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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.2012-10-09
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    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 $rank(A,b) = rank(A)$,2012-10-09

3 Answers 3

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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](http://math.stackexchange.com/a/209762).2012-10-09
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    @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.2012-10-09
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    Yes, and I need to close the other one now instead of this one :-)2012-10-09
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    @rob can't we just merge them all?2012-10-09
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    @robjohn: The other two are [here](http://math.stackexchange.com/questions/209770/find-the-value-of-c-wich-makes-it-possible-to-solve) and [here](http://math.stackexchange.com/questions/209763/linear-equation-systems), on the off chance that you’ve not seen them.2012-10-09
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    @draks: the one that was closed was *exactly* the same and Brian's answer was *exactly* the same, so there is no reason to merge it. However, I see that Brian has pointed out a couple of others.2012-10-09
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    @BrianM.Scott: thanks for the links. I merged the first and closed the second.2012-10-09
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    @robjohn: Sounds good. (It was a bit startling to see four of them almost at once.)2012-10-09
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    @robjohn: We just got another one of them, [here](http://math.stackexchange.com/questions/209806/find-the-value-of-c-wich-makes-it-possible-to-solve).2012-10-09
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    This must be an exam or contest problem somewhere.2012-10-09
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    @robjohn It may not be an exam or contest problem but just a single user reposting it many times - I can't imagine 7 different people posting the same question (4 of them in the span of a few minutes) even if it was a contest or exam problem.2012-10-09
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    @robjohn: And we just got four more. Right in a row.2012-10-09
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    I don't understand. None of the posters have the (exact) same IP. This is either an exam question, a homework question, or a strange form of malevolence.2012-10-09
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If you add the second and third equation you get 5u+7v=5+c ; if you multiply the second equation by 2 and then add it to the first then you get 4u+u+6v+v=12=5u+7v , so c=7