1. Algebra 2 Solving Systems Algebraically Lesson 3-2 Part 1

2. Goals Goal To solve a linear systems using substitution. Rubric Level 1 – Know the goals. Level 2 – Fully understand the goals. Level 3 – Use the goals to solve simple problems. Level 4 – Use the goals to solve more advanced problems. Level 5 – Adapts and applies the goals to different and more complex problems.

3. Vocabulary None

4. Essential Question Big Idea: Solving Equations and Inequalities When can you use substitution to solve a system?

5. The graph shows a system of linear equations. As you can see, without the use of technology, determining the solution from the graph is not easy. You can use the substitution method to find an exact solution. Substitution Method

6. Substitution is normally used when you know one of the variables in terms of the other variable (y = 2x + 3) or the coefficient of one of the variables is one so you can easily solve for it in terms of the other variable (3y – x = 4). Substitution is then used to reduce the system to one equation that has only one variable. Then you can solve this equation for the one variable and substitute again to find the other variable.

7. Solving Systems of Equations by Substitution Step 2 Step 3 Step 4 Step 5 Step 1 Solve for one variable in at least one equation, if necessary. Substitute the resulting expression into the other equation. Solve that equation to get the value of the first variable. Substitute that value into one of the original equations and solve for the other variable. Write the values from steps 3 and 4 as an ordered pair, (x, y), and check. Substitution Method

8. Solve the system by substitution. y = 3x y = x – 2 Step 1 y = 3x y = x – 2 Both equations are solved for y. Step 2 y = x – 2 3x = x – 2 Substitute 3x for y in the second equation. Now solve this equation for x. Subtract x from both sides and then divide by 2. Step 3–x 2x = –2 2 x = –1 Example: Substitution Use substitution because you know the variable y in terms of the variable x.

9. Solve the system by substitution. Step 4y = 3x Write one of the original equations. Substitute –1 for x. y = 3( – 1) y = –3 Step 5 ( – 1, –3) Check Substitute (–1, –3) into both equations in the system. Write the solution as an ordered pair. y = 3x –3 3(–1) –3 y = x – 2 –3 –1 – 2 –3 –3 Example: Continued

10. 5.1 - 10 Example: Solve the system. (1) (2) Solution Begin by solving one of the equations for one of the variables. We solve equation (2) for y. (2) Add x. Use substitution because the coefficient of y is one.

11. 5.1 - 11 Example: Now replace y with x + 3 in equation (1), and solve for x. (1) Let y = x + 3 in (1). Note the careful use of parentheses. Distributive property Combine terms. Subtract.

12. 5.1 - 12 Replace x with 1 in equation (3) to obtain y = 1 + 3 = 4. The solution of the system is the ordered pair (1, 4). Check this solution in both equations (1) and (2). Example: (1) Check: (2) ? True ? Both check; the solution set is {(1, 4)}.

13. You can substitute the value of one variable into either of the original equations to find the value of the other variable. Helpful Hint

14. Solve the system by substitution. x + 2y = –1 x – y = 5 Step 1 x + 2y = –1 Solve the first equation for x by subtracting 2y from both sides. Step 2 x – y = 5 (–2y – 1) – y = 5 Substitute – 2y – 1 for x in the second equation. –3y – 1 = 5 Simplify. −2y x = –2y – 1 Example: Substitution Use substitution because the coefficient of x is one. Note the careful use of parentheses.

15. Step 3 –3y – 1 = 5 Add 1 to both sides. +1 –3y = 6 –3 y = –2 Solve for y. Divide both sides by –3. Step 4 x – y = 5 x – (–2) = 5 x + 2 = 5 –2 –2 x = 3 Step 5 (3, –2) Write one of the original equations. Substitute –2 for y. Subtract 2 from both sides. Write the solution as an ordered pair. Example: Continued

16. When you solve one equation for a variable, you must substitute the value or expression into the other original equation, not the one that had just been solved. You have to use both equations. Caution

17. y + 6x = 11 3x + 2y = –5 Solve by substitution. Solve the first equation for y by subtracting 6x from each side. Step 1 y + 6x = 11 – 6x y = –6x + 11 Substitute –6x + 11 for y in the second equation. Distribute 2 to the expression in parentheses. 3x + 2(–6x + 11) = –5 3x + 2y = –5Step 2 3x + 2(–6x + 11) = –5 Example: Use substitution because the coefficient of y is one.

18. Step 3 3x + 2(–6x) + 2(11) = –5 –9x + 22 = –5 Simplify. Solve for x. Subtract 22 from both sides. –9x = –27 – 22 –22 Divide both sides by – 9. –9x = –27 –9 x = 3 3x – 12x + 22 = –5 3x + 2(–6x + 11) = –5 Continued

19. Step 4 y + 6x = 11 Substitute 3 for x. y + 6(3) = 11 Subtract 18 from each side. y + 18 = 11 –18 –18 y = –7 Step 5(3, –7) Write the solution as an ordered pair. Simplify. Write one of the original equations. Continued x = 3

20. Solve the system by substitution. Check your answer. y = x + 3 y = 2x + 5 Both equations are solved for y. Step 1y = x + 3 y = 2x + 5 Substitute 2x + 5 for y in the first equation. Solve for x. Subtract x and 5 from both sides. –x – 5 –x – 5 x = –2 Step 32x + 5 = x + 3 Step 2 2x + 5 = x + 3 y = x + 3 Your Turn:

21. Solve the system by substitution. Check your answer. Step 4y = x + 3 Write one of the original equations. Substitute –2 for x. y = –2 + 3 y = 1 Step 5 (–2, 1) Write the solution as an ordered pair. Check Substitute (–2, 1) into both equations in the system. y = x + 3 1 (–2) + 3 1 y = 2x + 5 1 2(–2) + 5 1 –4 + 5 1 Continued

22. Solve the system by substitution. x = 2y – 4 x + 8y = 16 The first equation is solved for x. Step 1 x = 2y – 4 Substitute 2y – 4 for x in the second equation. Simplify. Then solve for y. (2y – 4) + 8y = 16 x + 8y = 16Step 2 Step 310y – 4 = 16 Add 4 to both sides. +4 10y = 20 y = 2 10y 20 10 = Divide both sides by 10. Your Turn:

23. Step 4 x + 8y = 16 Write one of the original equations. Substitute 2 for y. x + 8(2) = 16 x + 16 = 16 x = 0 – 16 –16 Simplify. Subtract 16 from both sides. Step 5 (0, 2) Write the solution as an ordered pair. Continued

24. Solve the system by substitution. 2x + y = –4 x + y = –7 Solve the second equation for x by subtracting y from each side. Substitute –y – 7 for x in the first equation. Distribute 2. 2(–y – 7) + y = –4 x = –y – 7Step 2 Step 1 x + y = –7 – y x = –y – 7 2(–y – 7) + y = –4 –2y – 14 + y = –4 Your Turn:

25. Combine like terms. Step 3 +14 –y = 10 –2y – 14 + y = –4 Add 14 to each side. –y – 14 = –4 y = –10 Step 4 x + y = –7 Write one of the original equations. Substitute –10 for y. x + (–10) = –7 x – 10 = – 7 Continued

26. x – 10 = –7Step 5 +10 x = 3 Add 10 to both sides. Step 6(3, –10) Write the solution as an ordered pair. Continued

27. –2x + y = 8 3x + 2y = 9 Solve by substitution. Solve the first equation for y by adding 2x to each side. Step 1 –2x + y = 8 + 2x +2x y = 2x + 8 Substitute 2x + 8 for y in the second equation. 3x + 2(2x + 8) = 9 3x + 2y = 9Step 2 Distribute 2 to the expression in parentheses. 3x + 2(2x + 8) = 9 Your Turn:

28. Step 3 3x + 2(2x) + 2(8) = 9 7x + 16 = 9 Simplify. Solve for x. Subtract 16 from both sides. 7x = –7 –16 Divide both sides by 7. 7x = –7 7 x = –1 3x + 4x + 16 = 9 Continued 3x + 2(2x + 8) = 9

29. Step 4 –2x + y = 8 Substitute –1 for x. –2(–1) + y = 8 y + 2 = 8 –2 y = 6 Step 5(–1, 6) Write the solution as an ordered pair. Subtract 2 from each side. Simplify. Write one of the original equations. Continued x = –1

30. Example: A Linear System with No Solution (Inconsistent System) Show that this linear system has no solution. 2 x  y  5 Equation 1 2 x  y  1 Equation 2 Because Equation 2 can be revised to y  –2 x  1, you can substitute –2 x  1 for y in Equation 1. 2 x  y  5 2 x  (–2 x  1)  5 Substitute –2 x  1 for y. 1  5 Simplify. False statement! Once the variables are eliminated, the statement is not true regardless of the values of x and y. This tells you the system has no solution. Write Equation 1. M ETHOD : S UBSTITUTION

31. Substitute 2 x + 3 for y in Equation 2. Show that this linear system has infinitely many solutions. – 2 x  y  3 Equation 1 – 4 x  2y  6 Equation 2 M ETHOD : SUBSTITUTION You can solve Equation 1 for y. – 4 x  2( 2 x + 3)  6 – 4 x + 4x  6  6 Simplify. 6  6 True statement! The variables are eliminated and you are left with a statement that is true regardless of the values of x and y. This result tells you that the linear system has infinitely many solutions. Example: Linear Sys. with Infinite Solutions (Dependent System) y  2 x + 3

32. Your Turn: Solve the system. 1) 2) 30 = 30 (true statement) Infinite Solutions 6 = -9 (false statement) No Solution

33. Example: Application Lancaster Woodworkers Furniture Store builds two types of wooden outdoor chairs. A rocking chair sells for $265 and an Adirondack chair with footstool sells for $320. The books show that last month, the business earned $13,930 for the 48 outdoor chairs sold. How many of each chair were sold? Step 1 You are asked to find the number of each type of chair sold.

34. Example: continued Define variables and write the system of equations. Let x represent the number of rocking chairs sold and y represent the number of Adirondack chairs sold. x + y =48The total number of chairs sold was 48. 265x + 320y =13,930The total amount earned was $13,930. Step 2

35. Solve one of the equations for one of the variables in terms of the other. Since the coefficient of x is 1, solve the first equation for x in terms of y. x + y =48First equation x=48 – ySubtract y from each side. Example: continued Step 3

36. Step 4 Substitute 48 – y for x in the second equation. 265x + 320y =13,930Second equation 265(48 – y) + 320y =13,930Substitute 48 – y for x. 12,720 – 265y + 320y=13,930Distributive Property 55y=1210Simplify. y=22Divide each side by 55. Example: continued

37. Step 5 Now find the value of x. Substitute the value for y into either equation. x + y =48First equation x + 22 =48Replace y with 22. x=26Subtract 22 from each side. Answer:They sold 26 rocking chairs and 22 Adirondack chairs. Example: continued

38. Your Turn: At Amy’s Amusement Park, tickets sell for $24.50 for adults and $16.50 for children. On Sunday, the amusement park made $6405 from selling 330 tickets. How many of each kind of ticket was sold? Solution: 120 adult; 210 children Variables: x = # of adult tickets y = # of child tickets System:

39. Essential Question Big Idea: Solving Equations and Inequalities When can you use substitution to solve a system? Use substitution when one of the equations in the system is already solved for one of the variables. You can also use substitution when it is easy to isolate a variable in one of the equations.

40. Assignment: Section 3-2 Part 1, Pg. 156; # 1 – 3 all, 4 – 16 even.