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I have 2 simple equations but 3 unknowns and can't work it out. Can anyone help?

a + b = 150
b + c = 234

Many thanks in advance!

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    $\begingroup$ The equations are not enough to determine $a,b,c$. You can give one of $a,b,c$ any value you please - that then fixes the values of the other two. $\endgroup$
    – almagest
    Apr 6 '16 at 10:08
  • $\begingroup$ The system has an infinite number of solutions, for example $a = t-84, b = 234-t, c = t,\quad t \in R$ $\endgroup$
    – georg
    Apr 6 '16 at 10:16
  • $\begingroup$ Thanks. I guess it's impossible to solve then - this is a real life scenario I wanted the answer to, not an educational question :) $\endgroup$
    – B Bob
    Apr 6 '16 at 10:57
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For any value of $c$ you have: $$ b=234-c \quad \rightarrow \quad a+234-c=150 \quad \rightarrow \quad a=c-84 $$

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  • $\begingroup$ Thanks. I guess it's impossible to solve then - this is a real life scenario I wanted the answer to, not an educational question :) $\endgroup$
    – B Bob
    Apr 6 '16 at 10:58
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Let c = t.

Subsequently, b = 234 − c, so b = 234 − t. Then, a = c − 84, so a = t − 84. So the answer is a = t − 84, b = 234 − t, c = t, for any real number t.

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  • $\begingroup$ For any more complex equations which have all 3 unknowns included n the equations, I would prefer using the above method above all others. $\endgroup$
    – Human
    Dec 1 '19 at 17:37
  • $\begingroup$ In terms of t, so let one to be t, then put the other two in terms of t. $\endgroup$
    – Human
    Dec 1 '19 at 17:38
  • $\begingroup$ In some extreme cases, you may need to use a simple matrix to solve it, by row operations for augmented matrices. $\endgroup$
    – Human
    Dec 1 '19 at 17:39
  • $\begingroup$ Sorry if I may seem annoying about making tons of comments. $\endgroup$
    – Human
    Dec 1 '19 at 17:39

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