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Determine conditions on th b's if any in order to guarantee that the linear system is consistent. $$ \left[ \begin{array}{cccc|c} 1&-1&3&2&b1\\ -2&1&5&1&b2\\-3&2&2&-1&b3\\4&-3&1&3&b4 \end{array} \right] $$

The book says the answer is b1=b3+b4, b2= 2b3+b4.

However when I do it I come to the answer b3=b2-b1, b4=2b1-b2.

Am I doing this wrong? Is it the same answer? Are both answers acceptable?

If not please show how to get the answer.

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    Your solution implies $$b_3+b_4=b_1$$ (just add your equations). – 2017-02-23
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    You also get $$b_2=b_1+b_3=b_3+b_4+b_3=2b_3+b_4$$ Your solution is actually correct. – 2017-02-23
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    Both answers are equivalent. You should be able to verify this yourself at this point though. – 2017-02-23

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From book's answer.

b4=b2-2b3

Then put value of b4 in b1=b3+b4.

b1=b3+b2-2b3

b1=b2-b3

b3=b2-b1

Your answer.

Similarly other one.