous

Nope but i take whats coming lol

- across

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

- anonymous

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

- across

That's right thus far!

- anonymous

x+6=9
x=3

- anonymous

?

- anonymous

That's correct.

- across

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

- anonymous

Please put that in equation form

- across

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

- across

If you can do that without problems, then you would have mastered the solving of single-variable, linear equations. :P

- anonymous

Hmm

- anonymous

Im supposed to "equalize the fractions" right?

- anonymous

Wait!

- across

We're trying to solve for \(x\), so try and use the properties of addition, subtraction, multiplication and division to achieve that!\[\]

- anonymous

NONONONO

- anonymous

no clues :(

- anonymous

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

- anonymous

No that cant be it..

- anonymous

wait, just say yes or no, not why.

- across

You rewrote the equation. :P

- anonymous

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

- across

Yes

- anonymous

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

- anonymous

If so i have the answer

- across

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

- anonymous

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

- across

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

- anonymous

5-5
----
4-2?

- across

There's where we're getting stuck. :P
You need to learn how to simplify expressions of the form\[\frac{a}{b}\pm\frac{c}{d}.\]

- anonymous

:( i thought i knew this stuff...

- across

It's not too bad, really, there's a simple rule which states that\[\frac{a}{b}\pm\frac{c}{d}=\frac{ad\pm bc}{bd}.\]

- anonymous

Whats the logic in that?

- across

To put it simply, you cannot add nor subtract fractions with different bases. In other words,\[\frac{5}{4}-\frac{5}{2}\]cannot be simplified as it stands since they have different bases (4 and 2).

- anonymous

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

- across

However, if you have something like\[\frac{3}{2}+\frac{5}{2},\]they have similar bases and thus you can add them accordingly:\[\frac{3+5}{2}=\frac{8}{2}=4.\]

- across

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

- anonymous

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

- across

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

- anonymous

yeah

- anonymous

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

- across

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

- anonymous

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

- across

I saw something I liked: you know how to manipulate fractions. Could you manipulate\[\frac{5}{4}-\frac{5}{2}\]so that they have similar bases?

- anonymous

Im not sure i know what you mean by manipulate

- anonymous

You mean

- anonymous

make them higher so they match or something?

- across

With "manipulate" I mean to say "play with" or "rearrange," like this:\[\frac{5}{2}=\frac{10}{4}=\frac{15}{6}=etc.\]

- anonymous

Oh you did mean like that

- anonymous

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

- anonymous

Aw man.. im really worthless at this..

- anonymous

At this rate ill never learn quantum physics..

- across

That's good! But try and make the bases of the two fractions match and try to have only whole numbers in both the numerator and denominator.

- anonymous

Well how? i cant go upward.

- anonymous

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

- anonymous

Now i can have (15)/(4)

- across

I'll give you an example. I'm given\[\frac{7}{3}+\frac{4}{5}.\]I want their denominators match, so I multiply \(7/3\) by \(5/5\) to get \(35/15\), and I multiply \(4/5\) by \(3/3\) to get \(12/15\). They now have similar denominators and I can add them:\[\frac{35}{15}+\frac{12}{15}=\frac{47}{15}.\]

- anonymous

So i wasnt right?

- across

Very good! You converted\[\frac{5}{4}-\frac{5}{2}\]into\[\frac{5}{4}-\frac{10}{4}.\]What's the next step?

- anonymous

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

- across

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

- anonymous

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

- across

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

- anonymous

Can we multiply both sides by 3 now?

- across

Yes we can. :)

- anonymous

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

- anonymous

x=(-5)/4*3

- anonymous

Is it right? :D

- across

Well, let's look at the LHS only: when you multiplied\[\frac{2}{3}x\]by \(3\), you somehow got\[\frac{3}{3}x.\]How did this happen?

- anonymous

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

- anonymous

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

- anonymous

Please say its right..

- across

You could do that, yes, but let's do it one step at a time first so that you understand the process. When you have\[\frac{2}{3}x,\]to get rid of that \(3\) in the denominator, we DO multiply the fraction by \(3\) so that this happens:\[\frac{2}{3}x\implies3\cdot\frac{2}{3}x\implies\not{3}\cdot\frac{2}{\not{3}}x\implies2x.\]

- anonymous

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

- across

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

- anonymous

5/5?

- across

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

- anonymous

so essentially, i was right?

- across

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

- across

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

- anonymous

Both sides multiplied by 3/2?

- across

That's correct.

- anonymous

x=11/4

- anonymous

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

- across

I think you got it already, but you're trying to speed your way through it. Take your time! :)
Let's check again:\[-\frac{5}{4}\cdot\frac{3}{2}=?\]

- anonymous

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

- across

I see what you're trying to do, but remember that multiplication and division of fractions is totally different from addition and subtraction. :) They're easier, in fact!\[\frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd}.\]

- anonymous

-30/4?

- across

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

- anonymous

Wait, do you know what this means?

- across

What exactly?

- anonymous

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

- across

Hey, it's your first time doing this. Don't beat yourself too hard. We're taking it one step at a time, and that's the best way there is to go!

- across

Also, when I had your age, I knew much, much less than you do, and now look at me; I'm a math grad student! Your enthusiasm stretches beyond any of that I've seen in most students your age. Please keep at it; you will go really far. Take advantage of the fact that there are people out there that will gladly help you, and you can for sure include me in that group of people.

- wasiqss

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

- wasiqss

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

- anonymous

and you will never become a physicist

- anonymous

Ok, im done sulking. Wanna do some math?

- anonymous

Fumble you lagrangeson.

- anonymous

those word came out of your mouth, fumble yourself

- anonymous

You suck. At least across tries to help me.

- anonymous

your saying i havent helped you?

- anonymous

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

- anonymous

prove it then you little squirt

- anonymous

Screw you, ill get 2 phds

- anonymous

we'll see ...

- anonymous

Damn right well see!

- anonymous

across, 15/6?

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