1. Homework 12/15/2015 Solving Systems of linear Equations packet Page 1, 2, and 3 Note: I am not available after school =(

2. Systems of Linear Equations

3. 3 4 A solution to a system of equations is an ordered pair that satisfy all the equations in the system. 4 A system of linear equations can have: 1. Exactly one solution 2. No solutions 3. Infinitely many solutions

4. 4 ConsistentDependentInconsistent One solution Lines intersect No solution Lines are parallel Infinite number of solutions Coincide-Same line

5. 5 There are four ways to solve systems of linear equations: 4 1. By graphing 4 2. By substitution 4 3. By elimination

6. 6 4 Solving Systems by Graphing: 4 When solving a system by graphing: 1. Change the equations into y = mx + b so you can see the slope (m) and the y-intercept (b) 2. Graph both equations 3. Find the intersection point (break-even point) and that will determine the solution or solutions to the system of equations.

7. 4 Three possible solutions to a linear system in two variables: 4 One solution: coordinates of a point 4 No solutions: inconsistent case 4 Infinitely many solutions: dependent case

8. 8 2x – y = 2 x + y = -2 2x – y = 2 -y = -2x + 2 y = 2x – 2 x + y = -2 y = -x - 2 Different slope, different intercept!

9. 9 3x + 2y = 3 3x + 2y = -4 3x + 2y = 3 2y = -3x + 3 y = -3/2 x + 3/2 3x + 2y = -4 2y = -3x -4 y = -3/2 x - 2 Same slope, different intercept!!

10. x – y = -3 2x – 2y = -6 x – y = -3 -y = -x – 3 y = x + 3 2x – 2y = -6 -2y = -2x – 6 y = x + 3 Same slope, same intercept! Same equation!!

11. 11 4 Determine Without Graphing: 4 Once the equations are in slope-intercept form, compare the slopes and intercepts. 4 One solution – the lines will have different slopes. 4 No solution – the lines will have the same slope, but different intercepts. 4 Infinitely many solutions – the lines will have the same slope and the same intercept.

12. 12 Determine Without Graphing: 4 Given the following lines, determine what type of solution exists, without graphing. 4 Equation 1: 3x = 6y + 5 4 Equation 2: y = (1/2)x – 3 4 Writing each in slope-intercept form (solve for y) 4 Equation 1:y = (1/2)x – 5/6 4 Equation 2:y = (1/2)x – 3 4 Since the lines have the same slope but different y- intercepts, there is no solution to the system of equations. The lines are parallel.

13. Substitution Method: Procedure for Substitution Method 4 1. Solve one of the equations for one of the variables. 4 2. Substitute the expression found in step 1 into the other equation. 4 3. Now solve for the remaining variable. 4 4. Substitute the value from step 2 into the equation 4 written in step 1, and solve for the remaining variable.

14. Substitution Method: 1. Solve the following system of equations by substitution. Step 1 is already completed. Step 2:Substitute x+3 into 2 nd equation and solve. Step 3: Substitute –4 into 1 st equation and solve. The answer: ( -4, -1)

15. 1) Solve the system using substitution x + y = 5 y = 3 + x Step 1: Solve an equation for one variable. Step 2: Substitute The second equation is already solved for y! x + y = 5 x + (3 + x) = 5 Step 3: Solve the equation. 2x + 3 = 5 2x = 2 x = 1

16. 1) Solve the system using substitution x + y = 5 y = 3 + x Step 4: Plug back in to find the other variable. x + y = 5 (1) + y = 5 y = 4 Step 5: Check your solution. (1, 4) (1) + (4) = 5 (4) = 3 + (1) The solution is (1, 4). What do you think the answer would be if you graphed the two equations?

17. 2) Solve the system using substitution 3y + x = 7 4x – 2y = 0 Step 1: Solve an equation for one variable. Step 2: Substitute It is easiest to solve the first equation for x. 3y + x = 7 -3y x = -3y + 7 4x – 2y = 0 4(-3y + 7) – 2y = 0

18. 2) Solve the system using substitution 3y + x = 7 4x – 2y = 0 Step 4: Plug back in to find the other variable. 4x – 2y = 0 4x – 2(2) = 0 4x – 4 = 0 4x = 4 x = 1 Step 3: Solve the equation. -12y + 28 – 2y = 0 -14y + 28 = 0 -14y = -28 y = 2

19. 2) Solve the system using substitution 3y + x = 7 4x – 2y = 0 Step 5: Check your solution. (1, 2) 3(2) + (1) = 7 4(1) – 2(2) = 0

20. Solving Systems of Equations using Elimination Steps: 1. Place both equations in Standard Form, Ax + By = C. 2. Determine which variable to eliminate with Addition or Subtraction. 3. Solve for the remaining variable. 4. Go back and use the variable found in step 3 to find the second variable. 5. Check the solution in both equations of the system.

21. EXAMPLE #1: STEP 2:Use subtraction to eliminate 5x. 5x + 3y =11 5x + 3y = 11 -(5x - 2y =1) -5x + 2y = -1 5x + 3y = 11 5x = 2y + 1 Note: the (-) is distributed. STEP 3:Solve for the variable. 5x + 3y =11 -5x + 2y = -1 5y =10 y = 2 STEP1: Write both equations in Ax + By = C form. 5x + 3y =1 5x - 2y =1

22. STEP 4: Solve for the other variable by substituting into either equation. 5x + 3y =11 5x + 3(2) =11 5x + 6 =11 5x = 5 x = 1 5x + 3y = 11 5x = 2y + 1 The solution to the system is (1,2).

23. 5x + 3y= 11 5x = 2y + 1 Step 5:Check the solution in both equations. 5x + 3y = 11 5(1) + 3(2) =11 5 + 6 =11 11=11 5x = 2y + 1 5(1) = 2(2) + 1 5 = 4 + 1 5=5 The solution to the system is (1,2).

24. Example #2: x + y = 10 5x – y = 2 Step 1: The equations are already in standard form:x + y = 10 5x – y = 2 Step 2: Adding the equations will eliminate y. x + y = 10 +(5x – y = 2)+5x – y = +2 Step 3:Solve for the variable. x + y = 10 +5x – y = +2 6x = 12 x = 2

25. x + y = 10 5x – y = 2 Step 4: Solve for the other variable by substituting into either equation. x + y = 10 2 + y = 10 y = 8 Solution to the system is (2,8).

26. x + y = 10 5x – y = 2 x + y =10 2 + 8 =10 10=10 5x – y =2 5(2) - (8) =2 10 – 8 =2 2=2 Step 5: Check the solution in both equations. Solution to the system is (2,8).

27. NOW solve these using elimination: 1.2. 2x + 4y =1 x - 4y =5 2x – y =6 x + y = 3

28. Using Elimination to Solve a Word Problem: Two angles are supplementary. The measure of one angle is 10 degrees more than three times the other. Find the measure of each angle.

29. Using Elimination to Solve a Word Problem: Two angles are supplementary. The measure of one angle is 10 more than three times the other. Find the measure of each angle. x = degree measure of angle #1 y = degree measure of angle #2 Therefore x + y = 180

30. Using Elimination to Solve a Word Problem: Two angles are supplementary. The measure of one angle is 10 more than three times the other. Find the measure of each angle. x + y = 180 x =10 + 3y

31. Using Elimination to Solve a Word Problem: Solve x + y = 180 x =10 + 3y x + y = 180 -(x - 3y = 10) 4y =170 y = 42.5 x + 42.5 = 180 x = 180 - 42.5 x = 137.5 (137.5, 42.5)

32. Using Elimination to Solve a Word Problem: The sum of two numbers is 70 and their difference is 24. Find the two numbers.

33. Using Elimination to Solve a Word problem: The sum of two numbers is 70 and their difference is 24. Find the two numbers. x = first number y = second number Therefore, x + y = 70

34. Using Elimination to Solve a Word Problem: The sum of two numbers is 70 and their difference is 24. Find the two numbers. x + y = 70 x – y = 24

35. Using Elimination to Solve a Word Problem: x + y =70 x - y = 24 2x = 94 x = 47 47 + y = 70 y = 70 – 47 y = 23 (47, 23)

36. Now you Try to Solve These Problems Using Elimination. Solve 1.Find two numbers whose sum is 18 and whose difference is 22. 2.The sum of two numbers is 128 and their difference is 114. Find the numbers.