# Solution of 3 equations in 3 unknowns

Find the value of $c$ which makes it possible to solve:

$$u+v+2w=2,$$ $$2u+3v-w=5,$$ $$3u+4v+w=c$$

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And this is the fourth copy of this question in the last half hour. – Brian M. Scott Oct 9 '12 at 9:54
You have an equation of the form $Ax = b$ with $b = (2,5,c)$. Recall that this equation has a solution if, and only if $rank(A,b) = rank(A)$, – Stefan Oct 9 '12 at 9:55

HINT: Add the two first equations.

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Set up your augmented matrix in the usual way:

$$\left[\begin{array}{rrr|r} 1&1&2&2\\ 2&3&-1&5\\ 3&4&1&c \end{array}\right]\;.$$

Then row-reduce it; reducing the first column, for instance, yields

$$\left[\begin{array}{rrr|c} 1&1&2&2\\ 0&1&-5&1\\ 0&1&-5&c-6 \end{array}\right]\;.\tag{1}$$

Now you can either stop and think about the equations corresponding to the bottom two rows of $(1)$ (what does $c$ have to be in order for them to be consistent?), or finish the row-reduction and then think about what $c$ has to be to avoid having an inconsistent system.

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 This answer looks identical to this answer. – robjohn♦ Oct 9 '12 at 10:29 @robjohn: It is: I saw the later question first, answered, and then realized that it was a duplicate and copied the answer over to this one, which is the first of the four identical questions that arrived within just a half hour. – Brian M. Scott Oct 9 '12 at 10:32 Yes, and I need to close the other one now instead of this one :-) – robjohn♦ Oct 9 '12 at 10:34 @rob can't we just merge them all? – draks ... Oct 9 '12 at 10:38 @robjohn: The other two are here and here, on the off chance that you’ve not seen them. – Brian M. Scott Oct 9 '12 at 10:39