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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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closed as off-topic by Davide Giraudo, Casteels, Magdiragdag, TooTone, Eric Stucky Mar 6 '14 at 12:33

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Davide Giraudo, Casteels, Magdiragdag, TooTone, Eric Stucky
If this question can be reworded to fit the rules in the help center, please edit the question.

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
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. – mixedmath Oct 9 '12 at 18:44

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

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