SOLUTION: Given three positive integers a, b, and c, that satisfy both 2a + 3b + 4c = 25 and 4a + 3b + 2c = 35. Find all such ordered triples (a,b,c).

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Question 251665: Given three positive integers a, b, and c, that satisfy both 2a + 3b + 4c = 25 and
4a + 3b + 2c = 35. Find all such ordered triples (a,b,c).
Answer by jsmallt9(3758)   (Show Source): You can put this solution on YOUR website!
Graphically our two equations represent planes in 3 dimensional space. Two planes intersect each other in a line (if they are not parallel). All the points on this line have coordinates which satisfy both equations. Lines have an infinite number of points so there are an infinite set of ordered triples which will satisfy both equations. Since we can't list an infinite set of ordered triples we are looking for a way to express a formula which can be used to express them.

To find this formula we can
  1. Solve the first equation for a:
    Add negative 3b and negative 4c to each side:
    2a = -3b + (-4c) + 25
    Multiply both sides by 1/2:

    or
  2. Substitute this for a in the second equation:


  3. Solve this for b:


  4. Substitute this in for b in the equation where we had solved for a (Step #1):
  5. Simplify:


    6. We now have expressed two of the three coordinates, a and b, in terms of the third, c.

So the ordered pairs which satisfy both equations are of the form:
(c+5, -2c+5, c) where c can be any real number.

P.S. You also could have solved for a and c in terms of b or for b and c in terms of a.



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