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Daily Warm-Up 7.1

5.6. Daily Warm-Up 7.1. Daily Warm-Up Is the Ordered Pair a Solution of the equation 2x – 3y = 5? (1, 0) (-1, 1) (1, -1) Tonight’s Homework Graphing Systems Worksheet #1-10 + PREVIEW L 7.2. 7.1 Solving Linear Systems by Graphing.

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Daily Warm-Up 7.1

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  1. 5.6 Daily Warm-Up 7.1 • Daily Warm-Up • Is the Ordered Pair a Solution of the equation 2x – 3y = 5? • (1, 0) • (-1, 1) • (1, -1) • Tonight’s Homework • Graphing Systems Worksheet #1-10 + PREVIEW L 7.2

  2. 7.1 Solving Linear Systems by Graphing Objective: Solve a system of linear equations by graphing

  3. What is a System of Linear Equations? A system of linear equations is simply two or more linear equations using the same variables. We will only be dealing with systems of two equations using two variables, x and y. If the system of linear equations is going to have a solution, then the solution will be an ordered pair (x , y) where x and y make both equations true at the same time. We will be working with the graphs of linear systems and how to find their solutions graphically.

  4. y x (1 , 2) How to Use Graphs to Solve Linear Systems Consider the following system: x – y = –1 x + 2y = 5 Using the graph to the right, we can see that any of these ordered pairs will make the first equation true since they lie on the line. We can also see that any of these points will make the second equation true. However, there is ONE coordinate that makes both true at the same time… The point where they intersect makes both equations true at the same time.

  5. y x (1 , 2) How to Use Graphs to Solve Linear Systems Consider the following system: x – y = –1 x + 2y = 5 We must ALWAYS verify that your coordinates actually satisfy both equations. To do this, we substitute the coordinate (1 , 2) into both equations. x – y = –1 (1) – (2) = –1  x + 2y = 5 (1) + 2(2) = 1 + 4 = 5  Since (1 , 2) makes both equations true, then (1 , 2) is the solution to the system of linear equations.

  6. Solve the following system by graphing: 3x + 6y = 15 –2x + 3y = –3 Graphing to Solve a Linear System Start with 3x + 6y = 15 Subtracting 3x from both sides yields 6y = –3x + 15 Dividing everything by 6 gives us… While there are many different ways to graph these equations, we will be using the slope – intercept form. Similarly, we can add 2x to both sides and then divide everything by 3 in the second equation to get To put the equations in slope intercept form, we must solve both equations for y. Now, we must graph these two equations

  7. y Solve the following system by graphing: 3x + 6y = 15 –2x + 3y = –3 x (3 , 1) Label the solution! Since and , then our solution is correct! Graphing to Solve a Linear System Using the slope intercept forms of these equations, we can graph them carefully on graph paper. Start at the y – intercept, then use the slope. Lastly, we need to verify our solution is correct, by substituting (3 , 1).

  8. Graphing to Solve a Linear System Let's summarize! There are 4 steps to solving a linear system using a graph. Step 1: Put both equations in slope – intercept form Solve both equations for y, so that each equation looks like y = mx + b. Step 2: Graph both equations on the same coordinate plane Use the slope and y – intercept for each equation in step 1. Be sure to use a ruler and graph paper! Step 3: Estimate where the graphs intersect. This is the solution! LABEL the solution! Substitute the x and y values into both equations to verify the point is a solution to both equations. Step 4: Check to make sure your solution makes both equations true.

  9. y LABEL the solution! x Graphing to Solve a Linear System Let's do ONE more…Solve the following system of equations by graphing. 2x + 2y = 3 x – 4y = –1 Step 1: Put both equations in slope – intercept form Step 2: Graph both equations on the same coordinate plane Step 3: Estimate where the graphs intersect. LABEL the solution! Step 4: Check to make sure your solution makes both equations true.

  10. y x Graphing to Solve a Linear System It’s your turn! … Solve the following system of equations by graphing. 3x + 2y = 2 x = –1 Step 1: Put both equations in slope – intercept form Step 2: Graph both equations on the same coordinate plane Step 3: Estimate where the graphs intersect. LABEL the solution! Step 4: Check to make sure your solution makes both equations true.

  11. Summary • 1. When you are asked to solve the linear equation system by graphing, you only need to follow the four-step: • Step 1: Put both equations in slope – intercept form: y = mx + b • Step 2: Graph both equations on the same coordinate plane: use the y-intercept as the first point and then use slope to create the second point. • Step 3: Estimate where the graphs intersect, and also label the intersection which is the solution of the linear equation system. • Step 4: Check to make sure your solution makes both equations true because the graphing may contain some error. Substitute the x and y values into both equations to verify the point is a solution to both equations. • 2. Don’t forget the special line!!!

  12. Guided Practice – L 7.1 DHQ DAILY HOMEWORK QUIZ PAGE: 401 in Textbook Numbers: 8-10 on graph paper! * ALL ANSWERS ARE ORDERED PAIRS! Tonight’s Homework: Graphing Systems Worksheet #1-10 + PREVIEW L 7.2

  13. y x 5.6 Daily Warm-Up 7.2 Find the Solution to the System of Equations. Write your solution as an ordered pair (x, y). 2x – 3y = 9 X = -3 Tonight’s Homework: Substitution WS #1-10 + PREVIEW L 7.3

  14. 7.2 Solving Systems of Equations using Substitution Objective: Use substitution method to solve a linear system

  15. Solving Systems of Equations using Substitution Steps: 1. Solve one equation for one variable (y = ; x = ; a =) 2. Substitute the expression from step one into the other equation. 3. Simplify and solve the equation. 4. Substitute back into either original equation to find the value of the other variable. 5. Check the solution in both equations of the system.

  16. Example #1: y = 4x 3x + y = -21 Step 1:Solve one equation for one variable. y = 4x(This equation is already solved for y.) Step 2: Substitute the expression from step one into the other equation. 3x + y = -21 3x + 4x = -21 Step 3: Simplify and solve the equation. 7x = -21 x = -3

  17. y = 4x 3x + y = -21 Step 4: Substitute back into either original equation to find the value of the other variable. y = 4x y = 4(-3) = -12 or 3x + y = -21 3(-3) + y = -21 -9 + y = -21 y = -12 This substitution is easier Solution to the system is (-3, -12).

  18. y = 4x 3x + y = -21 Step 5: Check the solution in both equations. Solution to the system is (-3,-12). 3x + y = -21 3(-3) + (-12) = -21 -9 + (-12) = -21 -21= -21 y = 4x -12 = 4(-3) -12 = -12

  19. Example #2: x + y = 10 5x – y = 2 Step 1: Solve one equation for one variable. x + y = 10 y = –x +10 Step 2: Substitute the expression from step one into the other equation. 5x – y = 2 5x – (–x +10) = 2

  20. x + y = 10 5x – y = 2 Step 3: Simplify and solve the equation. 5x – (–x + 10) = 2 5x + x – 10 = 2 6x – 10 = 2 6x = 12 x = 2

  21. x + y = 10 5x – y = 2 Step 4: Substitute back into either original equation to find the value of the other variable. x + y = 10 2 + y = 10 y = 8 Solution to the system is (2,8).

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

  23. You try Exercise 1 Solve the system using the substitution method. y = –63 – 7x 3x + 7y = –73 A) (8, –8) B) (–8, –7) C) (–7, –8) D) (–7, 9)

  24. YOU TRY: Solve by substitution:

  25. Summary • 1. When solving the linear systems, it is very important to remember the 5 steps and follow them in the order. • When in a certain step the equation reduces to a contradiction, you could conclude that the system is inconsistent and has no solution. • When in a certain step the equation reduces to an identity, you could conclude that the system is dependent and has many solutions.

  26. Guided Practice – L 7.2 DHQ PAGE: 408 in Textbook # 7 • ANSWERS ARE ORDERED PAIRS! • MUST SHOW WORK Tonight’s Homework: Substitution WS #1-8, 10 (SKIP #9) + PREVIEW L 7.3

  27. y x Daily Warm-Up 7.3 Find the Solution to the System of Equations using 2 methods, GRAPHING and SUBSTITUTION. Write your solution as an ordered pair (x, y). y = 1 -2x + y = - 1 Tonight’s Homework: Elimination WS #1-10 + Study for QUIZ (L 1-3) – QUIZ IS THURSDAY (1/27)

  28. 7.3 Solving Linear Systems by Linear Combinations (Elimination) Method Objective: Use linear combinations to solve a system of linear equations.

  29. What is a System? A system of linear equations is: • A set of parabolas • A set of two or more lines • A stereo component

  30. Linear Combinations (Elimination) Method Linear Combinations (Elimination) method is used when it appears easy to eliminate one variable from the system through transformation. Remember that linear transformations do not change the solutions of a system.

  31. Linear Combinations (Elimination) Method Step 1 Make sure the variables are lined up properly by their names -2x + y = 4 -6x + y = 0 Step 2 Make the coefficients of one of variables opposites (Multiply one or both equations by appropriate numbers so that the coefficients in one variable are opposites). Correctly applying the distributive property is very important!!! In this question, notice that the y coefficients are 1, therefore we can multiply either equation by -1 and add the system, thus eliminating the y variable.

  32. Linear Combination (Elimination) Method Step 2 Let’s transform the second equation: -1(-6x + y = 0) 6x – y = 0 Step 3 Add the new two equations up: -2x + y = 4 + 6x – y = 0 4x = 4

  33. Linear Combination (Elimination) Method Step 4 Solving the resulted one variable equation for x 4x = 4 yields x = 1. Step 5 Back substitute to one of the equation in the system to find the value for the other variable. Once we have the x value, we can plug it into either of our original equations and solve for y: -2x + y = 4 -6x + y = 0

  34. Linear Combination (Elimination) Method Step 5 Plugging x = 1 into the first equation yields: -2 (1) + y = 4 -2 + y = 4, y = 6 Step 6 Check the answer. -2(1) + 6 = 4 -6(1) + 6 = 0 So our solution is (1, 6).

  35. Linear Combination (Elimination) Method Let’s Try One More Together! Use the elimination method to solve the system: x – 5y = –2 3x + 2y = 11 The first step is: • Add the equations together • Transform the first equation by multiplying it by –3 • I’m not sure. I need to review the elimination method.

  36. Linear Combination (Elimination) Method Yes, notice that if we multiply the first equation by -3, we obtain additive inverses for the x coefficient. Our system is now: -3x + 15y = 6 3x + 2y = 11 The next step is: • Add the two equations • Solve the second equation for y • I’m not sure. I need to review.

  37. Linear Combination (Elimination) Method Yes, by adding the two equations we get: – 3x + 15y = 6 + 3x + 2y = 11 17y = 17 Solving fory, we get y = 1. The next step is: • Plug y = 1 into either equation and solve for x • Plug x = 1 into either equation and solve for y • I’m not sure. I need to review.

  38. Linear Combination (Elimination) Method We found that y=1, not x=1. So we must plug y=1 into either equation to solve for x not for y.

  39. Linear Combination (Elimination) Method Yes, now we can plug y = 1 into either of the original equations, or the transformed equation. Let’s choose the first original equation: x – 5y = –2 x – 5(1) = –2 x – 5 = –2 x = 3 So our solution is (3, 1), which must be the only solution to a system of two lines. Next, we must check the solved order pair is the solution to the system. (omitted here)

  40. You Try Solve the system using elimination method: 4x + 8y = 20 -4x + 2y = -30 The solution is: • (1, 7) • (3, -2) • (7, -1) • (-7, -1)

  41. You Try Solve the system using elimination method: 2x + 5y = 7 3x + y = -9 The solution is: • (3, -4) • (-4, -3) • (-4, 3) • (-3, -4)

  42. Summary 1. The key concepts of the linear combinations (elimination) method are to (a) variables are lined up properly (b) find out if there are any opposite coefficients for the same variable. If yes, directly add two equations up and eliminate one variable. If not, (c) multiply (or divide) some number(s) to equation(s) so that two opposite coefficients for the same variable show up. Then follow (b). 2. Some problems can be applied both substitution and linear combination (elimination) methods. However, it is better to use linear combination (elimination) method than to use the substitution method to some other problems. Choosing an appropriate method is important.

  43. Guided Practice – L 7.3 DHQ PAGE: 414 in Textbook Number: 4 • ANSWERS ARE ORDERED PAIRS! • Tonight’s Homework: Elimination WS #1-10 + Study for QUIZ (L 1-3) – QUIZ IS THURSDAY (1/27)

  44. Daily Warm-Up 7.5 Rewrite the equations to Slope-Intercept Form and Graph. ( y = mx + b ) Tonight’s Homework: Textbook Pages 429-430 #12-29 (USE ANY METHOD)

  45. 7.5 Special Types of Liner Systems Objectives 1) Identify linear system algebraically as having one solution, no solution, and infinite many solutions. 2) Identify linear system geometrically as having one solution, no solution, and infinite many solutions. Slope-Intercept Form

  46. Special Types of Linear Systems Example 1 Solve the linear equation system Solution 1 Substitution x = –2y + 7 From (2) To (1) 3(–2y + 7) – 2y = 5 The solution is (3, 2). –6y + 21 – 2y = 5 –8y = –16 y = 2 then x = –2(2) + 7 = 3

  47. (3 , 2) Special Types of Linear Systems Example 1 Solve the linear equation system Solution 2 Graphing

  48. NUMBER OF SOLUTIONS OF A LINEAR SYSTEM y x Lines intersect one solution

  49. A Linear System with No Solution Show that this linear system has no solution. 2x y  5 Equation 1 2x y  1 Equation 2 METHOD 1:SUBSTITUTION BecauseEquation 2can be revised toy –2x 1, you can substitute–2x 1foryinEquation 1. WriteEquation 1. 2x y  5 2x (–2x 1)  5 Substitute–2x 1fory. 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.

  50. A Linear System with No Solution 6 5 y  2x 5 4 3 2 y  2x 1 1 5 4 3 2 1 0 1 2 3 4 5 1 Show that this linear system has no solution. 2x y  5Equation 1 2x y  1Equation 2 METHOD 2:GRAPHING y  –2x 5Revised Equation 1 Rewrite each equation in slope-intercept form. y  –2x 1Revised Equation 2 Graph the linear system. The lines are parallel; they have the same slope but different y-intercepts. Parallel lines never intersect, so the system has no solution.

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