3 by 3  Linear Systems     Geometry of 3X3 systems Recall that for lines, either they intersect in a point, are parallel, or are the same line.  Similarly, if we have three planes either they intersect in a point, a line, don't intersect at all, or are the same planes.  Therefore when we solve three linear equations and three unknowns, we can expect that the solution is a point, a line, no solution, or a plane.    Step by step rules for the elimination method for solving a three by three linear system Example:           x + 2y - z  = 2                Equation 1         2x + 2y + 2z = 12          Equation2         x - y + 2z = 5                 Equation 3   Step 1:  Choose the most convenient variable to eliminate     In our example any seem convenient.   Step 2:  Use any two of the equation to eliminate the variable.  Then use a different pair to eliminate the same variable.  The result is a 2 by 2 system. Multiply equation 1 by two and subtract from equation 2:         2x + 2y + 2z  = 12         2x + 4y - 2z   =  4  -                                                                   -2y + 4z  = 8 and Multiply equation 3 by 2 and subtract from equation 2         2x + 2y + 2z  = 12         2x - 2y + 4z   =  10  -                                                                    4y - 2z   =   2   Step 3:  Use elimination to solve the system of the two equations that you found.         2y + 4z  = 8         4y - 2z = 2 We multiply the second equation by 2 and add it to the first equation:         -2y + 4z  =   8          8y - 4z    =  4  +                                                                   6y    =  12         y = 2 Re-substituting:         -2(2) + 4z  =  8  so that          4z = 12,     z = 3   Step 4:  Substitute the two values into any of the three equations the get the third value.         x + 2(2) - (3)  =  2  so          x = 1   Step 5:  Substitute all three values into the three equations to check your work.   Step 6:  Reread the question and answer it.   Exercise Solve         3x -2y + 5z  =  2         4x - 7y - z  =  19         5x - 6y + 4z  =  13 Application Example:   Chris invests \$2,200 into three accounts that pay 6%, 8% and 9% in annual interest. He has three times as much invested at 9% as he does at 6%.  If his total interest for the year is \$178, how much is invested at each rate?    Solution: Let          x = the amount in the 6% account         y = the amount in the 8% account         z = the amount in the 9% account then         x + y + z   =  2,200         3x - z  =  0         .06x + .08y + .09z  =  178   Eliminate y y is already eliminated in equation 2, then take          third - .08(first)  gives         .06x + .08y + .09z   =  178         .08x + .08y + .08z   =  176   -                                                                    -.02x           + .01z   =   2  or          -2x + z = 200 Adding equation 2 to the result gives                  3x - z   =   0                 -2x + z  =   200   +                                                                         x      =     200           3x - z  =  0  gives          z   =   600         200 + y + 600 = 2200  gives         y = 1400 ok. Chris has 200 invested in the 6% account, 1400 invested in the 8% account, and 600 invested in the 9% account.   Back to the Intermediate Algebra (Math 154) Home Page e-mail Questions and Suggestions