My Math Forum Equations with variables

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 May 24th, 2011, 07:46 PM #1 Newbie   Joined: May 2011 Posts: 28 Thanks: 0 Equations with variables Is there a good resource for rules of solving equations with multiple variables? I would benefit from more worked out examples and discussion of the rules, I seem to be missing bits and pieces in my understanding. Ie: 2/3x-3=8 and 10x=3x-14, in the first question, 3 is added to both sides, but in the second it is subtracted. Why is this when both questions indicate subtraction from the variable, shouldn't addition be used to solve in both cases? I know I am just missing a rule somewhere, interested in more resources. Thanks!!
 May 25th, 2011, 12:12 AM #2 Senior Member   Joined: Apr 2011 From: USA Posts: 782 Thanks: 1 Re: Equations with variables You're missing the basic rule that says we do the opposite operation in order to get rid of something from a side. While "something is subtracted from a variable" each time is not relevant. Let's look at your first one: $\frac{2}{3}x - 3= 8$ You have to remember that the object is to isolate the term with the variable. We only have one variable, on the left side. We have no need to move it to the right side. But we want to isolate the term in which the variable exists. If we want to isolate that term, we must get rid of the 3. If we were to add 3 to the left side, it would net out the 3. That is, a -3 and a +3 make 0. And anything plus or minus 0 is still the same thing. So we have effectively wiped out the 3 from that side by doing this. But as you probably know, you cannot do something to one side without doing the same thing to the other side. When have an equation (contains an = sign), both side must remain equal. We can manipulate the sides and move stuff, but we have to keep them "equivalent" to what we already have. So if I'm going to add 3 to the left side, I must also add 3 to the right side. So we end up with: $\frac{2}{3}x - 3 + 3= 8 + 3$ $\frac{2}{3}x - 0= 8 + 3$ $\frac{2}{3}x= 11$ I'm not going to take it further than that, just so that you can absorb what you specifically asked about with these two equations as a comparison. Now to this equation: $10x=3x - 14$ The fact that we happen to be subtracting something from a term with variable is not relevant, at least not in the way you stated it. That is, our purpose is not to automatically add on anything that happens to be subtracted from a variable. Our purpose is to isolate the variable, by whatever means we can which still mathematically work. Now, you can add the 14 to both sides, which would get rid of the 14 (the way we got rid of the 3 in the first equation), but it serves no purpose. (Unless you just like playing around. ) The reason is that it would simply move the 14 to the other side, and we'd still have a variable on both sides, and not have isolated the variable to one side. This is different than the first equation because there's a variable on both sides. So in this case, we want to get these variables together on the same side. There's two ways to do this: move the 10x to the right side, or move the 3x to the left side. You will find there are many times when you can do certain moves, but they don't really get you anywhere. So let's think about that. If we move the 10x to the right, we still have something on the left (it'll be a 0, and we have to leave it cause we can't just "drop" the left side out), and we'll still have the 14 on the right. So that doesn't accomplish our goal. What if instead we move the 3x to the left side? Well, that will put our 2 x's together, isolates the x's to one side only, and leaves our constant (14) by itself, with no variable on that side. Now, we have to keep the 3x as a "set." We cannot subtract the 3 off there, nor the x. We have to keep "3x" together, at least for now. So I'm going to subtract the whole thing. I'm trying to do the opposite operation. The 3x is positive. Since it's a +, I have to do the opposite and minus it. That will net it out to zero, which we can then toss out. But of course, I have to do this to both sides: $10x - 3x=3x - 3x - 14$ $10x - 3x=0 - 14$ $10x - 3x=-14$ Again, I'm going to stop so you can absorb one thing at a time. In reality, getting down the basics of adding and subtracting from a side should be cleared up for you before you attempt doing other steps of problem solving. That is, starting with something like x - 10 = 5 would be a better choice so that you only have to deal with that one rule first. That's just my opinion. You can always ask further questions as you go. This is one site I'm aware of - note that I don't use it much and therefore have no opinion about it, and I'm sure there's a million more. http://www.math.com/homeworkhelp/Algebra.html
 May 25th, 2011, 04:52 PM #3 Newbie   Joined: May 2011 Posts: 28 Thanks: 0 Re: Equations with variables That's great, thanks a lot! The book I have for learning math gives a limited amount of examples and doesn't integrate the rules with the demonstration of the operations, so a lot of the time I am left to guess why a certain operation was done and it always helps to know why. Not just that it needs to be done
 May 25th, 2011, 06:55 PM #4 Senior Member   Joined: Apr 2011 From: USA Posts: 782 Thanks: 1 Re: Equations with variables Sounds like that book kinda stinks. But then, a lot of textbooks do unfortunately.
 June 5th, 2011, 11:33 AM #5 Newbie   Joined: May 2011 Posts: 28 Thanks: 0 Re: Equations with variables I am wondering what step I am missing in solving this expression: x+3/ = 3/ 10 5 I cross multiply, then solve for 15X=30, but this is wrong. Is this wrong because the variable is the numerator? Should I multiply both sides by the LCM? And just use cross multiplication for the cases where the variable is the denominator? Thanks!
 June 5th, 2011, 11:35 AM #6 Newbie   Joined: May 2011 Posts: 28 Thanks: 0 Re: Equations with variables That auto edited my numbers out of order, on either side of that equation the fractions are x+3/10 = 3/5.
 June 5th, 2011, 11:45 AM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Equations with variables We have: $\frac{x+3}{10}=\frac{3}{5}$ Cross multiplication yields: $5(x+3)=30$ Divide through by 5: $x+3=6$ Subtract through by 3: $x=3$ Perhaps simpler is: $\frac{x+3}{10}=\frac{3}{5}$ Multiply through by 10: $x+3=3\cdot\frac{10}{5}=6$ Subtract through by 3: $x=3$
 June 5th, 2011, 09:02 PM #8 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Since 10 = 5*2, x = 3*2 - 3 = 3.
June 5th, 2011, 09:19 PM   #9
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Re: Equations with variables

Quote:
 Originally Posted by Russell368 I cross multiply, then solve for 15X=30, but this is wrong. Is this wrong because the variable is the numerator? Should I multiply both sides by the LCM? And just use cross multiplication for the cases where the variable is the denominator?
You might want to show how you came up with the 15x = 30. It looks like when you multiplied the left side by the 5 that you did not distribute it over the whole thing. The left side's numerator is (x + 3). You have to think of that as a "set" with parenthesis around it. That is, when you multiply by 5, you are multiplying by the result of (x + 3), not just the 3. But since you have a variable, you don't know what (x + 3) is and can't multiply by "the result" of it. So, you have to distribute the 5 over it instead. Any time you have something in a fraction, assume parenthesis around it. Also, looks like you took what used to be adding an x and turned it into a multiplication. (i.e. making it 15x which was originally an addition)

So the variable has nothing to do with why it's wrong. It wouldn't matter where the variable was. Quite personally, to me, cross multiplication is kind of a pre-algebra thing, learned for things like proportions and such. But that may be cause that's how things were when I was learning algebra. I would never use it for something like this. I would multiply both sides by the LCM as you suggest, and as MarkFL has done at the bottom.

(Now that I think about it, I probably would do a "cross multiplication" for something with a variable in the denominator, but I wouldn't be looking at it that way. Like if that (x + 3) were in the denominator instead, I'd multiply both sides by the (x + 3) and then multiply both sides by the 5. That is, as two steps -- but I guess that comes down to cross multiplication, and I just don't look at it like that. )

 June 6th, 2011, 10:56 AM #10 Newbie   Joined: May 2011 Posts: 28 Thanks: 0 Re: Equations with variables Ah yes, thanks. I neglected to assume brackets around the variable. Of course then as brackets are first, if they cannot be solved, they must be done later and the equation isolated through first. So either way I would have run into the same problem whichever method I used. [/quote] You might want to show how you came up with the 15x = 30. It looks like when you multiplied the left side by the 5 that you did not distribute it over the whole thing. The left side's numerator is (x + 3). You have to think of that as a "set" with parenthesis around it. That is, when you multiply by 5, you are multiplying by the result of (x + 3), not just the 3. But since you have a variable, you don't know what (x + 3) is and can't multiply by "the result" of it. So, you have to distribute the 5 over it instead. Any time you have something in a fraction, assume parenthesis around it. Also, looks like you took what used to be adding an x and turned it into a multiplication. (i.e. making it 15x which was originally an addition) [/quote]

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