## Beth331178 2 years ago How do I solve proportions with variable in them, and fraction/equations with variables?

1. theEric

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

Thank You for the reply, I'll try to draw an example of the problem |dw:1359433211281:dw|

3. theEric

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

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

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

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

Well, in your problem, you want x! Not $x+9$and also not $(x+9)\div18$!

8. theEric

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

Do you know that? It's just the way mathematicians set our math-writing up.

10. Beth331178

Ok yes, I understand order of operations

11. theEric

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

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

More technically, it's operation that is the last to modify your variable.

14. theEric

Last in the order of operations in the expression.

15. Beth331178

um do you mean subtract 9 from the equation? I don't know what you mean by "negate the outside layer"

16. theEric

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

18. theEric

By negate, I mean, take something that affects your expression and use an operation to make that something's effect be nothing.

19. theEric

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

Let me better convey what I mean by outside layer, though.

21. theEric

|dw:1359435707707:dw|

22. theEric

|dw:1359435884058:dw|

23. theEric

It's like an onion! It has layes! And the most inside layer is the "x".

24. Beth331178

Oh I see! but where did you get the 4 and 5 from?

25. theEric

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

I didn't realize my one "layer" miss the 4! Huh! Sorry about that..

27. theEric

I hope that helps. That's just my attempt at an explanation about how I see it.

28. Beth331178

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

Here, being my example? Or yours? Yours is simpler :)

30. Beth331178

Mine :)

31. Beth331178

Thank you so much for your help! I have to get going now though

32. theEric

Ah! Sorry! Take care! Good luck!

33. theEric

I'll type more anyway!

34. Beth331178

You too...I'll figure them out..bye!

35. theEric

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

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!