Chapter 2 Linear Systems in Two Variables and Inequalities with Applications Section 2.2

1. Chapter 2 Linear Systems in Two Variables and Inequalities with Applications Section 2.2

2. Section 2.2Substitution and Elimination • Solving a Linear System in Two Variables by Substitution • Solving a Linear System in Two Variables by Elimination • Algebraic Modeling with Linear Systems • Identifying Inconsistent and Dependent Systems Algebraically

3. Solving a Linear System with Two Variables by Substitution • Choose one of the equations and solve it for one of the variables in terms of the other. • Substitute the expression from step (1) into the other equation. This will result in an equation with just one variable. Solve for the variable. • Substitute the value found in step (2) into either of the original equations to find the remaining variable. • Write the answer as an ordered pair and check the solution to the system in both original equations. • Note: When solving an application problem, state the answer in the context of the problem.

4. Solve the system by substitution: • 5 + y = –4x • 3y – 6x = 21 • Solve “5 + y = –4x” for y. • y = –4x – 5 • (2) Substitute y = –4x – 5 into 3y– 6x = 21 and solve for x: • 3(–4x – 5)– 6x = 21 • –12x – 15 – 6x = 21 • –18x = 36 • x = –2 • (continued on the next slide)

5. (Contd.) • Solve the system by substitution: • 5 + y = –4x • 3y – 6x = 21 • (3) Back-substitute x = –2 into any of the original equations to find the value of y. • Hint: Using “y = –4x – 5” is a good choice because “y” is already isolated. • y = –4x – 5 • y = –4(–2) – 5 • y = 3 • (4) The solution to this system is (–2, 3). • Check the solution: • 5 + y = –4x 5 + 3 = –4(–2) True • 3y – 6x = 21 3(3) – 6(–2) = 21 True

6. Vlad does not manage his finances well, and during the last years the bank has been increasing the interest rates on his 2 credit cards. His credit cards interest rates are 18% and 26.5%. Vlad has a total balance of $5400 on both cards and he carries $1167.50 in interest, excluding other fees. How much balance does he have on each card? Let x = amount owed on the 18% interest rate card y = amount owed on the 26.5% interest rate card Equation for total balance: x + y = 5400 Equation for interest charged: 0.18x + 0.265y = 1167.50 (1) Solve the first equation for x. (Note: Can solve for y instead) x + y = 5400 x = 5400 – y (continued on next slide)

7. (Contd.) (2) Substitute “x = 5400 – y " into the second equation and solve for y. 0.18x + 0.265y = 1167.50 0.18(5400 – y) + 0.265y = 1167.50 972 – 0.18y + 0.265y = 1167.50 972 + 0.085y = 1167.50 0.085y = 195.5 y = 2300 (3) Back-substitutey = 2300 into any of the original equations to find x. x + 2300 = 5400 x = 3100 (4) The solution to the system is (3100, 2300).Vlad has a balance of $3,100 on the 18% interest rate card and $2,300 on the 26.5% interest rate card.

8. Solving a Linear System with Two Variables by Elimination • Eliminate one of the variables by using the concept of "opposites." If necessary, multiply one or both equations by a nonzero constant that will produce coefficients of either x or y that are opposites of each other. • Add the two equations, making sure that one of the variables drops out, leaving one equation and only one unknown. • Solve the resulting equation for the variable. • Back-substitute the value found in (3) into one of the original equations to find the value of the remaining variable. • Write the answer as an ordered pair and check the solution to the system in both original equations. • Note: When solving an application problem, state the answer in the context of the problem.

9. Solve the system by elimination: 3y – 6x = 15 2y + 4x = –14 Step 1: Eliminate one of the variables by using the concept of "opposites.” Multiply both sides of the first equation by 2. Multiply both sides of the second equation by –3. (Note: Can choose to multiply first equation by 4 and second equation by 6 instead.) 2(3y – 6x = 15)  6y – 12x = 30 –3(2y + 4x = –14)  –6y – 12x = 42 Make sure you multiply all the terms throughout the equations! (continued on next slide)

10. (Contd.) Solve the system by elimination: 3y – 6x = 15 2y + 4x = –14 Steps 2-3: Add the equivalent equations to eliminate one of the variables; solve for the remaining variable. 6y – 12x = 30 –6y – 12x = 42 0y – 24x = 72 x = –3 Step 4: Back-substitute x = –3 into any of the original equations to find the value of y. 3y – 6(–3) = 15 3y + 18 = 15 y = –1 Step 5: The solution to the system is (–3, –1). You can check the answer.

11. A group of fitness club members felt like "breaking the rules" for one day, and enjoyed delicious desserts after their workout. Two had chocolate cake and the other four ordered special lemon pie. Two chocolate cakes and four special lemon pies cost $27.90. If four club members had ordered chocolate cake and two had chosen the special lemon pie, the cost would have been $29.70. Establish a system of linear equations and solve it by elimination to find the price of each of these desserts. Let c = chocolate cake p = special lemon pie Equation for 2 cakes and 4 pies: 2c + 4p = 27.90 Equation for 4 cakes and 2 pies: 4c + 2p = 29.70 (continued on next slide)

12. (Contd.) 2c + 4p = 27.90 4c + 2p = 29.70 Step 1: Multiply both sides of the first equation by –2. – 2(2c + 4p = 27.90)  –4c – 8p = –55.80 Make sure you multiply all the terms throughout the equation! Steps 2-3: Add the equivalent equation from (1) to the second equation of the system, and solve for p. –4c – 8p = –55.80 4c + 2p = 29.70 0c – 6p = –26.10 p = 4.35 (continued on next slide)

13. (Contd.) 2c + 4p = 27.90 4c + 2p = 29.70 Step 4: Back-substitute p = 4.35 into any of the original equations to find c. 2c + 4(4.35) = 27.90 2c + 17.4 = 27.90 2c = 10.5 c = 5.25 Step 5: The solution to the system is (5.25, 4.35). This means that each chocolate cake was $5.25 and each special lemon pie was $4.35.

14. Analyze the following system: • 4x + 5y = 97 • 8x – 5y = –35 • Which of the following statements is correct? • a. Solving by substitution is more convenient. • b. Solving by elimination is more convenient. • Since 5y and -5y are opposites, these y-terms would be • eliminated immediately, allowing us to solve for x. • Answer: b • Note: Solving by substitution requires to leave a variable by • itself on one side of the equation; in this case it would result in • fractions, making the solution process more laborious.

15. Identifying Inconsistent and Dependent Systems Algebraically If both variables of a 2 x 2 linear system are eliminated when solving algebraically, this implies one of the following: Case 1:  If the resulting equation represents a false statement, the system is inconsistent; the graph will display parallel lines, thus the system has no real solution. Example 2x – 6y = 1 –2x + 6y = 8 0 = 9 False statement! No x and y values will satisfy both equations simultaneously.

16. Identifying Inconsistent and Dependent Systems Algebraically Case 2:  If the resulting equation represents an identity, the system is dependent (coincident lines, infinitely many solutions). Example 3x – 5y = –8 –3x + 5y = 8 0 = 0 Identity! All x and y values that satisfy one equation will also satisfy the other equation.

17. Using your textbook, practice the problems assigned by your instructor to review the concepts from Section 2.2.