Systems of Linear Equations in Three Variables

Systems of Linear Equations in Three Variables and Their Solutions An equation such as x + 2y - 3z = 9 is called a linear equation in three variables. In general, any equation of the form Ax + By + Cz = D where A, B, C, and D are real numbers such that A, B, and C are not all 0, is a linear equation in the variables x, y, and z. The graph of this linear equation in three variables is a plane in three-dimensional space.

Text Example Show that the ordered triple (-1, 2, -2) is a solution of the system: x + 2y – 3z = 9 2x – y + 2z = -8 -x + 3y – 4z = 15. Solution Because -1 is the x-coordinate, 2 is the y-coordinate, and -2 is the z-coordinate of (-1, 2, -2), we replace x by -1, y by 2, and z by -2 in each of the three equations. x + 2y – 3z = 9 -1 + 2(2) – 3(-2) = 9 -1 + 4 + 6 = 9 9 = 9 ? true 2x – y + 2z = -8 2(-1) – 2 + 2(-2) = -8 -2 – 2 – 4 = -8 -8 = -8 ? true -x + 3y – 4z = 15 -(-1) + 3(2) – 4(-2) = 15 1 + 6 + 8 = 15 15 = 15 ? true The ordered triple (-1, 2, -2) satisfies the three equations. Thus, the solution set is {(-1, 2, -2)}.

Solving Linear Systems in Three Variables by Eliminating Variables Reduce the system to two equations in two variables. This is usually accomplished by taking two different pairs of equations and using the addition method to eliminate the same variable from each pair. Solve the resulting system of two equations in two variables using addition or substitution. The result is an equation in one variable that gives the value of that variable. Back-substitute the value of the variable found in step 2 into either of the equations in two variables to find the value of the second variable. Use the values of the two variables from steps 2 and 3 to find the value of the third variable by back-substituting into one of the original equations. Check the proposed solution in each of the original equations.

Text Example Solve the system: 3 = 3z + 5y -2x -3 2z 3y 3x 4z – 2y 5x Equation 1 Equation 2 Equation 3 Solution There are many ways to proceed. Because our initial goal is to reduce the system to two equations in two variables, the central idea is to take two different pairs of equations and eliminate the same variable from each pair. Step 1 Reduce the system to two equations in two variables. We choose any two equations and use the addition method to eliminate a variable. Let's eliminate z from Equations 1 and 2. We do so by multiplying Equation 2 by 2. Then we add equations. Equation 1 No change. Multiply by 2. -3 = 2z + 3y 3x 3 4z – 2y 5x -3 = 4y + 11x -6 4z 6y 6x 3 – 2y 5x Equation 2 Add: Equation 4

Text Example cont. Solution Now we must eliminate the same variable from another pair of equations. We can eliminate z from Equations 2 and 3. First, we multiply Equation 2 by –3. Next, we multiply Equation 3 by 2. Finally, we add equations. Equation 2 Multiply by -3. Multiply by 2. 3 = 3z + 5y -2x -3 2z 3y 3x 15 = y + -13x 6 6z 10y -4x 9 – 9y -9x Equation 3 Add: Equation 5 Equations 4 and 5 give us a system of two equations in two variables. Step 2 Solve the resulting system of two equations in two variables. We choose any two equations and use the addition method to eliminate a variable. Let's eliminate z from Equations 1 and 2. We do so by multiplying Equation 2 by 2. Then we add equations. Equation 4 No change. Multiply by -4. 15 = y + -13x -3 4y 11x -1 = x -63 63x -60 4y – 52x -3 + 11x Equation 5 Add: Divide both sides by 63.

Text Example cont. Solution Step 3 Use back-substitution in one of the equations in two variables to find the value of the second variable. We back-substitute -1 for x in either Equation 4 or 5 to find the value of y. -13x + y = 15 Equation 5  -13(-1) + y = 15 Substitute -1 for x.   13 + y = 15 Multiply.   y = 2 Subtract 13 from both sides. Step 4 Back-substitute the values found for two variables into one of the original equations to find the value of the third variable. We can now use any one of the original equations and back-substitute the values of x and y to find the value for z. We will use Equation 2. 3x + 3y + 2z = -3 Equation 2 3(-l) + 3(2) + 2z = -3 Substitute -1 for x and 2 for y.

Text Example cont. Solution 3 + 2z = -3 Multiply and then add. 2z = -6 Subtract 3 from both sides. z = -3 Divide both sides by 2.   With x = -1, y = 2, and z = -3, the proposed solution is the ordered triple (-1,2,-3). Step 5 Check. Check the proposed solution, (-1, 2, -3), by substituting the values for x, y, and z into each of the three original equations. These substitutions yield three true statements. Thus, the solution set is {(-1, 2, -3)}.

Text Example Solve the system: Solution 16 = z + 2y x 17 2z y 8 Equation 1 Equation 2 Equation 3 Solution Step 1 Reduce the system to two equations in two variables. Because Equation 1 contains only x and z, we could eliminate y from Equations 2 and 3. This will give us two equations in x and z. To eliminate y from Equations 2 and 3, we multiply Equation 2 by -2 and add Equation 3. Equation 2 Multiply by -2. No change. -3 = z + 2y x 17 2z y -18 = 3z – -x 16 z + 2y x -34 4z -2x Equation 3 Add: Equation 4 Equation 4 and the given Equation 1 provide us with a system of two equations in two variables.

Text Example cont. Solution Step 2 Solve the resulting system of two equations in two variables. We will solve Equations 1 and 4 for x and z. 5 = z -10 -2z -18 3z – -x 8 + x Equation 1 Equation 4 Add: Divide both sides by -2. Step 3 Use back-substitution in one of the equations in two variables to find the value of the second variable. To find x, we back-substitute 5 for z in either Equation 1 or 4. We will use Equation 1. x + z = 8 x + 5 = 8 x = 3 Equation 1 Substitute 5 for z. Subtract 5 from both sides.

Text Example cont. Solution Step 4 Back-substitute the values found for two variables into one of the original equations to find the value of the third variable. To find y, we back-substitute 3 for x and 5 for z into Equation 2 or 3. We can't use Equation 1 because y is missing in this equation. We will use Equation 2. x + y + 2z = 17 3 + y + 2(5) = 17 y + 13 = 17 y = 4 We found that z = 5, x = 3, and y = 4. Thus, the proposed solution is the ordered triple (3, 4, 5). Step 5 Check. Substituting 3 for x, 4 for y, and 5 for z into each of the three original equations yields three true statements. Consequently, the solution set is {(3,4,5)}.

Example Reduce the system to 2X2 -2(x - 3y + z = 1) 2x -3z = -8 Solve the system: x - 3y + z = 1 2x -3z = -8 3x +8y +2z = 1 Solution: Reduce the system to 2X2 -2(x - 3y + z = 1) 2x -3z = -8 -2x+6y-2z= -2 2x -3z= -8 6y - 5z = -10 -3(x - 3y + z = 1) 3x +8y +2z = 1 -3x +9y -3z = -3 3x +8y +2z = 1 17y - z = -2

Example cont. Solution: (-1, 0, 2) 6y - 5z = -10 -5(17y - z = -2) 6y - 5z = -10 -85y +5z = 10 -79y = 0 y = 0 6y - 5z = -10 17y - z = -2 6(0) - 5z = -10 -5z = -10 z = 2 x - 3(0) + 2 = 1 x+2 = 1 x = -1 Solution: (-1, 0, 2)

Systems of Linear Equations in Three Variables