I'm in a little bit of a pickle here, can someone tell me if it is actually possible to solve these? From the general rule of thumb, 4 variables, 4 equations should be able to right?
Is it possible to solve: xa = 4200, xb = 4410, ya=4000, yb = 4200 for x, y, a and b?
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linear-algebra
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0@Rainbow: offhand question, have you taken up calculus already? The "nuke" solution becomes less of a "nuke" if you have a lot of simultaneous equations. – 2010-11-15
1 Answers
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The best you can do is $20x = 21y$ and $21a = 20b.$ If you write your second pair of equations as $a \left( \frac{20}{21} x \right) = 4000,$ $b \left( \frac{20}{21} x \right) = 4200$ we note that they are equivalent to your first pair $ax = 4200$ $bx = 4410,$ simply by multipling up by the $20/21.$
As for your general rule of thumb, note that a system of equations can be inconsistent.
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0Okay, thanks for your help Derek. – 2010-11-15