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1. Section 1 Review: Systems of Linear Equations in Two Variables Chapter 3 Systems of Linear Equations; Matrices

2. Review: Systems of Linear Equations in Two Variables If 2 adult tickets and 1 child ticket cost $32 and if 1 adult ticket and 3 child tickets cost $36, what is the price of each? Solution If x = price of an adult ticket and y = price of a child ticket, then 2 x + y = 32 x + 3 y = 36 This is a system of two linear equations in two variables. Finding ordered pairs ( x , y ) that satisfy one of the equations is not difficult. The solution to the system of equations is the set of all ordered pairs that satisfy both equations at the same time.

3. Definition Systems of Two Linear Equations in Two Variables Given the linear system ax + by = h cx + dy = k where a, , b , c , d , h , and k are real constants, a pair of numbers x = x and y = y [also written as the ordered pair ( x 0, y )] is a solution of this system if each equation is satisfied by the pair. The set of all such ordered pairs is called the solution set for the system. To solve a system is to find its solution set.

4. Methods for Solving Systems of Two Linear Equations in Two Variables We consider three methods of solving these systems of equations. Graphing Substitution Elimination by addition Each method has its advantage, depending on the situation.

5. Example 1 Solving a System by Graphing Solve the ticket problem by graphing: 2 x + y = 32 x + 3 y = 36 Solution Graphing a line by finding x and y intercepts is an easy process (when the line does not pass through the origin). First equation: 2 x + y = 32 Substitute y = 0 to find the x intercept. This gives x = 16 and the point (16, 0). Substitute x = 0 to find the y intercept. This gives y = 32 and the point (0, 32).

6. Example 1 Solving a System by Graphing continued Solve the ticket problem by graphing: 2 x + y = 32 x + 3 y = 36 Second equation: x + 3 y = 36 Substitute y = 0 to find the x intercept. This gives x = 36 and the point (36, 0). Substitute x = 0 to find the y intercept. This gives y = 12 and the point (0, 12).

7. Example 1 Solving a System by Graphing continued Graph the intercepts for the first line (16, 0) and (0, 32). Draw the line through these points. Graph the intercepts for the second line (36, 0) and (0, 12). Draw the line through these points. The point of intersection appears to be at (12, 8). (12, 8)

8. Example 1 Solving a System by Graphing continued We estimated the intersection point to be (12, 8) which would mean that the adult tickets ( x ) would cost $12 and the child tickets ( y ) would cost $8. We check these results. The values x = 12 and y = 8 check and are the solution to the system. 2 x + y = 32 2(12) + 8 = 24 + 8 = 32 x + 3 y = 36 12 + 3(8) = 12 + 24 = 36

9. Solving by Graphing Insights Graphing a system of two linear equations in two variables gives useful information about the solution set. In general, any two lines in a coordinate plane must either: intersect in exactly one point, be parallel (and never intersect), or coincide (have identical graphs). The following examples illustrate these possibilities.

10. Example 2A Solving a System by Graphing Solve the system by graphing: A. x – 2 y = 2 x + y = 5 The graph shows the single intersection point at (4, 1). There is exactly one solution to this system of equations.

11. Example 2B Solving a System by Graphing Solve the system by graphing: B. x + 2 y = –4 2 x + 4 y = 8 The lines in this system are shown in the graph and are parallel. Each line has slope –1/2. There is no solution to this system.

12. Example 2C Solving a System by Graphing Solve the system by graphing: C. 2 x + 4 y = 8 x + 2 y = 4 The lines in this system coincide as shown in the graph. There are an infinite number of solutions.

13. Definition Systems of Linear Equations: Basic Terms A system of linear equations is consistent if it has one or more solutions and inconsistent if no solutions exist. Furthermore, a consistent system is said to be independent if it has exactly one solution (often referred to as the unique solution ) and dependent if it has more than one solution. Two systems of equations are equivalent if they have the same solution set.

14. Theorem 1 Possible Solutions to a Linear System The linear system ax + by = h cx + dy = k must have (A) Exactly one solution (Consistent and independent) or (B) No solution (Inconsistent) or (C) Infinitely many solutions (Consistent and dependent) There are no other possibilities.

15. Example 3 Solving a System Using a Graphing Calculator Use a graphing calculator and graphing approximation techniques to solve the system to two decimal places. 5 x + 2 y = 15 2 x – 3 y = 16 Solution Solve each equation for y : 5 x + 2 y = 15 gives y = –2.5 x + 7.5 2 x – 3 y = 16 gives

16. Example 3 Solving a System Using a Graphing Calculator continued Enter each equation into the graphing calculator. Graph in a viewing window that shows the intersection point. Round the intersection values to two decimal places. The solution is x = 4.05 and y = –2.63, or ( 4.05, –2.63) When these values are checked, the checks are sufficiently close, but due to rounding are not exact.

17. Substitution Substitution is an algebraic method that provides exact solutions to a system of two equations in two variables, provided a solution exists. We first choose one of the equations and solve for one variable in terms of the other, making a choice that avoids fractions, if possible. We then substitute the result into the other equation and solve the resulting equation in one variable. This result is substituted into the results of the first step to find the second variable.

18. Example 4 Solving a System by Substitution Solve by substitution: 5 x + y = 4 2 x – 3 y = 5 Solution Solve the first equation for y in terms of x . 5 x + y = 4 gives y = 4 – 5 x Substitute into the second equation. 2 x – 3(4 – 5 x ) = 5 2 x – 12 + 15 x = 5 17 x = 17 x = 1

19. Example 4 Solving a System by Substitution continued Replace x with 1 in y = 4 – 5 x to find y . y = 4 – 5(1) y = 4 – 5 = –1 The solution is x = 1, y = –1 or (1, –1). Check 5 x + y = 4 2 x – 3 y = 5

20. Elimination by Addition Solving systems using graphing and substitution work well for systems with two variables. These processes do not extent easily to larger systems. Elimination by addition is an important method of solution. It generalizes to larger system and is the basis for computer-based solution methods. Solving a system using elimination involves a series of operations that transform the system into an equivalent system where the solution is obvious.

21. Theorem 2 Operations That Produce Equivalent Equations Theorem 2 A system of linear equations is transformed into an equivalent system if Two equations are interchanged. An equation is multiplied by a nonzero constant. A constant multiple of on equation is added to another equation.

22. Example 5 Solving a System Using Elimination by Addition Solve the system using elimination by addition: 3 x – 2 y = 8 2 x + 5 y = –1 Solution Use Theorem 2 to eliminate one of the variables. Multiply the top equation by 5 and the bottom equation by 2. 5(3 x – 2 y ) = 5·(8) 2(2 x + 5 y ) = 2·(–1) 15 x – 10 y = 40 4 x + 10 y = –2 19 x = 38 Add the top equation to the bottom equation eliminating the y terms. Solve for x. x = 2

23. Example 5 Solving a System Using Elimination by Addition continued With x = 2, substitute into either original equations and solve for y . 3 x – 2 y = 8 3·2 – 2 y = 8 6 – 2 y = 8 –2 y = 2 y = –1 The solution is x = 2, y = –1 or (2, –1).

24. System of Equations-No Solution The system of equations x + y = 10 x + y = 5 has no solution. (Think about two numbers that when added give 10 and simultaneously add to 5. No such numbers exist.) Multiply the bottom equation by –1, add the two equations to obtain x + y = 10 – x – y = – 5 0 = 5 This contradiction is not possible and the assumption that the original system has solutions is false. This system has no solutions, the graphs of the equations are parallel lines, and the system is inconsistent.

25. System of Equations-Infinitely Many Solutions Consider the system of equations 2 x + 2 y = 10 x + y = 5 Multiply the bottom equation by –2, add the two equations to obtain 2 x + 2 y = 10 –2 x – 2 y = – 10 0 = 0 The result 0 = 0 implies that the equations are equivalent, their graphs coincide, and the system is dependent. If we let x = k , where k is any real number, and solve either equation for y , we obtain y = 5 – k , so ( k , 5 – k ) is a solution to this system for any real number k . The variable k is called a parameter and replacing k with a real number produces a particular solution to the system.