§ 3.3 Systems of Linear Equations in Three Variables

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. Blitzer, Intermediate Algebra, 5e – Slide #2 Section 3.3

Solving Linear Systems in 3 Variables by Eliminating Variables 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. Blitzer, Intermediate Algebra, 5e – Slide #3 Section 3.3

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

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

Systems of Equations in 3 Variables EXAMPLE Solve the system: x + y + z = 4 Equation 1 x - 2y - z = 1 Equation 2 2x - y - 2z = -1 Equation 3 SOLUTION 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. Blitzer, Intermediate Algebra, 5e – Slide #6 Section 3.3

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

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

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

Systems of Equations in 3 Variables CONTINUED 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. x - 2y - z = 1 An equation in three variables x – 2(-1) – (3) = 1 Replace z with 3 and y with -1 x + 2 – 3 = 1 Multiply x -1 = 1 Simplify x = 2 Add 1 to both sides Blitzer, Intermediate Algebra, 5e – Slide #10 Section 3.3

Systems of Equations in 3 Variables CONTINUED 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)}. Blitzer, Intermediate Algebra, 5e – Slide #11 Section 3.3

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

Systems of Equations in 3 Variables EXAMPLE 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”. Blitzer, Intermediate Algebra, 5e – Slide #13 Section 3.3

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

Systems of Equations in 3 Variables CONTINUED The third sentence says: 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. The corresponding equation is: (2x + 3y) – z = 1 2x + 3y - z = 1 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. Blitzer, Intermediate Algebra, 5e – Slide #15 Section 3.3

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

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

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

Systems of Equations in 3 Variables CONTINUED 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. x - 3y + 3z = 2 An equation in three variables x – 3(2) + 3(3) = 2 Replace y with 2 and z with 3 x – 6 + 9 = 2 Multiply x + 3 = 2 Add x = -1 Subtract 3 from both sides Blitzer, Intermediate Algebra, 5e – Slide #19 Section 3.3

Systems of Equations in 3 Variables CONTINUED 5) Check. Therefore, the potential solution is (-1,2,3) To check, we will plug the point into each equation. 3x + y + 2z = 5 x - 3y + 3z = 2 2x + 3y - z = 1 ? ? ? 3(-1) + (2) + 2(3) = 5 (-1) – 3(2) + 3(3) = 2 2(-1) + 3(2) – (3) = 1 ? ? ? -3 + 2 + 6 = 5 -1 – 6 + 9 = 2 -2 + 6 – 3 = 1 true 5 = 5 true 2 = 2 true 1 = 1 Therefore, the solution is (-1,2,3). Blitzer, Intermediate Algebra, 5e – Slide #20 Section 3.3