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Beth331178

  • one year ago

How do I solve proportions with variable in them, and fraction/equations with variables?

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  1. theEric
    • one year ago
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    Since fractions and proportions are described with division or multiplictation, you can often divide or multiply both sides by whatever your variable is being multiplied to or dvided by! I don't know what this is for; or what you mean exactly, but if you come up with any example, I can try to help!

  2. Beth331178
    • one year ago
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    Thank You for the reply, I'll try to draw an example of the problem |dw:1359433211281:dw|

  3. theEric
    • one year ago
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    Like.... three-fifths of my toys are action figures, or something like that -\[\frac{3}{5}toys=actionFigures\] If I told you I have 10 toys, so the "toys" variable is now known to you, you can find the next variable, "actionFigures" by easy substitution. Example:\[\frac{3}{5}(10)=\frac{30}{5}=6=actionFigures\] But, if I just say... "and I have 6 action figures", then you have to solve if you wanted to know how many toys I have. Why would you want to know that? I don't know... Maybe you're assesing my personality, or something.. You can just flip the fraction and put it on the other side, if it's simple as that. But you can also work through it, so you ready to handle any proportion thrown your way. If you understand how doing operations on all sides of a true equation produce a new equation, this will be simple. If not, then just think about this. If two things, one on each side of an equation, are equal, then they are the same amount. If you change all the same amounts you have in the same way, then everything is still the same! Going back to\[\frac{3}{5}toys=actionFigures\]you see the \[\frac{3}{5}\] there. That's just like \[(3\div5)\] so you have\[(3\div5)toys=actionFigures\]. You don't want your amount of toys being divided by 5, or multiplied by three - you just want the number of toys (no alterations of it). You can alter this side, but the resulting equation will be true only if you alter the other sides in the same way. So, you don't want to divide by 5? Multiply by 5! This way, there is really nothing happening. And you don't want to multiply by three, so divided by three.\[(3\div5\times toys)\div3\times5=\]\[3\div5\times toys\div3\times5=\]\[3\div3\times5\div5\times toys=\]\[1\times1\times toys=\]\[toys\]and you must do the same to the other side,\[actionFigures\rightarrow actionFigures\times5\div3\]\[actionFigures\times5\div3=\]\[\frac{5}{3}actionFigures\]so there's the flipped fraction. And since I said I have 6 action figures,\[\frac{5}{3}actionFigures=\frac{5}{3}(6)=\frac{30}{3}=10=toys\] Feel free to respond with any questions. I don't know if that was even what you were looking for!

  4. theEric
    • one year ago
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    Ah, I see you relpied!! :) Did you want to read through that? It's a general explanation. If you don't want to read it, I say don't force yourself. But it is relevant. If you try to read it and decide you'd rather just work through your problem with me, let me know! that's fine too!

  5. theEric
    • one year ago
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    What's most important in my long explanation is the part about doing the same thing to all sides of an equation. Once you've got that, you'll be on your way to solving your example! I can help with that, if you'd like! I'll quote myself... "If you understand how doing operations on all sides of a true equation produce a new equation, this will be simple. If not, then just think about this. If two things, one on each side of an equation, are equal, then they are the same amount. If you change all the same amounts you have in the same way, then everything is still the same!"

  6. Beth331178
    • one year ago
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    Thank you, I read it but I'm still a little confused with how to solve for the variable. Should I cross multiply? I'm not sure how to set it up!

  7. theEric
    • one year ago
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    Well, in your problem, you want x! Not \[x+9\]and also not \[(x+9)\div18\]!

  8. theEric
    • one year ago
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    In \[\frac{x+9}{18}\]you know you have to change it, but not how, right? You know how order of operations matters in math expressions, right? Like\[6\div3+1\]is\[(6\div3)+1\] but not\[6\div(3+1)\]

  9. theEric
    • one year ago
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    Do you know that? It's just the way mathematicians set our math-writing up.

  10. Beth331178
    • one year ago
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    Ok yes, I understand order of operations

  11. theEric
    • one year ago
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    Good! :) So in\[\frac{x+9}{18}\] you know that subtracting 9 won't make it any better. You sort of have to negate what feels like the outside layer of the term. Like multiplying by 18. You get it? Or am I too vague? Be honest - I can take it!

  12. theEric
    • one year ago
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    and when I say "what feels like the outside layer of the term", I mean "what feels like the outside layer of the term to me", because that's how it feels to me.

  13. theEric
    • one year ago
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    More technically, it's operation that is the last to modify your variable.

  14. theEric
    • one year ago
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    Last in the order of operations in the expression.

  15. Beth331178
    • one year ago
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    um do you mean subtract 9 from the equation? I don't know what you mean by "negate the outside layer"

  16. theEric
    • one year ago
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    I was wondering if I should have been more specific - I should've been. I wrote a little response to explain what I meant about how I feel about it, but I don't think that would help. Di you understand the more technical definition? I can explain how I think about it - in layers - if you want.

  17. Beth331178
    • one year ago
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    yes please :)

  18. theEric
    • one year ago
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    By negate, I mean, take something that affects your expression and use an operation to make that something's effect be nothing.

  19. theEric
    • one year ago
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    Okay! By the outside layer, I mean the number/variable that is in the last operation modifying the variable but does not contain the variable. By negate, I mean, take something that affects your expression and use an operation to make that something's effect be nothing.

  20. theEric
    • one year ago
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    Let me better convey what I mean by outside layer, though.

  21. theEric
    • one year ago
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    |dw:1359435707707:dw|

  22. theEric
    • one year ago
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    |dw:1359435884058:dw|

  23. theEric
    • one year ago
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    It's like an onion! It has layes! And the most inside layer is the "x".

  24. Beth331178
    • one year ago
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    Oh I see! but where did you get the 4 and 5 from?

  25. theEric
    • one year ago
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    From x, the next layer is the next modification, which you'll have to discover by using your knowledge of order of operations. But When I see all that, I ca quickly find that there is a big chunk that contains x, and then something that modifies it. You have to take the the last modification, or layer, off before you can go to the next one. I just added them on for fun...

  26. theEric
    • one year ago
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    I didn't realize my one "layer" miss the 4! Huh! Sorry about that..

  27. theEric
    • one year ago
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    I hope that helps. That's just my attempt at an explanation about how I see it.

  28. Beth331178
    • one year ago
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    oh I see, you were just trying to show the different 'layers' of it with the numbers. It's okay, It helps a little bit. What would I do from here?

  29. theEric
    • one year ago
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    Here, being my example? Or yours? Yours is simpler :)

  30. Beth331178
    • one year ago
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    Mine :)

  31. Beth331178
    • one year ago
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    Thank you so much for your help! I have to get going now though

  32. theEric
    • one year ago
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    Ah! Sorry! Take care! Good luck!

  33. theEric
    • one year ago
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    I'll type more anyway!

  34. Beth331178
    • one year ago
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    You too...I'll figure them out..bye!

  35. theEric
    • one year ago
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    Your example was\[\frac{x+9}{18}=\frac{12}{72}\]The "outer layer", and I mean last operation on the chunk that has "x" in it, is the dividing by 18. So you negate it - muliply all sides by 18! So \[\frac{x+9}{18}\times 18=x+9\]\[=\frac{12}{72}\times 18 = \frac{12\times 18}{72}\]so\[x+9=\frac{12\times18}{72}\] Now the "outer layer" is adding 9. So subtract 9, from all sides.\[x+9-9=x\]\[=\frac{12\times18}{72}-9=...calculation...=-6\]so\[x=-6\]"x" is all y itself - we've peeled off all the layers, in my mind - so this is want we wanted!

  36. theEric
    • one year ago
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    I don't get on too often, but if you'd like me to help more, just put post @Eric and it'll let me know!

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