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Here is the example case. You might recognize this to be from Lewis Carroll's "A Tangled Tale". If you're already familiar, you can scroll down to the question.

Example

Given that one glass of lemonade, 3 sandwiches, and 7 biscuits cost \$14 and that one glass of lemonade, 4 sandwiches, and 10 biscuits cost \$19, find the cost of the following:

(a) a glass of lemonade, a sandwich, and a biscuit
(b) 2 glasses of lemonade, 3 sandwiches, and 5 biscuits.

Lewis' approach is substitution, whereas I use an approach involving determinants.

I can rewrite the two cases as $$a+3b+7c=14$$ $$a+4b+10c=17$$ where I want to find $$a+b+c$$ and $$2a+3b+5c.$$

I'll demonstrate the second case.

Then I want to find $x,y$ such that $$x\langle1,3,7\rangle+y\langle1,4,10\rangle=\langle2,3,5\rangle$$ and this gives me three more equations $$x+y=2$$ $$3x+4y=3$$ $$7x+10y=5$$ However, we have three equations and two unknowns so using Cramer's rule with any two of the three equations will give the correct values for $x,y$. I'll use the first two equations. $$x=\frac{\begin{vmatrix}2&1\\3&4\end{vmatrix}}{\begin{vmatrix}1&1\\3&4\end{vmatrix}}=5$$ $$y=\frac{\begin{vmatrix}1&2\\3&3\end{vmatrix}}{\begin{vmatrix}1&1\\3&4\end{vmatrix}}=-3$$

So then $$2a+3b+5c=14(5)+17(-3)=19.$$

Problem

Given $$a+2b+3c=4$$ $$5a+6b+7c=8$$ $$9a+c=2$$ $$-3a+14b+5c=6,$$ find $$7a+8b+9c.$$

However, I believe that has no solution.

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1 Answer

up vote 1 down vote accepted

Yes, your system of four equations is inconsistent (i.e. has no solution). The unique solution of the first three is $a=0,b=-1,c=2$, which violates the fourth.

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Yeah I thought so. Since there were enough equations to solve for the cost of each item individually, there wasn't any real need to try and find a combination of them that would sum up to the final state. Thanks for the sanity check. –  miles t Oct 25 '12 at 10:28
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