Simultaneous Equation (1 Viewer)

budj

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we needto find x+y+z

These are the equations (there are four unknowns)

x+y-z=6
x-w+z=3
xyz=140
 

Archman

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by inspection, 16 is one of the trivial answers :p (x=4, y=7, z=5, w=6)
but honestly, if your looking for non-integer solutions as well, its a pretty crappy question. because from the second line, x+z=3+w (which don't mean crap cuz w don't appear anywhere else.)
so the question is to find 3+w+y with x+y-z=6 and xyz=140
that has infinite solutions. (just set y as something that looks nice and u'll find x and z)
so 3+w+y can really be anything

bottom line is, in conventional cases, u need about 4 conditions (not 3) in order to have a decent set of solutions for 4 variables.
 

theVirus

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<FONT=Courier New><COLOR=DarkOrange>Yeah, the answer you said by inspection is the correct answer. But you only did it by trial and error. I presume that means that it can only be found out by Trial and Error? the only solutions that we are looking for are Integers. It is a question out of a Mensa book.</COLOR></FONT>
 

Affinity

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There is usually some trial and error involved in these kinds of equations for integer solutions, though archman probably solved it by atmospheric extraction. However there are some methods that tell one what to look for.
x * y * z = 140 = 2 * 2 * 5 * 7, and combined with the first equation, there aren't many cases to consider.
 
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Slidey

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Drink some ascorbic acid, then.
 

Arkad

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theVirus said:
<FONT=Courier New><COLOR=DarkOrange>Yeah, the answer you said by inspection is the correct answer. But you only did it by trial and error. I presume that means that it can only be found out by Trial and Error? the only solutions that we are looking for are Integers. It is a question out of a Mensa book.</COLOR></FONT>
Now since y=(3-w+2z), x=(3+w-z), xy=140/z (look hard at the original equations and you'll see these facts)

Hence (3-w+2z)(3+w-z) = 140/z,

Now if you're good you can solve this polynomial of degree 3 and find the solution set in terms of w, the only free variable in this question. And after you did that you can just set w as anything and find infinite amount of solutions.

Or even better just use the trial and error method if you are just trying to get a integer solution. This question is very hard I think if you're trying to go for a solution set.

Note: Ascorbic acid does not destroy free radicals, not in math anyway and schrodinger is a cat.
 
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