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    Joined: Mar 2008
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    OK math people, I understand substitution but this question is about elimination. It must be obvious why you can add or subtract the equations to solve them because none of our sources tell us. But it escapes me. Note: I know HOW to do it, and I know you can put the answer into your equations to prove it works. I want to know in logical words why a person would expect it to work.

    Anyone want to take a stab at it?

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    If two quantities are equal, they are still equal if the same amount is added to each quantity. Algebraically, if x = y, and b = c, then x+b = y+c .

    The two sides of an equation are equal by definition, so if they are added to the two sides of another equation, the sums are still equal.


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    I'll give it a try. Suppose you have two equations known to be true, say x=y and u=v. For either equation, I should be able to add the same thing to both sides of the equal sign and still have a true equation, i.e., x+u=y+u. Agreed? But since u=v, we could also write this as x+u=y+v, and that must also be true. The last statement is equivalent to adding the 2 equations, but we could have also subtracted, or taken any linear combination.

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    With any equation, the value of the things on the left, and the value of the things on the right of the equals sign are the same. You may not know what they are, but you do know that they have the same value.

    You also know that you can add an equal amount to both sides of any equation, while maintaining equality.

    So if X = 5, then X + 1 = 5 + 1.

    Adding an equation to an existing equation is the same as adding the 1s above, except that you don't know the value of what you added yet. You just know that you added the same amount to both sides, because your equal sign tells you the values are the same.

    What might make this completely obvious would be to solve everything out like you normally do. Then after you're done, go back and plug your values into the equation that you added to the other equation. What you'll find is that both sides evaluate to the same value. Thus when you added each of those sides to the other equation, you were simply adding some unknown amount (known now that you solved it.)

    I hope that helps. It might be easier to work through with an example.

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    I should be able to explain if I understand what you're asking, but I'm not sure I do. An example will help: let me try to give one and explain it, and if it isn't an example of the right thing, then please give a better example!

    Suppose we know
    3a + b = 7
    and
    a - b = 1

    If I understand correctly, you're saying that you know the thing to do is to add up these equations to get 4a = 8, telling you that a = 2, and then substitute back to get b = 1, but you're not sure why that is a legitimate thing to do - is that right?

    Technically it works because the + operator is a congruence with respect to = (that's computer science terminology - you almost have to program a computer that doesn't know this to notice that it needs to be said!) but I think it's useful. That is, if you know that a = b then, for any c, a + c = b + c, etc. (And this is so universally true of anything for which we use these symbols that I don't even have to say whether I mean natural numbers, complexes, processes in some weird process calculus or what!) It's a special case of "it's OK to do the same thing to both sides [oh, except divide by something that might be zero or take roots, but, err, don't worry about that at school]". In this case, since we know 3a + b = 7 we can deduce that 3a + b + 1 = 7 + 1 = 8, and since we know a - b = 1 we can deduce that 3a + b + 1 = 3a + b + a - b = 4a, and oh look, we're also using the fact that = is an equivalence relation (that is, three things: for any x and y and z (1) x = x, (2) x = y if and only if y = x (3) if x = y and y = z then x = z) to deduce 4a = 8.

    I'll stop there in case I'm barking up the wrong tree entirely...
    ETA: gosh, you have to be fast round here. Well, there are more long words in my version ;-)

    Last edited by ColinsMum; 02/21/12 07:59 AM.

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    HMM. I must be dense today. So, when you are adding two equations to find a solution that satisfies both, you are really saying that the equations are equivalent for that solution (or set of solutions) and it's OK to add equivalent things.

    So if you have x + 2y = 35 and 3x + 14 y = 120 then you've decided they are equivalent so adding the left sides and the right sides makes them stay equivalent?

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    No: if these two equations were equivalent, then having the second one would be giving you no new information, and you wouldn't get a single solution at all.

    (These two are equivalent:
    a + b = 2
    2a + 2b = 4
    and you'll see that the second one adds nothing to what you know.)


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    You can make any linear combinations of equations in a linear sytem to change it but of course the number of equations has to remain the same (if not you will have more unknown than equations and you lose the unicity of solution). It will not change the results.


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    Originally Posted by master of none
    HMM. I must be dense today. So, when you are adding two equations to find a solution that satisfies both, you are really saying that the equations are equivalent for that solution (or set of solutions) and it's OK to add equivalent things.

    So if you have x + 2y = 35 and 3x + 14 y = 120 then you've decided they are equivalent so adding the left sides and the right sides makes them stay equivalent?

    Equation 1:
    x + 2y = 35

    Equation 2:
    3x + 14 y = 120

    Let's say we multiply both sides of equation 1 by 3 so we can subtract it from equation 2:

    3*(x + 2y) = 3*35
    3x + 6y = 105

    Let's alter equation 2:
    3x + 14y - 105 = 120 - 105

    Subtracting 105 from both sides didn't change anything, right? It's perfectly valid to do?

    But we know that 105 is equal to (3x + 6y).

    So we can re-write that as 3x + 14y - (3x + 6y) = 120 - 105
    Which becomes 8y = 15.

    This is the same as subtracting the equations directly
    3x + 14 y = 120
    -3x -6y -105
    8y = 15

    I hope that makes it clear.

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    Thanks DAD22.
    Perfectly clear! Wow, it actually makes sense! It is usually the simple things that stump me.






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