Please help me with this. The question was not given from a book. Solve for X, Y, and Z in the following sysyems of three equations: I'll just do (b), and give the answers for the first and third. The first and third are done the same way. The first one is shorter because z has already been eliminated from the second equation so you only need to eliminate z from the 1st and 3rd equations. If you get stuck on (a) or (c), post again. a. x + 2y + z = 6 x + y = 4 3x + y + z = 8 Answer: (x, y, z) = (2, 2, 0) b. 10x + y + z = 12 8x + 2y + z = 11 20x - 10y - 2z = 8 Pick any of the three letters to eliminate first. I'll pick y. Now pick any pair of equations that contain y, to use to eliminate y. I'll pick the first two equations: 10x + y + z = 12 8x + 2y + z = 11 To eliminate y multiply the first equation thru by -2 and add the two equations term by term: -20x - 2y - 2z = -24 8x + 2y + z = 11 ----------------------- -12x - z = -13 Now pick a different pair of equations that contain y, to use to eliminate the SAME letter y. I'll pick the first and third equations: 10x + y + z = 12 20x - 10y - 2z = 8 To eliminate y multiply the first equation thru by 10 and add the two equations term by term: 100x + 10y + 10z = 120 20x - 10y - 2z = 8 ------------------------ 120x + 8z = 128 Now you have two equations in two unknowns -12x - z = -13 120x + 8z = 128 Eliminate x by multiplying the first by 10 and adding the the second: -120x - 10z = -130 120x + 8z = 128 -------------------- -2z = -2 z = 1 Substitute z = 1 in -12x - z = -13 -12x - (1) = -13 -12x = -12 x = 1 Now pick any original equation and substitute z = 1 and x = 1 I'll pick 8x + 2y + z = 11 8x + 2(1) + 1 = 11 8x + 2 + 1 = 11 8x + 3 = 11 8x = 8 x = 1 Answer: (x, y, z) = (1, 1, 1) c. 22x + 5y + 7z = 12 10x + 3y + 2z = 5 9x + 2y + 12z = 14 This one is done the same way as b. Use the above for a model to solve this one by. Answer: (x, y, z) = (0, 1, 1) Edwin