6.2 Solving Systems of Linear Equations 455 We have investigated three ways to solve a system of equations. First, when there are only two variables, it is often simplest to use the substitution method. However, when there are more than two variables, it is generally easier to convert the system of equations into an augmented matrix and apply elementary row operations. Second, we could perform Gaussian elimination and stop when the matrix is in row- echelon form. The solution is then found using back-substitution. Third, we could use Gauss-Jordan elimination. In this case, we continue performing row operations until the matrix is in reduced row-echelon form. We can then read off the solution as the augmentation. Exercise Set 6.2 Solve the following system of equations using the sub- stitution method. 1. x + 2y = 5 2x - y = 0 z-14 3.5w- 2w+32=26 5. st=7 -2s+2t -5 2.3x-2y=20 4x - y = 35 5s -8 3r2s=1 4.-2r 6.) 4a 12a - b=2 3b=6 22. x+2y z 6 -2x +2= 3 x-y =-4 7. x-y- 7 -2x+2y=-14 2 8. 4xy 12x-3y = 5 9.-16. Solve Exercises 1-8 using Gaussian elimination. Solve using Gaussian elimination. 17. x y + 3z=-1 +6x=37 2y + z = -2 18. -1.5yz= 5 -x-1.5y-2.52 = 7 y - 0.52 = -2 19. 7x-y-9z = 1 2x -4z=-4 -4x + 6z=-3 20. y + 32 = 2 x + y + 6z=0 x +2z = 4 21. 2x 2y+32= 3 4x-3y+32 2 -x+y2=4 23.-28. Solve the systems in Exercises 17-22 using Gauss-Jordan elimination. Solve using Gaussian elimination. 29. x+y+2z- 5 x+y+ z = -10 2x + 3y+4x= 2 30. 3xy 2z--10 -2xy +32= 14 5x 3y 3z=-29 31. x + y-2z = 4 4x + 7y+3z 3 14x+23y+5z - 17 32. x + y - 2z = 4 4x + 7y+32= 3 14x+23y+5z = 10 33. xy+ 32 = 2 2x + 3y z 5 -x-9y+11z = 1 34. xy+3z= 2 35. 2x + 3y - z = 5 -x-9y+11z=-4 x-2y 52 0 2x+3y+15z = 0 -2xy8z= 1