Download presentation
Presentation is loading. Please wait.
Published byBertram Fletcher Modified over 6 years ago
1
Solving Systems of Equations in Three Variables
Section 12.2 Solving Systems of Equations in Three Variables
2
Objectives Determine whether an ordered triple is a solution of a system Solve systems of three linear equations in three variables Solve systems of equations with missing variable terms Identify inconsistent systems and dependent equations
3
Objective 1: Determine Whether an Ordered Triple Is a Solution of a System
A linear equation in three variables is an equation that can be written in the form Ax + By + Cz = D where A, B, C, and D are real numbers and A, B, and C are not all 0. The equation x − 5y + 7z = 10, where each variable is raised to the first power, is an example of a linear equation in three variables. A solution of a linear equation in three variables is an ordered triple of numbers of the form (x, y, z) whose coordinates satisfy the equation. For example, (2, 0, 1) is a solution of x + y + z = 3 because a true statement results when we substitute 2 for x, 0 for y, and 1 for z: = 3. A solution of a system of three linear equations in three variables is an ordered triple that satisfies each equation of the system.
4
Objective 1: Determine Whether an Ordered Triple is a Solution of a System
The graph of an equation of the form Ax + By + Cz = D is a flat surface called a plane. A system of three linear equations with three variables is consistent or inconsistent, depending on how the three planes corresponding to the three equations intersect. The following illustration shows some of the possibilities. A system of three linear equations in three variables can have exactly one solution, no solution, or infinitely many solutions.
5
EXAMPLE 1 Determine whether (–4, 2, 5) is a solution of the system: Strategy We will substitute the x-, y-, and z-coordinates of (–4, 2, 5) for the corresponding variables in each equation of the system. Why If each equation is satisfied by the x-, y-, and z-coordinates, the ordered triple is a solution of the system.
6
EXAMPLE 1 Determine whether (–4, 2, 5) is a solution of the system: Solution We substitute – 4 for x, 2 for y, and 5 for z in each equation. Since (–4, 2, 5) satisfies each equation, it is a solution of the system.
7
Objective 2: Solve Systems of Three Linear Equations in Three Variables
To solve a system of three linear equations in three variables means to find all of the solutions of the system. Solving such a system by graphing is not practical because we would need a coordinate system with three axes. The best way to solve systems of three linear equations in three variables is usually the elimination method. Write each equation in standard form Ax + By + Cz = D and clear any decimals or fractions. Pick any two equations and eliminate a variable. Pick a different pair of equations and eliminate the same variable as in step 1. Solve the resulting pair of two equations in two variables. To find the value of the third variable, substitute the values of the two variables found in step 4 into any equation containing all three variables and solve the equation. Check the proposed solution in all three of the original equations. Write the solution as an ordered triple.
8
Solve the system: EXAMPLE 2 Strategy Since the coefficients of the z-terms are opposites in the second and third equations, we will add the left and right sides of those equations to eliminate z. Then we will choose another pair of equations and eliminate z again. Why The result will be a system of two equations in x and y that we can solve by elimination.
9
Solve the system: EXAMPLE 2 Solution Step 1: We can skip step 1 because each equation is written in standard form and there are no fractions or decimals to clear. We will number each equation and move to step 2. Step 2: If we pick equations 2 and 3 and add them, the variable z is eliminated.
10
Solve the system: EXAMPLE 2 Solution Step 3: We now pick a different pair of equations (equations 1 and 3) and eliminate z again. If each side of equation 3 is multiplied by 2, and the resulting equation is added to equation 1, z is eliminated. Step 4: Equations 4 and 5 form a system of two equations in x and y.
11
Solve the system: EXAMPLE 2 Solution To solve this system, we multiply equation 4 by –5 and add the resulting equation to equation 5 to eliminate y. To find y, we substitute 3 for x in any equation containing x and y (such as equation 5) and solve for y:
12
Solve the system: EXAMPLE 2 Solution Step 5: To find z, we substitute 3 for x and 2 for y in any equation containing x, y, and z (such as equation 1) and solve for z: Step 6: To verify that the solution is (3, 2, 1), we substitute 3 for x, 2 for y, and 1 for z in the three equations of the original system. The solution set is written as {(3, 2, 1)}. Since this system has a solution, it is a consistent system.
13
Objective 3: Solve Systems of Equations with Missing Variable Terms
When one or more of the equations of a system is missing a variable term, the elimination of a variable that is normally performed in step 2 of the solution process can be skipped.
14
Solve the system: EXAMPLE 3 Strategy Since the third equation does not contain the variable y, we will work with the first and second equations to obtain another equation that does not contain y. Why Then we can use the elimination method to solve the resulting system of two equations in x and z.
15
Solve the system: EXAMPLE 3 Solution Step 1: We use the addition property of equality to write each equation in the standard form Ax + By + Cz = D and number each equation. Step 2: Since equation 3 does not have a y-term, we can skip to step 3, where we will find another equation that does not contain a y-term.
16
Solve the system: EXAMPLE 3 Solution Step 3: If each side of equation 2 is multiplied by 2 and the resulting equation is added to equation 1, y is eliminated. Step 4: Equations 3 and 4 form a system of two equations in x and z:
17
Solve the system: EXAMPLE 3 Solution To solve this system, we multiply equation 3 by –5 and add the resulting equation to equation 4 to eliminate x: To find x, we substitute 1 for z in equation 3.
18
Solve the system: EXAMPLE 3 Solution Step 5: To find y, we substitute –1 for x and 1 for z in equation 1: The solution of the system is (−1, 5, 1) and the solution set is {(−1, 5, 1)}. Step 6: Check the proposed solution in all three of the original equations.
19
Objective 4: Identify Inconsistent Systems and Dependent Equations
When the equations in a system of two equations with two variables are dependent, the system has infinitely many solutions. This is not always true for systems of three equations with three variables. In fact, a system can have dependent equations and still be inconsistent. The following illustration shows the different possibilities.
20
Solve the system: EXAMPLE 6 Strategy Since the coefficients of the b-terms are opposites in the second and third equations, we will add the left and right sides of those equations to eliminate b. Then we will choose another pair of equations and eliminate b again. Why The result will be a system of two equations in a and c that we can attempt to solve by elimination.
21
Solve the system: EXAMPLE 6 Solution We add equations 2 and 3 of the system to eliminate b. We can multiply both sides of equation 1 by 2 and add the resulting equation to equation 2 to eliminate b again: Equations 4 and 5 form a system in a and c.
22
Solve the system: EXAMPLE 6 Solution If we multiply both sides of equation 5 by –1 and add the result to equation 4, the terms involving a and c are both eliminated. The false statement 0 = –1 indicates that the system has no solution and is, therefore, inconsistent. The solution set is Ø.
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.