1. Chapter Seven Linear Systems and Matrices

2. Ch. 7 Overview Non-linear Systems of Equations Systems of Linear Equations in Two Variables Multivariable Linear Systems Matrices and Systems of Equations

3. Ch. 7 Overview (cont.) Operations with Matrices The Inverse of a Square Matrix The Determinant of a Square Matrix Applications of Matrices and Determinants

4. 7.1 – Solving Systems including Non- linear & Applications Systems of equations The number of equations must equal the number of unknown variables if a solution is to be found. A solution of a system is an ordered pair that satisfies each equation

5. A system of equations is a collection of two or more equations, each containing one or more variables. A solution of a system of equations consists of values for the variables that reduce each equation of the system to a true statement. When a system of equations has at least one solution, it is said to be consistent; otherwise it is called inconsistent. To solve a system of equations means to find all solutions of the system.

6. 7.1 – The Method of Substitution 1.Solve one of the equations for one variable. 2.Substitute into the other equation. 3.Solve the equation you substituted into (it will now only be in one variable) 4.Back-substitute into the first equation. 5.Check the solution(s).

7. Example 1 Solve the system algebraically. 2x + y = 6 -x + y = 0

8. Other things that can happen 1.No-Solution Case The graphs don’t intersect. 2.More than One Solution Case The graphs intersect in more than one place.

9. Example 2 Solve the system with substitution. y = x + 4 y= -x 2 + 3

10. Solution Example 2 Substituting x + 4 for y in the second equation gives: x + 4 = -x 2 + 3 x 2 + x +1 = 0 2 Imaginary solutions means that there is no REAL solution to the system. Don’t bother finding y because solutions to systems are intersection points and must be real. Graph on next page

12. Example 3 Solve the system: x 3 – y = 0 x – y = 0

13. Solution Example 3 Solving for y in the second equation yields: y = x Substituting for y in the first equation: x 3 – x = 0 x(x 2 – 1) = 0 x = 1, -1, and 0. Solutions: (1, 1), (-1, -1), (0, 0)

14. Example 4 Solve the system of equations graphically. y = ln(x) x + y = 1 (1,0) Make sure to check graphical solutions in the equations!

15. Slide 6.1 - 15 Copyright © 2010 Pearson Education, Inc. Example 5 Solve the system of equations: Solution: If we multiply the first equation by 2 and the second equation by –1, then add, the x variable is eliminated.

16. Slide 6.1 - 16 Copyright © 2010 Pearson Education, Inc. Example 5 Solution continued Solve 2y 2 + y – 1 = 0 for y:

17. Slide 6.1 - 17 Copyright © 2010 Pearson Education, Inc. Example 5 Solution continued Solving x 2 + y 2 = 4 for x results in

18. Slide 6.1 - 18 Copyright © 2010 Pearson Education, Inc. Example 5 Solution continued Thus there are four solutions:

19. You Try: 1. Solve the system. 3x + 2y = 14 and x – 2y = 10.

20. You Try: Ex. Find all points of intersection. 4x – y – 5 = 0 and 4x 2 – 8x + y + 5 = 0

21. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 21

22. You Try: 3. Solve the system graphically. 3x + 2y = 6 and y = ln(x-1)

23. Homework Day 1 7.1 pg.459 9-49 odd, skip #13

24. 7.1 Day 2 Applications

25. Section 7-1 Application Problems Write a system of equations to represent the situation. Solve the system algebraically. Find the dimensions of a rectangle if its perimeter is 280 centimeters and its width is 20 centimeters less than its length.

26. P=280 P=2l+2w W=l-20 280=2l+2(l-20) solve for l, substitute to find w. length = 80 cm, width = 60 cm.

27. Section 7-1 Application Problems Write a system of equations to represent the situation. Solve the system algebraically. Find the dimensions of a rectangular tract of land if its perimeter is 40 miles and its area is 96 square miles. Dimensions: 8x12

28. Section 7-1 Application Problems The sum of two numbers is 31. Their product is 234. Write and solve a system to find the numbers. 13 and 18

29. Section 7-1 Application Problems A business invests $10,000 in equipment. Each unit manufactured costs 65 cents to produce and sells for $1.20. How many units need to be sold for the business to break even? *Break even is the point where revenue = costs. 18,182 units

30. Section 7-1 Application Problems $30,000 is invested in 2 funds paying 4% and 6% simple interest. The investor wants a yearly interest check of $1560. How much should be invested in each fund? $12,000 in the 4% fund, 18,000 in the 6% fund.

31. Section 7-1 Application Problems You have been offered 2 jobs selling electronics. Job A offers a salary of $2,000 per month plus 2% commission on sales. Job B offers $1400 per month plus 5% commission on sales. If you know you can sell at least $25,000 of electronics per month, which job should you take? What amount would you have to sell for it to make no difference? Job B, $20,000

32. Section 7-1 Application Problems A company has fixed monthly manufacturing costs of $12,000, and it costs $0.95 to produce each unit. The company then sells each unit for $1.25. How many units must be sold before this company breaks even? 40,000 units

33. Section 7-1 Application Problems At each game, a ballpark earns $800 total from concession sales. The ballpark must pay a total of $2500 to the teams and $1000 to ballpark employees. Each fan at the game is given a free bobble-head which costs the ballpark $3 per bobble. Each ticket costs $12. How many tickets must the ballpark sell in order to break even? 300 tickets

34. Homework Day 2 7.1 pg.460 51-77 odd, skip #73

35. Section 7-1 Day 2 - HWQ Solve the system algebraically:

36. HWQ Solution