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# Help me do sums.

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# Help me do sums.

September 24, 2008 8:17 AM Subscribe

Stupidly simple math question.

X = A + B

Y = C + B

Assuming you know X and Y, is it possible to determine the values for A, B, and C?

If not, which values can you determine?

X = A + B

Y = C + B

Assuming you know X and Y, is it possible to determine the values for A, B, and C?

If not, which values can you determine?

Two equations and three unknowns. You can't determine A, B, or C.

Simple example:

10 = A + B

12 = C + B

Possible solutions: (B = 1; A = 9; C = 11), (B = 2; A = 8; C = 10), etc etc etc. Unlimited.

You can determine the value of (C - A); it's (Y - X). That's it.

posted by Perplexity at 8:23 AM on September 24, 2008

Simple example:

10 = A + B

12 = C + B

Possible solutions: (B = 1; A = 9; C = 11), (B = 2; A = 8; C = 10), etc etc etc. Unlimited.

You can determine the value of (C - A); it's (Y - X). That's it.

posted by Perplexity at 8:23 AM on September 24, 2008

on first glace no.. because it would produce the simultanteous equation:

x - a = y - c

for which there will be infinite values for x, y, a and c. I can't see how any amount of substitution and re-arranging the equation is going to escape that.

But I am no mathematician... so tell me I am wrong :)

posted by TheOtherGuy at 8:25 AM on September 24, 2008

x - a = y - c

for which there will be infinite values for x, y, a and c. I can't see how any amount of substitution and re-arranging the equation is going to escape that.

But I am no mathematician... so tell me I am wrong :)

posted by TheOtherGuy at 8:25 AM on September 24, 2008

Unless there is other information such as "all variables are limited to positive, nonzero integers" or some such limitation, there is no unique solution and you can't determine anything at all about A, B or C.

posted by rocket88 at 8:26 AM on September 24, 2008

posted by rocket88 at 8:26 AM on September 24, 2008

There is no way to solve this. Reasoning it out with example values:

A + B = 10

C + B = 20

How exactly would you go about solving this? B could be 9. Or 10. Or 0.5. Or -50, whatever you want. You can pick any normal number for B and find values which will solve the equation.

There is no one solution, there is an infinity of solutions.

posted by splice at 8:27 AM on September 24, 2008

A + B = 10

C + B = 20

How exactly would you go about solving this? B could be 9. Or 10. Or 0.5. Or -50, whatever you want. You can pick any normal number for B and find values which will solve the equation.

There is no one solution, there is an infinity of solutions.

posted by splice at 8:27 AM on September 24, 2008

No. Too many variables. The only meaningful rearrangement you can get is that X - Y = A - C. Which isn't very helpful, as even if you plug in known constants for X and Y, you still have a two-variable equation with no way of solving.

Say, for example, X is 4 and Y is 10. B could be 1, 2, 3, or 4, with values for A being 3, 2, 1, 0 and C 9, 8, 7, 6 respectively. X - Y = A - C for all four of those cases, and regardless of the values of X and Y, that remains true.

Unless my algebra-fu has failed me, I don't think you can solve for either A, B, or C.

posted by valkyryn at 8:29 AM on September 24, 2008

Say, for example, X is 4 and Y is 10. B could be 1, 2, 3, or 4, with values for A being 3, 2, 1, 0 and C 9, 8, 7, 6 respectively. X - Y = A - C for all four of those cases, and regardless of the values of X and Y, that remains true.

Unless my algebra-fu has failed me, I don't think you can solve for either A, B, or C.

posted by valkyryn at 8:29 AM on September 24, 2008

To repeat what everyone else is saying, you can make A any value you want, then let B = X - A and C = Y - X + A

posted by DanSachs at 8:39 AM on September 24, 2008

posted by DanSachs at 8:39 AM on September 24, 2008

To not repeat what everyone else has said:

Once you fix X and Y, each of those two equations describes a plane in 3-dimensional space (A-B-C space). The intersection of two planes is a line (or a plane), but not a point.

The equation of the line they intersect at is X-A=B=Y-C

posted by vacapinta at 8:49 AM on September 24, 2008

Once you fix X and Y, each of those two equations describes a plane in 3-dimensional space (A-B-C space). The intersection of two planes is a line (or a plane), but not a point.

The equation of the line they intersect at is X-A=B=Y-C

posted by vacapinta at 8:49 AM on September 24, 2008

You may also be interested in Diophantine equations, which is about solving sets of equations for integer solutions, which greatly limits the # of solutions.

posted by smackfu at 7:37 PM on September 24, 2008

posted by smackfu at 7:37 PM on September 24, 2008

This thread is closed to new comments.

posted by martinX's bellbottoms at 8:22 AM on September 24, 2008