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In a problem my teacher did the following:

5a - 2b = 3m 
5b - 2c = 3n    +
-----------------
5a - 2c + 3b = 3m + 3n

I tried solving for b in the first equation and then plugging it in to the second but could not get the same result.

When is adding full equations like this valid?

Thanks!

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5  
I'm not sure what your trouble is, but I can answer your question: if it actually is an equation, then it is always valid. This is because it is the same as adding the same quantity to both sides of the first equation; the LHS of the second equation on the left, and the RHS of the second equation on the right. (It's expressed differently on the two sides, but it is the same quantity, after all.) –  Niel de Beaudrap Oct 29 '10 at 15:51
    
"(It's expressed differently on the two sides, but it is the same quantity, after all.) " was exactly what I was looking for! Thank you so much! –  Bradley Barrows Oct 29 '10 at 16:04

2 Answers 2

up vote 1 down vote accepted

If A = B and C = D then A+C = B+D.

More generally, if $f$ is any function, then $x = y$ implies $f(x) = f(y)$.

In this case we have used the function f(x) = x + C. to go from A = B to A + C = B + C as well as g(x) = B + x to go from C = D to B + C = B + D. Then we paste the two equations together (transitivity) to get A + C = B + D.

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LEMMA $\rm\quad\ \: A = A',\ \ \: B = B'\ \Rightarrow\ A+B\ =\:\ A'+B'$

Proof $\rm\ \ 0 = A-A' + B-B' \: =\ (A+B) - (A'+B')$

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edit; Oh I get it, you define $A = A' \iff 0 = A - A'$. –  anon Oct 30 '10 at 16:57
1  
@muad: It's not a definition but, rather, a consequence of axioms. In essence it shows that in an Abelian group the proof reduces simply to 0 + 0 = 0, i.e. that 0 is a subgroup. Generalizing from 0 to any subgroup G yields the proof that the corresponding equivalence relation $a \equiv b \iff a-b \in G$ is a congruence, i.e. the group operations preserve such equivalences, so they can be added and subtracted. –  Bill Dubuque Oct 30 '10 at 17:47
    
Thank you! I like that idea. –  anon Oct 30 '10 at 17:51

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