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§ 3.3 Systems of Linear Equations in Three Variables.

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Presentation on theme: "§ 3.3 Systems of Linear Equations in Three Variables."— Presentation transcript:

1 § 3.3 Systems of Linear Equations in Three Variables

2 Blitzer, Intermediate Algebra, 5e – Slide #2 Section 3.3 Systems of Equations in 3 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, C, and D are not all 0, is a linear equation in three variables x, y, and z. The graph of this linear equation in three variables in a plane in three dimensional space. A solution of a system of linear equations in three variables is an ordered triple of real numbers that satisfies all equations in the system.

3 Blitzer, Intermediate Algebra, 5e – Slide #3 Section 3.3 Systems of Equations in 3 Variables Solving Linear Systems in 3 Variables by Eliminating Variables 1) 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 both pairs. 2) 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. 3) 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. 4) 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. 5) Check the proposed solution in each of the original equations.

4 Blitzer, Intermediate Algebra, 5e – Slide #4 Section 3.3 Systems of Equations in 3 VariablesEXAMPLE Show that the ordered triple (2,-1,3) is a solution of the system: First Equation: SOLUTION x + y + z = 4 x - 2y - z = 1 2x - y - 2z = -1 Substitute the given values x + y + z = (-1) + 3 = 4 4 = 4Simplifytrue

5 Blitzer, Intermediate Algebra, 5e – Slide #5 Section 3.3 Systems of Equations in ApplicationCONTINUED Second Equation: Substitute the given values2 - 2(-1) - 3 = 1 1 = 1Simplify x - 2y - z = 1 Substitute the given values2(2) - (-1) – 2(3) = -1 1 = 1Simplify Third Equation: 2x - y - 2z = -1 Therefore, the ordered triple (2,-1,3) satisfies all three equations and is therefore a solution for the system of equations. true

6 Blitzer, Intermediate Algebra, 5e – Slide #6 Section 3.3 Systems of Equations in 3 VariablesEXAMPLE Solve the system: SOLUTION x + y + z = 4 x - 2y - z = 1 2x - y - 2z = -1 Equation 1 Equation 2 Equation 3 1) Reduce the system to two equations in two variables. There are multiple ways that we could proceed. We will use the addition method on equation’s 1 and 2 to eliminate x. Then we’ll eliminate the same variable, x, using equation’s 1 and 3.

7 Blitzer, Intermediate Algebra, 5e – Slide #7 Section 3.3 Systems of Equations in 3 Variables 2x - y - 2z = -1 x + y + z = 4Equation 1 Equation 3 No change Multiply by -2 -2x - 2y - 2z = -8 Add: -3y - 4z = -9 CONTINUED 2x - y - 2z = -1 x + y + z = 4 x - 2y - z = 1 Equation 1 Equation 2 No change Multiply by -1 x + y + z = 4 -x + 2y + z = -1 Add: 3y + 2z = 3

8 Blitzer, Intermediate Algebra, 5e – Slide #8 Section 3.3 Systems of Equations in 3 VariablesCONTINUED Now I have a system of equations in two variables that I can solve. 3y + 2z = 3 -3y - 4z = -9 Add: - 2z = -6 z = 3 Divide both sides by -2 Now we can use one of the two equations in two variables to solve for y. 2) Solve the resulting system of two equations in two variables.

9 Blitzer, Intermediate Algebra, 5e – Slide #9 Section 3.3 Systems of Equations in 3 VariablesCONTINUED 3y + 2z = 3 3y + 2(3) = 3 3y + 6 = 3 3y = -3 y = -1 An equation in two variables Replace z with 3 Multiply Subtract 6 from both sides Divide both sides by 3 3) Use back-substitution in one of the equations in two variables to find the value of the second variable.

10 Blitzer, Intermediate Algebra, 5e – Slide #10 Section 3.3 Systems of Equations in 3 VariablesCONTINUED 4) Back-substitute the values found for two variables into one of the original equations to find the value for the third variable. Now we can use any of the original equations to find x. Let’s use equation 2. An equation in three variables Replace z with 3 and y with -1 Multiply Simplify Add 1 to both sides x - 2y - z = 1 x – 2(-1) – (3) = 1 x + 2 – 3 = 1 x -1 = 1 x = 2

11 Blitzer, Intermediate Algebra, 5e – Slide #11 Section 3.3 Systems of Equations in 3 VariablesCONTINUED 5) Check. The previous example shows this step in its entirety. Therefore, the solution is (2,-1,3) and the solution set is {(2,-1,3)}.

12 Blitzer, Intermediate Algebra, 5e – Slide #12 Section 3.3 Systems of Equations Types of Systems of Equations ConsistentA system that has at least one solution InconsistentA system that is not consistent (has no solutions) DependentA system that has infinitely many solutions Independent A system that is not dependent (has one or no solutions)

13 Blitzer, Intermediate Algebra, 5e – Slide #13 Section 3.3 Systems of Equations in 3 VariablesEXAMPLE The following is known about three numbers: Three times the first number plus the second number plus twice the third number is 5. If 3 times the second number is subtracted from the sum of the first number and 3 times the third number, the result is 2. If the third number is subtracted from the sum of 2 times the first number and 3 times the second number, the result is 1. Find the numbers. SOLUTION First, we need to establish our variables. Let x = the “first number”. Let y = the “second number”. Let z = the “third number”.

14 Blitzer, Intermediate Algebra, 5e – Slide #14 Section 3.3 Systems of Equations in 3 VariablesCONTINUED Three times the first number plus the second number plus twice the third number is 5. Now we need to set up the equations. The first sentence says: The corresponding equation is: 3x + y + 2z = 5 The second sentence says: The corresponding equation is: (x + 3z) – 3y = 2 If 3 times the second number is subtracted from the sum of the first number and 3 times the third number, the result is 2. x + 3z – 3y = 2 x - 3y + 3z = 2

15 Blitzer, Intermediate Algebra, 5e – Slide #15 Section 3.3 Systems of Equations in 3 VariablesCONTINUED 1) Reduce the system to two equations in two variables. There are multiple ways that we could proceed. We will use the addition method on equation’s 1 and 2 to eliminate x. Then we’ll eliminate the same variable, x, using equation’s 2 and 3. The third sentence says: The corresponding equation is: (2x + 3y) – z = 1 2x + 3y - z = 1 If the third number is subtracted from the sum of 2 times the first number and 3 times the second number, the result is 1.

16 Blitzer, Intermediate Algebra, 5e – Slide #16 Section 3.3 Systems of Equations in 3 VariablesCONTINUED Equation 2 Equation 3 No change Multiply by -2 Add: 9y - 7z = -3 Equation 1 Equation 2 No change Multiply by -3 Add: 10y - 7z = -1 Equation 1 Equation 2 Equation 3 3x + y + 2z = 5 x - 3y + 3z = 2 2x + 3y - z = 1 3x + y +2z = 5 x - 3y + 3z = 2 2x + 3y - z = 1 3x + y + 2z = 5 -3x + 9y - 9z = -6 -2x + 6y - 6z = -4 2x + 3y - z = 1

17 Blitzer, Intermediate Algebra, 5e – Slide #17 Section 3.3 Systems of Equations in 3 VariablesCONTINUED Now I have a system of equations in two variables that I will solve. Add: -y = -2 Now I can use one of the two equations in two variables to solve for z. No Change Multiply by y + 7z = 1 y = 2 2) Solve the resulting system of two equations in two variables. 9y - 7z = -3 10y - 7z = -1 9y - 7z = -3

18 Blitzer, Intermediate Algebra, 5e – Slide #18 Section 3.3 Systems of Equations in 3 VariablesCONTINUED An equation in two variables Replace y with 2 Multiply Subtract 20 from both sides Divide both sides by -7 3) Use back-substitution in one of the equations in two variables to find the value of the second variable. 10y - 7z = -1 10(2) - 7z = z = z = -21 z = 3

19 Blitzer, Intermediate Algebra, 5e – Slide #19 Section 3.3 Systems of Equations in 3 VariablesCONTINUED 4) Back-substitute the values found for two variables into one of the original equations to find the value for the third variable. Now I can use any of the original equations to find c. I’ll use equation 2. An equation in three variables Replace y with 2 and z with 3 Multiply Add Subtract 3 from both sides x - 3y + 3z = 2 x – 3(2) + 3(3) = 2 x – = 2 x + 3 = 2 x = -1

20 Blitzer, Intermediate Algebra, 5e – Slide #20 Section 3.3 Systems of Equations in 3 VariablesCONTINUED 5) Check. To check, we will plug the point into each equation. (-1,2,3) true Therefore, the solution is (-1,2,3). ? ?? ?? ? Therefore, the potential solution is 3x + y + 2z = 5x - 3y + 3z = 22x + 3y - z = 1 3(-1) + (2) + 2(3) = 5(-1) – 3(2) + 3(3) = 22(-1) + 3(2) – (3) = = 5-1 – = – 3 = 1 5 = 5 2 = 2 1 = 1


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