1. Solving Systems of Equations Modeling real-world problems

2. A system of equations is a set of two or more equations that have variables in common. The common variables relate to similar quantities. You can think of an equation as a condition imposed on one or more variables, and a system as several conditions imposed simultaneously. Remember, when solving systems of equations, you are looking for a solution that makes each equation true.

3. A system of linear equations is a set of two or more equations with the same variable. The solution of a system in x and y is any ordered pair (x, y) that satisfies each of the equations in the system. The solution of a system of equations is the intersection of the graphs of the equations.

4. Summary of Solutions of Systems of Linear Equations The lines intersect so there is one solution. The lines are parallel so there are no solutions. The lines are the same so there are infinitely many solutions. x + 2y = 7 x = y + 4 y – 2x = 7 Y = 2x + 3 -3x = 5 – y 2y = 6x + 10 One Solution No Solution Infinitely Many Solutions

5. Solving Systems of Equations Graphing Method

6. If you can graph a straight line, you can solve systems of equations graphically! The process is very easy. Simply graph the two lines and look for the point where they intersect (cross). Remember using the graphing method many times only approximates the solution, so sometimes it can be unreliable.

7. Solving Systems of Equations by Graphing. To solve a system of equations graphically, graph both equations and see where they intersect. The intersection point is the solution. 4x – 6y = 12 4x = 6y + 12 4x – 12 = 6y 6y = 4x – 12 6 6 6 y = 2/3x – 2 slope = 2/3 y-intercept = -2 2x + 2y = 6 2y = -2x + 6 2 2 2 y = -x + 3 slope = -1/1 y-intercept = 3

8. Graph the equations. The slope-intercept method of graphing was used in this example. The point of intersection of the two lines (3, 0) is the solution to the system of equations. This means that (3, 0), when substituted into either equation, will make them both true.

9. Locate the point where the lines intersect. From the graph, the solution appears to (2, 3). Check to be sure that (2, 3) is the solution, substitute 2 for x and 3 for y into each equation. x + y = 5y = 2x – 1 2 + 3 = 53 = 2(2) - 1

10. Note Some systems of equations may be very difficult to solve using the graphing method. The exact solution would be hard to determine from a graph because the coordinates are not integers. Solving a system algebraically is better than graphing when you need an accurate solution. Take for example the system of equations: 3x + 2y = 12 x – y = 3 x = 3/8 and y = 15/4

11. Solving Systems of Equations Algebraically Substitution Method

12. The substitution method is used to eliminate one of the variables by replacement when solving a system of equations. Think of it as “grabbing” what one variable equals from one equation and “plugging” it into the other equation.

13. Solve this system of equations using the substitution method. Step 1 Put it in Slope-Intercept Form 3y – 2x = 11 Y + 2x = 9 Solve one of the equations for either “x” or “y”. In this example it is easier to solve the second equation for “y”, since it only involves one step. Y = 9 – 2x

14. Step 2 Replace the “y” value in the first equation by what “y” now equals (y = 9 – 2x). Grab the “y” value and plug it into the other equation. 3y – 2x = 11 3(9 – 2x) – 2x = 11

15. Step 3 Solve this new equation for “x”. 3(9 – 2x) – 2x = 11 27 – 6x – 2x = 11 27 – 6x – 2x = 11 27 – 8x = 11 -8x = -16 x = 2

16. Step 4 Now that we know the “x” value (x = 2), we place it into either of the ORIGINAL equations in order to solve for “y”. Pick the easier one to work with! Y + 2x = 9 y + 2(2) = 9 y + 4 = 9 y = 5

17. Step 5 Check: substitute x = 2 and y = 5 into BOTH ORIGINAL equations. If these answers are correct BOTH equations will be true! 3y – 2x = 11 3(5) – 2(2) = 11 15 – 4 = 11 11 = 11 True? Y + 2x = 9 5 + 2(2) = 9 5 + 4 = 9 9 = 9 True?

18. The Substitution Method Step 1 Solve one equation for x (or y). Step 2 Substitute the expression from Step 1 into the other equation. Step 3 Solve for y (or x). Step 4 Take the value of y (or x) found in Step 3 and substitute it into one of the original equations. Then solve for the other variable. Step 5 The ordered pair of values from Steps 3 and 4 is the solution. If the system has no solution, a contradictory statement will result in either Step 3 or 4.

19. Solving Systems of Equations Elimination method

20. You can use the Addition and Subtraction Properties of Equality to solve a system by the elimination method. You can add or subtract equations to eliminate (getting rid of) a variable. Step 1 5x – 6y = -32 3x + 6y = 48 Eliminate y because the sum of the coefficients of y is zero 5x – 6y = -32 3x + 6y = 48 8x + 0 = 16 x = 2 Addition Property of Equality Solve for x If you add the two equations together, the +6y and -6y cancel each other out because of the Property of Additive Inverse

21. Step 2 Solve for the eliminated variable y using either of the original equations. 3x + 6y = 48 3(2) + 6y = 48 6 + 6y = 48 6y = 42 y = 7 Choose the 2 nd equation Substitute 2 for x Simplify. Then solve for y. Remember x = 2

22. Since x = 2 and y = 7, the solution is (2, 7) Check 5x – 6y = -323x + 6y = 48 5(2) – 6(7) = -323(2) + 6(7) = 48 10 – 42 = -32 6 + 42 = 48 -32 = -32 48 = 48 Remember, the order pair (2, 7) must make both equations true. True

23. Solving systems of equations in real-world problems. April sold 75 tickets to a school play and collected a total of $495. If the adult tickets cost $8 each and child tickets cost $5 each, how many adult tickets and how many child tickets did she sell? Solution: Let a represent the adult tickets and c represent the child tickets. Individual tickets sold equaled 75, so a + c = 75 All total April sold $495 in tickets, since adult tickets are $8 and child tickets are $5, so 8a + 5c = 495. System of Equations a + c = 75 8a + 5c == 495

24. Solution a + c = 75 8a + 5c = 495 a = 75 – c 8(75 – c) + 5c = 495 600 – 8c + 5c = 495 600 - 3c = 495 105 = 3c 35 = c a + c = 75 a + 35 = 75 a = 40 There were 40 adult tickets and 35 child tickets sold. 40 + 35 = 75 8(40) + 5(35) = 495 320 + 175 = 495

25. Write a system of equations and solve. At a baseball game, Jose bought five hot dogs and three sodas for $17. At the same time, Allison bought two hot dogs and four sodas for $11. Find the cost of one hot dog and one soda.

26. Suppose your community center sells a total of 292 tickets for a basketball game. An adult ticket cost $3. A student ticket cost $1. The sponsors collected $470 in ticket sales. Write and solve a system to find the number of each type of ticket sold.

27. Let a = number of adult tickets Let s = number of student tickets total number of ticket total number of sales a + s = 292 3a + 1s = 470 Solve by elimination (get rid of s) because the difference of the coefficients of s is zero. a + s = 292 3a + s = 470 -2a + 0 = -178 a = 89 That means you must subtract the two equations so, -3a – a = -470 is what you must subtract. Next Step This is the number of adult tickets sold.

28. Solve for the eliminated variable using either of the original equations. a + s = 292 89 + s = 292 s = 203 There were 89 adult tickets sold and 203 student tickets sold. Is the solution reasonable? The total number of tickets is 89 + 203 = 292. The total sales is $3(89) + $1(203) = $470. The solution is correct. This is the number of student tickets sold.

29. If you have noticed in the last few examples that to eliminate a variable its coefficients must have a sum or difference of zero. Sometime you may need to multiply one or both of the equations by a nonzero number first so that you can then add or subtract the equations to eliminate one of the variables. 2x + 5y = 177x + 2y = 102x + 5y = -22 6x – 5y = -19-7x + y = -1610x + 3y = 22 If you notice the systems of equations above, two of them have something in common. The third doesn’t. We can add these two equations together to eliminate the y variable. We can add these two equations together to eliminate the x variable. What are we going to do with these equations, can’t eliminate a variable the way they are written?

30. Multiplying One Equation Solve by Elimination 2x + 5y = -22 10x + 3y = 22 Step 1 2x + 5y = -225(2x + 5y = -22)10x + 25y = -110 10x + 3y = 2210x + 3y = 22-(10x + 3y = 22) 0 + 22y = -132 y = -6 Start with the given system. To prepare for eliminating x, multiply the first equation by 5. Subtract the equations to eliminate x. NEXT Be careful when you subtract. All the signs in the equation that is being subtracted change. -10x – 3y = -22 Ask: Is one coefficient a factor of the other coefficient for the same variable?

31. Step 2 Solve for the eliminated variable using either of the original equations. 2x + 5y = -22 Choose the first equation. 2x +5(-6) = -22 Substitute -6 for y. 2x – 30 = -22 Solve for x. 2x = 8 x = 4 The solution is (4, -6).

32. Solve by elimination. -2x + 5y = -32 7x – 5y = 17 3x – 10y = -25 4x + 40y = 20 2x – 3y = 61 2x + y = -7 Ask: Is one coefficient a factor of the other coefficient for the same variable?

33. Multiplying Both Equations To eliminate a variable, you may need to multiply both equations in a system by a nonzero number. Multiply each equation by values such that when you write equivalent equations, you can then add or subtract to eliminate a variable. 4x + 2y = 14 7x + 3y = -8 In these two equations you cannot use graphing or substitution very easily. However ever if we multiply the first equation by 3 and the second by 2, we can eliminate the y variable. Find the least common multiple LCM of the coefficients of one variable, since working with smaller numbers tends to reduce the likelihood of errors. 4 x 7 = 28 2 x 3 = 6 NEXT

34. 4X + 2Y = 143(4X + 2Y = 14)12X + 6Y = 42 7X – 3Y = - 82(7X – 3Y = -8)14X – 6Y = -16 26X + 0 = 26 26X = 26 X = 1 Solve for the eliminated variable y using either of the original equations. 4x + 2y = 14 4(1) + 2y = 14 4 + 2y = 14 2y = 10 y = 5The solution is (1, 5). Start with the given system. To prepare to eliminate y, multiply the first equation by 3 and the second equation by 2. Add the equations to eliminate y.

35. Practice and Problem Solving Solve by elimination: 1) 2x + 5y = 172) 7x + 2y = 10 6x + 5y = -9 -7x + y = -16 3) 2x – 3y = 614) 24x + 2y = 52 2x + y = -7 6x – 3y = -36 5) y = 2x6) 9x + 5y = 34 y = x – 1 8x – 2y = -2 Word Problems You choose what method you what to use to solve question 5 thru 10.

36. 7) The sum of two numbers is 20. Their difference is 4. Write and solve a system of equations. 8) Your school sold 456 tickets for a school play. An adult ticket cost $3.50. A student ticket cost $1. Total ticket sales equaled $1131. Let a = adult tickets sold, and s = student tickets sold. How many tickets of each were sold? 9) Suppose the band sells cans of popcorn for $5 each and mixed nuts for $8 each. The band sells a total of 240 cans and makes a total of $1614. Find the number of cans of each sold. One more problem, my favorite.

37. 10) A farmer raises chicken and cows. He has a total of 34 animals in his barnyard. His six- year son came in one day all excited saying, “Daddy, daddy, did you know all your animals have a total of 110 legs.” Write a system of equations to represent this situation. How many chickens and how many cows does the farmer have?

38. When you solve systems using elimination, plan a strategy. The flowchart like the one below can help you decide how to eliminate a variable. When you solve systems using elimination, plan a strategy. Here are a few hints to help you decide how to eliminate a variable. Answer these questions. Can I eliminate a variable by adding or subtracting the given equations? YES Or NO Can I multiply one of the equations by a number, and then add or subtract the equations? YES or NO Multiply both equations by different numbers. then add or subtract the equations Do It.