Solving Systems of Linear Equations Three Equations…Three Unknowns

Solving a System of Three Linear Equations Problem 12 Page 317 (1) 4y + 3z = -2 (2) 5x - 4y = 1 (3) -5x + 4y + z = -3 Find Two Equations to Add Together to Make a Variable Drop Out. You may have to Multiply One Or Two of the Equations by A Factor. In this case you do no. Just Add Equations 2 and 3 together.

Solving a System of Three Linear Equations Problem 12 Page 317 (2) 5x - 4y = 1 (3) -5x + 4y + z = -3 (4) Z = -2 We Got Lucky Here, Both x and y Dropped Out, So We Already have z = -2 Now Substitute z = -2 into Equation (1) Which Only Has y and z. We will be able to solve for y

Solving a System of Three Linear Equations Problem 12 Page 317 (1) 4y + 3z = -2 (4) z = -2 Substituting z = -2 into Equation (1) we get: (1) 4y + 3(-2) = -2 4y + -6 = -2 4y = 4 y = 1 Now Substitute y = 1 into Equation (2) to solve for x

Solving a System of Three Linear Equations Problem 12 Page 317 Substiituting y = 1 into Equation (2) (2) 5x - 4y = 1 5x - 4(1) = 1 5x - 4 = 1 5x = 5 x = 1 Now we have all three variables lets use Equation (3) to check:

Solving a System of Three Linear Equations Problem 12 Page 317 x= 1, y = 1, z = -2 (3) -5x + 4y + z = -3 -5(1) + 4(1) + (-2) = -3 -5 + 4 - 2 = -3 -1 - 2 = -3 -3 = -3 CHECK!

Solving a System of Three Linear Equations Problem 24 Page 318 (1) x + y + z = 9 (2) 3x - y + z = -1 (3) -2x + 2y - 3z = -2 Two Times We Will Have To Find Two Equations to Add Together to Make The Same Variable Drop Out. You may have to Multiply One or Two of the Equations by A Factor. Let’s Make y’s Drop Out…

Solving a System of Three Linear Equations Add Equations (1) and (2), and y’s will drop out (1) + (2) (1) x + y + z = 9 (2) 3x - y + z = -1 (4) 4x + 2z = 8 So we have Equation 4 with just x and z in it. We will go back to our original 3 equations and use two other equations to make y drop out.

Solving a System of Three Linear Equations (1) x + y + z = 9 (2) 3x - y + z = -1 (3) -2x + 2y - 3z = -2 If we multiply equation (2) by 2 we can add it with equation (3)

Solving a System of Three Linear Equations 2*(2) + (3) 2*(2) 6x - 2y + 2z = -2 (3) -2x + 2y - 3z = -2 (5) 4x - z = -4 Now Let’s Look at Equations (4) and (5) and Solve the System of 2 Linear Equations

Solving a System of Three Linear Equations (4) 4x + 2z = 8 (5) 4x - z = -4 Again we have a choice whether to get rid of x or z. If I wanted to get rid of x I just have to multiply one of the equations by -1 if I want to get rid of z I would just have to multiply equation (5) by 2. Guess I will get rid of x by multiplying equation (5) by -1

Solving a System of Three Linear Equations (4) 4x + 2z = 8 -1*(5) -4x + z = 4 (6) 3z = 12 z = 4 Substituting z = 4 into Equation (4), We will solve for x

Solving a System of Three Linear Equations Substitute z = 4 into Equation (4) to Solve for x: (4) 4x + 2z = 8 4x + 2(4) = 8 4x + 8 = 8 4x = 0 x = 0 Now we will substitute x = 0 and z = 4 into Equation (1) to solve for y

Solving a System of Three Linear Equations Substitute x = 0 and z = 4 into Equation (1) to solve for y (1) x + y + z = 9 (0) + y + (4) = 9 y = 5 Looks like we have solved it! Now Use Equation (2) to check by substituting x = 0, y = 5 and z = 4

Solving a System of Three Linear Equations Substitute x = 0, y = 5 and z = 4 into Equation (2) to check: (2) 3x - y + z = -1 3(0) - (5) + (4) = -1 -5 + 4 = -1 -1 = -1 CHECK We have found a solution! (0,5,4)