anonymous
  • anonymous
Can someone give me a only letter equation to solve?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
across
  • across
2x+3=11
anonymous
  • anonymous
take 3 from both sides getting 2x=8 divide both sides by 2 gettin x=4
anonymous
  • anonymous
Btw, grats on Master rank

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

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