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58x/3 + 77/3 =11
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58x/3 + 77/3 =11
Is there any nice way to simplify this? Am I nuts? Returning to school after years of being away from it :'(
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>>36512
first multiply both sides by 3
2. subtract 77
3. divide by 584
It's not hard
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File: Untitled.png (7 KB, 484x499) Image search: [iqdb] [SauceNao] [Google]
7 KB, 484x499
Is this what you're talking about?
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>>36516
>divide by 584
*divide by 58
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http://www.themathpage.com/

This guy's math tutorials are pretty good
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>>36512
Well first you should know that having 1 equation with 1 unknown means you can actually "solve" it. Have 2 equations with 2 unknowns means you can solve for both unknown, etc..
So yes you can simplify it down to x= a number.

It's ok if actually doing algebra is hard for you, but conceptually it should be piss easy.
You have one thing is equal to another, as long as you're changing them both in the same way then you ensure that they're still equal.

e.g. every step of an algebra problem actually gives you a NEW equation, but you know it's true because the previous equation was. When given an algebra problem you're supposed to assume the unknowns take on values that make the starting equation true.
If 2x=4 is given, then you can manipulate both sides to get a new equation (x=2) which must be true because you got there by doing the same thing to both sides of a true equation.
(I only mention this because I've seen enough dumb shit people do in algebra to think a lot of people are under the impression that it's the same equation throughout every step, and that leads to a lot of confusion)

>>36517
That's right. But you can go further. At this point multiplying or dividing anything would be a bad idea because then you'd have to do it to both terms on the left side.
You have 77 and 33 as isolated terms, so it makes the most sense to combine them.
You can either subtract 33 from both sides, or subtract 77 from both sides. Then see if you can finish figuring out the value of x.