Inopeki
Can someone give me a only letter equation to solve?
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across
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2x+3=11
Inopeki
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take 3 from both sides getting 2x=8
divide both sides by 2 gettin x=4
Inopeki
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Btw, grats on Master rank
Inopeki
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Can i have a harder one now?
jimmyrep
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good job
FoolForMath
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that's called only a letter equation? :D
Inopeki
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Nope but i take whats coming lol
across
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Let's do a slightly trickier one: (1/3)x+2=3
Inopeki
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Ill take (1/3)x+2=3
1x+6=9, right sofar?
across
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That's right thus far!
Inopeki
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x+6=9
x=3
Inopeki
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?
FoolForMath
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That's correct.
across
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Very good! Let's do this one now: (2/3)x+(5/2)=(5/4)
(I love fractions.)
Inopeki
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Please put that in equation form
across
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\[\frac{2}{3}x+\frac{5}{2}=\frac{5}{4}\]
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If you can do that without problems, then you would have mastered the solving of single-variable, linear equations. :P
Inopeki
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Hmm
Inopeki
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Im supposed to "equalize the fractions" right?
Inopeki
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Wait!
across
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We're trying to solve for \(x\), so try and use the properties of addition, subtraction, multiplication and division to achieve that!\[\]
Inopeki
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NONONONO
Inopeki
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no clues :(
Inopeki
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(2/3)x+(5/2)=(5/4)----->(4/6)x+(10/4)=(10/8)?
Inopeki
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No that cant be it..
Inopeki
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wait, just say yes or no, not why.
across
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You rewrote the equation. :P
Inopeki
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(2/3)x+(5/2)=(5/4)
(2/3)x=(5/4)-(5/2)
Right so far?
across
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Yes
Inopeki
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Now multiply both sides by 3 to get (3/3)x=(5/4)-(5/2)*3?
Inopeki
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If so i have the answer
across
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You have the right idea, but you're overlooking a few steps. :)
Inopeki
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(3/3)x=(5/4)*3-(5/2)*3??
across
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Let's first focus on trying to simplify this:\[\frac{5}{4}-\frac{5}{2}.\]Can you do that?
Inopeki
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5-5
----
4-2?
across
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There's where we're getting stuck. :P
You need to learn how to simplify expressions of the form\[\frac{a}{b}\pm\frac{c}{d}.\]
Inopeki
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:( i thought i knew this stuff...
across
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It's not too bad, really, there's a simple rule which states that\[\frac{a}{b}\pm\frac{c}{d}=\frac{ad\pm bc}{bd}.\]
Inopeki
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Whats the logic in that?
across
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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).
Inopeki
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So i need to find the greates common (something i cant remember)?
across
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However, if you have something like\[\frac{3}{2}+\frac{5}{2},\]they have similar bases and thus you can add them accordingly:\[\frac{3+5}{2}=\frac{8}{2}=4.\]
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You can do that OR you could use the fact that\[\frac{a}{b}\pm\frac{c}{d}=\frac{ad\pm bc}{bd}.\]
Inopeki
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How about if they have similar (the ones we have over the base)(whatever they call them)
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The numerator? Do you mean something like\[\frac{3}{5}+\frac{3}{6}?\]
Inopeki
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yeah
Inopeki
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Oh wait, never mind i think i see why that wont work
across
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It doesn't matter: they have different bases and you can't add them.
Inopeki
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(2/3)x=(5/4)-(5/2) 5*2-4*5/2*5?
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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?
Inopeki
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Im not sure i know what you mean by manipulate
Inopeki
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You mean
Inopeki
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make them higher so they match or something?
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With "manipulate" I mean to say "play with" or "rearrange," like this:\[\frac{5}{2}=\frac{10}{4}=\frac{15}{6}=etc.\]
Inopeki
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Oh you did mean like that
Inopeki
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(5/2)
(5/4)-------->2.5/2?
Inopeki
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Aw man.. im really worthless at this..
Inopeki
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At this rate ill never learn quantum physics..
across
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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.
Inopeki
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Well how? i cant go upward.
Inopeki
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(5/2)---->(10/4)!!!!
(5/4)
Inopeki
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Now i can have (15)/(4)
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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:\[\frac{35}{15}+\frac{12}{15}=\frac{47}{15}.\]
Inopeki
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So i wasnt right?
across
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Very good! You converted\[\frac{5}{4}-\frac{5}{2}\]into\[\frac{5}{4}-\frac{10}{4}.\]What's the next step?
Inopeki
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(2/3)x=(5/4)-(10/4)
(2/3)x=(10-5)/4?
across
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That's very close! But you switched the signs:\[\frac{5}{4}-\frac{10}{4}=\frac{5-10}{4}.\]
Inopeki
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Oh right! so its (2/3)x=(-5)/4
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That's correct! Now, let's solve for x. :)
Inopeki
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Can we multiply both sides by 3 now?
across
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Yes we can. :)
Inopeki
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(3/3)x=(-5)/4*3
Inopeki
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x=(-5)/4*3
Inopeki
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Is it right? :D
across
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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?
Inopeki
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Wait, should i divide both sides by 2/3?
Inopeki
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x=(-5)/4*(2/3)
Inopeki
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Please say its right..
across
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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:\[\frac{2}{3}x\implies3\cdot\frac{2}{3}x\implies\not{3}\cdot\frac{2}{\not{3}}x\implies2x.\]
Inopeki
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Oh right, i knew it had something to do with multiplying with the denominator!
across
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Now, let's do a cool little trick: what happens if I multiply\[\frac{2}{3}x\]by\[\frac{3}{2}?\]:)
Inopeki
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5/5?
across
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Close!\[\frac{3}{2}\cdot\frac{2}{3}x=\frac{6}{6}x=x.\]
Inopeki
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so essentially, i was right?
across
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Now, can you apply this to solve\[\frac{2}{3}x=-\frac{5}{4}?\]
across
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Yes, you were right, but I don't think you understood the whole process, which now you do. :)
Inopeki
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Both sides multiplied by 3/2?
across
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That's correct.
Inopeki
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x=11/4
Inopeki
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If thats not right, im gonna go give up on physics and go cry in a corner..
across
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I think you got it already, but you're trying to speed your way through it. Take your time! :)
Let's check again:\[-\frac{5}{4}\cdot\frac{3}{2}=?\]
Inopeki
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-5/4*6/4 so 1/4?
across
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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!\[\frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd}.\]
Inopeki
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-30/4?
across
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Look at our final expression:\[-\frac{5}{4}\cdot\frac{3}{2}=-\frac{5\cdot3}{4\cdot2}=?\]
Inopeki
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Wait, do you know what this means?
across
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What exactly?
Inopeki
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That im a complete and utter failure and that i will never become a physicist.
across
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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
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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
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susanne its really sweet of u to help sumone like this with this devotion
wasiqss
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n yehh that kid is damn focused on the job for sure!
LagrangeSon678
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and you will never become a physicist
Inopeki
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Ok, im done sulking. Wanna do some math?
Inopeki
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Fumble you lagrangeson.
LagrangeSon678
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those word came out of your mouth, fumble yourself
Inopeki
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You suck. At least across tries to help me.
LagrangeSon678
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your saying i havent helped you?
Inopeki
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You just said i wasnt gonna become a physicist! Mark my words, I WILL!
LagrangeSon678
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prove it then you little squirt
Inopeki
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Screw you, ill get 2 phds
LagrangeSon678
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we'll see ...
Inopeki
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Damn right well see!
Inopeki
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across, 15/6?