Assignment Questions? Pg. 141-144 #13- (-2, 1) #14- (3, -4) #16- (-4, 6) #29- 250 T-Shirts #50- (3, 3) #53- No Solution #54- (-5, -3) #58- Infinite Solutions #67- (16, -8) #76- Gloria is Correct

Unit 1-5: Systems of Equations with Three Variables

What is Systems of Equations with Three Variables???? https://www.boundless.com/algebra/textbooks/boundless-algebra-textbook/systems-of-equations-and-matrices-6/systems-of-equations-in-three-variables-41/solving-systems-of-equations-in-three-variables-194-5835/

Warm Up Solve the system of equations. 5x + 3y + 2z = 2 2x + y = 18 A System with One Solution Solve the system of equations. 5x + 3y + 2z = 2 2x + y = 18 x = 6

Example 1 Solve the system of equations. 5x + 3y + 2z = 2 A System with One Solution Solve the system of equations. 5x + 3y + 2z = 2 2x + y – z = 5 x + 4y + 2z = 16 Step 1 Use elimination to make a system of two equations in two variables. 5x + 3y + 2z = 2 5x + 3y + 2z = 2 First equation Multiply by 2. 2x + y – z = 5 (+)4x + 2y – 2z = 10 Second equation 9x + 5y = 12 Add to eliminate z.

Example 1 5x + 3y + 2z = 2 First equation A System with One Solution 5x + 3y + 2z = 2 First equation (–) x + 4y + 2z = 16 Third equation 4x – y = –14 Subtract to eliminate z. Notice that the z terms in each equation have been eliminated. The result is two equations with the same two variables, x and y.

Example 1 Step 2 Solve the system of two equations. A System with One Solution Step 2 Solve the system of two equations. 9x + 5y = 12 9x + 5y = 12 Multiply by 5. 4x – y = –14 (+) 20x – 5y = –70 29x = –58 Add to eliminate y. x = –2 Divide by 29.

Example 1 A System with One Solution Substitute –2 for x in one of the two equations with two variables and solve for y. 4x – y = –14 Equation with two variables 4(–2) – y = –14 Replace x with –2. –8 – y = –14 Multiply. y = 6 Simplify. The result is x = –2 and y = 6.

Example 1 A System with One Solution Step 3 Solve for z using one of the original equations with three variables. 2x + y – z = 5 Original equation with three variables 2(–2) + 6 – z = 5 Replace x with –2 and y with 6. –4 + 6 – z = 5 Multiply. z = –3 Simplify. Answer: The solution is (–2, 6, –3). You can check this solution in the other two original equations.

Example 2 What is the solution to the system of equations shown below? 2x + 3y – 3z = 16 x + y + z = –3 x – 2y – z = –1 A. B. (–3, –2, 2) C. (1, 2, –6) D. (–1, 2, –4)

Example 3 A. Solve the system of equations. 2x + y – 3z = 5 No Solution and Infinite Solutions A. Solve the system of equations. 2x + y – 3z = 5 x + 2y – 4z = 7 6x + 3y – 9z = 15 Eliminate y in the first and third equations. 2x + y – 3z = 5 6x + 3y – 9z = 15 Multiply by 3. 6x + 3y – 9z = 15 (–)6x + 3y – 9z = 15 0 = 0

Example 3 No Solution and Infinite Solutions The equation 0 = 0 is always true. This indicates that the first and third equations represent the same plane. Check to see if this plane intersects the second plane. x + 2y – 4z = 7 6x + 12y – 24z = 42 Multiply by 6. 6x + 3y – 9z = 15 (–)6x + 3y – 9z = 15 9y – 15z = 27 Divide by the GCF, 3. 3y – 5z = 9 Answer: The planes intersect in a line. So, there is an infinite number of solutions.

Example 4 No Solution and Infinite Solutions B. Solve the system of equations. 3x – y – 2z = 4 6x – 2y – 4z = 11 9x – 3y – 6z = 12 Eliminate x in the first two equations. 3x – y – 2z = 4 6x – 2y + 4z = 8 Multiply by 2. 6x – 2y – 4z = 11 (–) 6x – 2y – 4z = 11 0 = –3 Answer: The equation 0 = –3 is never true. So, there is no solution of this system.

Example 5 A. What is the solution to the system of equations shown below? x + y – 2z = 3 –3x – 3y + 6z = –9 2x + y – z = 6 A. (1, 2, 0) B. (2, 2, 0) C. infinite number of solutions D. no solution

Example 6 B. What is the solution to the system of equations shown below? 4x + 5y – 6z = 2 –3x – 2y + 7z = -15 -x + 4y + 2z = -13 A. (-4, -3, -2) B. (-3, -2, 4) C. (-3, -2, -4) D. no solution

Assignment Pg. 165-166 #1, 4, 5, 8, 10, 12, 17, 18

Example 3 Write and Solve a System of Equations SPORTS There are 49,000 seats in a sports stadium. Tickets for the seats in the upper level sell for $25, the ones in the middle level cost $30, and the ones in the bottom level are $35 each. The number of seats in the middle and bottom levels together equals the number of seats in the upper level. When all of the seats are sold for an event, the total revenue is $1,419,500. How many seats are there in each level? Explore Read the problem and define the variables. u = number of seats in the upper level m = number of seats in the middle level b = number of seats in the bottom level

Example 3 Plan There are 49,000 seats. u + m + b = 49,000 Write and Solve a System of Equations Plan There are 49,000 seats. u + m + b = 49,000 When all the seats are sold, the revenue is 1,419,500. Seats cost $25, $30, and $35. 25u + 30m + 35b = 1,419,500 The number of seats in the middle and bottom levels together equal the number of seats in the upper level. m + b = u

Example 3 Write and Solve a System of Equations Solve Substitute u = m + b in each of the first two equations. (m + b) + m + b = 49,000 Replace u with m + b. 2m + 2b = 49,000 Simplify. m + b = 24,500 Divide by 2. 25(m + b) + 30m + 35b = 1,419,500 Replace u with m + b. 25m + 25b + 30m + 35b = 1,419,500 Distributive Property 55m + 60b = 1,419,500 Simplify.

Example 3 Now, solve the system of two equations in two variables. Write and Solve a System of Equations Now, solve the system of two equations in two variables. m + b = 24,500 55m + 55b = 1,347,500 Multiply by 55. 55m + 60b = 1,419,500 (–) 55m + 60b = 1,419,500 –5b = –72,000 b = 14,400

Example 3 Write and Solve a System of Equations Substitute 14,400 for b in one of the equations with two variables and solve for m. m + b = 24,500 Equation with two variables m + 14,400 = 24,500 b = 14,400 m = 10,100 Subtract 14,400 from each side.

Example 3 Write and Solve a System of Equations Substitute 14,400 for b and 10,100 for m in one of the original equations with three variables. m + b = u Equation with three variables 10,100 + 14,400 = u m = 10,100, b = 14,400 24,500 = u Add. Answer:

Example 3 Write and Solve a System of Equations Substitute 14,400 for b and 10,100 for m in one of the original equations with three variables. m + b = u Equation with three variables 10,100 + 14,400 = u m = 10,100, b = 14,400 24,500 = u Add. Answer: There are 24,500 upper level, 10,100 middle level, and 14,400 bottom level seats.

Example 3 Check Check to see if all the criteria are met. Write and Solve a System of Equations Check Check to see if all the criteria are met. 24,500 + 10,100 + 14,400 = 49,000 The number of seats in the middle and bottom levels equals the number of seats in the upper level. 10,100 + 14,400 = 24,500 When all of the seats are sold, the revenue is $1,419,500. 24,500($25) + 10,100($30) + 14,400($35) = $1,419,500

Example 2 BUSINESS The school store sells pens, pencils, and paper. The pens are $1.25 each, the pencils are $0.50 each, and the paper is $2 per pack. Yesterday the store sold 25 items and earned $32. The number of pens sold equaled the number of pencils sold plus the number of packs of paper sold minus 5. How many of each item did the store sell? A. pens: 5; pencils: 10; paper: 10 B. pens: 8; pencils: 7; paper: 10 C. pens: 10; pencils: 7; paper: 8 D. pens: 11; pencils: 2; paper: 12

Example 2 BUSINESS The school store sells pens, pencils, and paper. The pens are $1.25 each, the pencils are $0.50 each, and the paper is $2 per pack. Yesterday the store sold 25 items and earned $32. The number of pens sold equaled the number of pencils sold plus the number of packs of paper sold minus 5. How many of each item did the store sell? A. pens: 5; pencils: 10; paper: 10 B. pens: 8; pencils: 7; paper: 10 C. pens: 10; pencils: 7; paper: 8 D. pens: 11; pencils: 2; paper: 12

End of the Lesson