# Solution of 3 equations in 3 unknowns [closed]

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 StuckyMar 6 '14 at 12:33

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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
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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