1. Chapter 3 Systems of Linear Equations

2. § 3.1 Systems of Linear Equations in Two Variables

3. Blitzer, Intermediate Algebra, 5e – Slide #3 Section 3.1 Systems of Equations We know that an equation of the form Ax + By = C is a line when graphed. Two such equations is called a system of linear equations. A solution of a system of linear equations is an ordered pair that satisfies both equations in the system. For example, the ordered pair (2,1) satisfies the system 3x + 2y = 8 4x – 3y = 5

4. Blitzer, Intermediate Algebra, 5e – Slide #4 Section 3.1 Systems of Equations Since two lines may intersect in exactly one point, may not intersect at all, or may intersect in every point; it follows that a system of linear equations will have exactly one solution, will have no solution, or will have infinitely many solutions.

5. Blitzer, Intermediate Algebra, 5e – Slide #5 Section 3.1 Systems of EquationsEXAMPLE SOLUTION Determine whether (3,2) is a solution of the system Because 3 is the x-coordinate and 2 is the y-coordinate of the point(3,2), we replace x with 3 and y with 2. Since the result is false, (3,2) is NOT a solution for the system. Also, I need not check the other equation since the first one failed. ? ? ? false

6. Blitzer, Intermediate Algebra, 5e – Slide #6 Section 3.1 Systems of Equations Solve Systems of Two Linear Equations in Two Variables, x and y, by Graphing 1) Graph the first equation. 2) Graph the second equation on the same set of axes. 3) If the lines representing the two graphs intersect at a point, determine the coordinates of this point of intersection. The ordered pair is the solution to the system. 4) Check the solution in both equations. NOTE: In order for this method to be useful, you must graph the lines very accurately.

7. Blitzer, Intermediate Algebra, 5e – Slide #7 Section 3.1 Systems of EquationsEXAMPLE SOLUTION Solve by graphing: 1) Graph the first equation. I first rewrite the equation in slope-intercept form. Now I can graph the equation. m = -4 = -4/1, b = 4

8. Blitzer, Intermediate Algebra, 5e – Slide #8 Section 3.1 Systems of Equations 2) Graph the second equation on the same set of axes. I first rewrite the equation in slope-intercept form. CONTINUED Now I can graph the equation. m = 3 = 3/1, b = -3

9. Blitzer, Intermediate Algebra, 5e – Slide #9 Section 3.1 Systems of Equations 3) Determine the coordinates of the intersection point. This ordered pair is the system’s solution. Using the graph below, it appears that the solution is the point (1,0). We won’t know for sure until after we check this potential solution in the next step. CONTINUED

10. Blitzer, Intermediate Algebra, 5e – Slide #10 Section 3.1 Systems of Equations 4) Check the solution in both equations. CONTINUED Because both equations are satisfied, (1,0) is the solution and {(1,0)} is the solution set. ? ?? ? true

11. Blitzer, Intermediate Algebra, 5e – Slide #11 Section 3.1 Substitution Method Solving Linear Systems by Substitution 1) Solve either of the equations for one variable in terms of the other. (If one of the equations is already in this form, you can skip this step) 2) Substitute the expression found in step 1 into the other equation. This will result in an equation in one variable! (Hence, it’s called the “Substitution Method”) 3) Solve the equation containing one variable. (the resultant equation from step 2) 4) Back-Substitute the value found in step 3 into one of the original equations. Simplify and find the value of the remaining variable. 5) Check the proposed solution in both of the system’s given equations.

12. Blitzer, Intermediate Algebra, 5e – Slide #12 Section 3.1 Substitution MethodEXAMPLE SOLUTION Solve by the substitution method: 1) Solve either of the equations for one variable in terms of the other. We’ll isolate the variable y from the first equation. Solve for y by subtracting 4x from both sides

13. Blitzer, Intermediate Algebra, 5e – Slide #13 Section 3.1 Substitution Method 2) Substitute the expression from step 1 into the other equation. -4x + 4 is the ‘expression from step 1’ and ‘the other equation’ is 3x – y = 3. Therefore: Replace y with -4x+4 CONTINUED Distribute Add like terms Add 4 to both sides Divide both sides by 7 3) Solve the resulting equation containing one variable.

14. Blitzer, Intermediate Algebra, 5e – Slide #14 Section 3.1 Substitution Method 4) Back-substitute the obtained value into one of the original equations. We back-substitute 1 for x into one of the original equations to find y. Let’s use the first equation. CONTINUED Replace x with 1 Multiply Subtract 4 from both sides Therefore, the potential solution is (1,0).

15. Blitzer, Intermediate Algebra, 5e – Slide #15 Section 3.1 Substitution Method 5) Check. Now we will show that (1,0) is a solution for both of the original equations. CONTINUED ? ?? ? true Because both equations are satisfied, (1,0) is the solution and {(1,0)} is the solution set.

16. Blitzer, Intermediate Algebra, 5e – Slide #16 Section 3.1 Addition (Elimination) Method Solving Linear Systems by Addition 1) If necessary, rewrite both equations in the form Ax + By = C. 2) If necessary, multiply either equation or both equations by appropriate nonzero numbers so that the sum of the x-coefficients or the sum of the y- coefficients is 0. 3) Add the equations in step 2. The sum should be an equation in one variable. 4) Solve the equation in one variable (the result of step 3). 5) Back-substitute the value obtained in step 4 into either of the given equations and solve for the other variable. 6) Check the solution in both of the original equations. NOTE: As you now know, there is more than one method to solve a system of equations. The reason for learning more than one method is because sometimes one method will be preferable or easier to use over another method.

17. Blitzer, Intermediate Algebra, 5e – Slide #17 Section 3.1 Addition (Elimination) MethodEXAMPLE SOLUTION Solve by the addition method: 1) Rewrite both equations in the form Ax + By = C. We first arrange the system so that variable terms appear on the left and constants appear on the right. We obtain Add 2y to both sides 4x = -2y + 4 -y = -3x + 3 4x + 2y = 4 3x - y = 3Add 3x to both sides

18. Blitzer, Intermediate Algebra, 5e – Slide #18 Section 3.1 Addition (Elimination) Method 2) If necessary, multiply either equation or both equations by appropriate numbers so that the sum of the x- coefficients is 0. We can eliminate the y’s by multiplying the second equation by 2. Or we can eliminate the x’s by multiplying the first equation by -3 and the second equation by 4. Let’s use the first method. Multiply by 2 4x + 2y = 4 3x - y = 3 CONTINUED 4x + 2y = 4 6x - 2y = 6 No Change

19. Blitzer, Intermediate Algebra, 5e – Slide #19 Section 3.1 Addition (Elimination) Method 3) Add the equations. CONTINUED 4x + 2y = 4 6x - 2y = 6 Add: 10x + 0y = 10 4) Solve the equation in one variable. Now I solve the equation 10x = 10. 10x = 10 x = 1

20. Blitzer, Intermediate Algebra, 5e – Slide #20 Section 3.1 Addition (Elimination) Method 5) Back-substitute and find the value of the other variable. Now we will use one of the original equations and replace x with 1 to determine y. I’ll use the second equation. CONTINUED -y = -3x + 3 -y = -3(1) + 3 -y = -3 + 3 -y = 0 y = 0 Replace x with 1 Multiply Add Multiply by -1 Therefore, the potential solution is (1,0).

21. Blitzer, Intermediate Algebra, 5e – Slide #21 Section 3.1 Addition (Elimination) Method 6) Check. I now check the potential solution (1,0) in both original equations. CONTINUED ? ? ? ? true 4x = -2y + 4 -y = -3x + 3 4(1) = -2(0) + 4 4 = 0 + 4 4 = 4 -(0) = -3(1) + 3 0 = -3 + 3 0 = 0 Because both equations are satisfied, (1,0) is the solution and {(1,0)} is the solution set.

22. Blitzer, Intermediate Algebra, 5e – Slide #22 Section 3.1 Solving Systems of Equations Comparing Solution Methods MethodAdvantagesDisadvantages Graphing You can see the solutions.If the solutions do not involve integers or are too large to be seen on the graph, it’s impossible to tell exactly what the solutions are. Substitution Gives exact solutions. Easy to use if a variable is on one side by itself. Solutions cannot be seen. Introduces extensive work with fractions when no variable has a coefficient of 1 or -1. Addition Gives exact solutions. Easy to use if a variable has a coefficient of 1 or - 1. Solutions cannot be seen.

23. Blitzer, Intermediate Algebra, 5e – Slide #23 Section 3.1 Solving Systems of Equations The Number of Solutions to a System of Two Linear Equations Number of SolutionsWhat This Means Graphically Graphical Examples Exactly one ordered-pair solution The two lines intersect at one point. No solutionThe two lines are parallel. Infinitely Many Solutions The two lines are identical.

24. Blitzer, Intermediate Algebra, 5e – Slide #24 Section 3.1 Solving Systems of Equations NOTE: It is extremely helpful to understand these relationships as well as any other relationship between an equation and it’s graph. NOTE: To determine that a system has exactly one solution, solve the system using one of the methods. A single solution will occur as in the previous examples. NOTE: To determine that a system has no solution, solve the system using one of the methods. Eventually, you’ll get an obviously false statement, like 3 = 4. NOTE: To determine that a system has infinitely many solutions, solve the system using one of the methods. Eventually, you’ll get an obviously true statement, like -2 = -2.

25. Blitzer, Intermediate Algebra, 5e – Slide #25 Section 3.1 Solving Systems of EquationsEXAMPLE At a price of p dollars per ticket, the number of tickets to a rock concert that can be sold is given by the demand model N = -25p + 7800. At a price of p dollars per ticket, the number of tickets that the concert’s promoters are willing to make available is given by the supply model N = 5p + 6000. (a) How many tickets can be sold and supplied for $50 per ticket? (b) Find the ticket price at which supply and demand are equal. At this price, how many tickets will be supplied and sold?

26. Blitzer, Intermediate Algebra, 5e – Slide #26 Section 3.1 Solving Systems of EquationsSOLUTION CONTINUED The number of tickets that can be sold for $50 per ticket is found using the demand model: (a) How many tickets can be sold and supplied for $50 per ticket? N = -25(50) + 7800 = -1250 + 7800 = 6550 tickets sold. The number of tickets that can be supplied for $50 per ticket is found using the supply model: N = 5(50) + 6000 = 250 + 6000 = 6250 tickets supplied.

27. Blitzer, Intermediate Algebra, 5e – Slide #27 Section 3.1 Solving Systems of EquationsCONTINUED N = -25p + 7800 N = 5p + 6000 1) Solve either of the equations for one variable in terms of the other. The system of equations already has N isolated in both equations. (b) Find the ticket price at which supply and demand are equal. At this price, how many tickets will be supplied and sold?

28. Blitzer, Intermediate Algebra, 5e – Slide #28 Section 3.1 Substitution MethodCONTINUED 2) Substitute the expression from step 1 into the other equation. I’ll replace the N in the second equation with - 25p + 7800. -25p + 7800 = 5p + 6000 Add 25p to both sides 3) Solve the resulting equation containing one variable. Divide both sides by 30 Subtract 6000 from both sides -25p + 7800 = 5p + 6000 7800 = 30p + 6000 1800 = 30p 60 = p

29. Blitzer, Intermediate Algebra, 5e – Slide #29 Section 3.1 Substitution Method Replace p with 60 CONTINUED Multiply Add N = -25p + 7800 N = -25(60) + 7800 4) Back-substitute the obtained value into one of the original equations. We back-substitute 60 for p into one of the original equations to find N. Let’s use the first equation. N = -1500 + 7800 N = 6300

30. Blitzer, Intermediate Algebra, 5e – Slide #30 Section 3.1 Substitution MethodCONTINUED Therefore, the solution is (60,6300). Therefore, supply and demand will be equal when the ticket price is $60 and 6300 tickets are sold. 5) Check. Now we will show that (60,6300) is a solution for both of the original equations. ?? ?? true N = -25p + 7800N = 5p + 6000 6300 = -25(60) + 78006300 = 5(60) + 6000 6300 = -1500 + 78006300 = 300 + 6000 6300 = 6300