1 / 116

Warm Up Evaluate each expression for x = 1 and y =–3. 1. x – 4 y 2. –2 x + y

Warm Up Evaluate each expression for x = 1 and y =–3. 1. x – 4 y 2. –2 x + y Write each expression in slope-intercept form. 3. y – x = 1 4. 2 x + 3 y = 6 5. 0 = 5 y + 5 x. 13. –5. y = x + 1. y = x + 2. y = – x. Objectives.

zuri
Télécharger la présentation

Warm Up Evaluate each expression for x = 1 and y =–3. 1. x – 4 y 2. –2 x + y

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Warm Up Evaluate each expression for x = 1 and y =–3. 1.x – 4y2. –2x + y Write each expression in slope-intercept form. 3.y –x = 1 4. 2x + 3y =6 5. 0 = 5y + 5x 13 –5 y = x + 1 y =x + 2 y = –x

  2. Objectives Identify solutions of linear equations in two variables. Solve systems of linear equations in two variables by graphing.

  3. A system of linear equations is a set of two or more linear equations containing two or more variables. A solution of a system of linear equations with two variables is an ordered pair that satisfies each equation in the system. So, if an ordered pair is a solution, it will make both equations true.

  4. 3x – y 13 3(5) – 2 13 0 2 – 2 0 15 – 2 13 0 0  13 13  Example 1A: Identifying Systems of Solutions Tell whether the ordered pair is a solution of the given system. (5, 2); 3x – y = 13 Substitute 5 for x and 2 for y in each equation in the system. The ordered pair (5, 2) makes both equations true. (5, 2) is the solution of the system.

  5. Helpful Hint If an ordered pair does not satisfy the first equation in the system, there is no reason to check the other equations.

  6. –x + y = 2 x + 3y = 4 –(–2) + 2 2 –2 + 3(2) 4 4 2 –2 + 6 4 4 4 Example 1B: Identifying Systems of Solutions Tell whether the ordered pair is a solution of the given system. x + 3y = 4 (–2, 2); –x + y = 2 Substitute –2 for x and 2 for y in each equation in the system.  The ordered pair (–2, 2) makes one equation true but not the other. (–2, 2) is not a solution of the system.

  7. y = 2x – 1 y = –x + 5 All solutions of a linear equation are on its graph. To find a solution of a system of linear equations, you need a point that each line has in common. In other words, you need their point of intersection. The point (2, 3) is where the two lines intersect and is a solution of both equations, so (2, 3) is the solution of the systems.

  8. Helpful Hint Sometimes it is difficult to tell exactly where the lines cross when you solve by graphing. It is good to confirm your answer by substituting it into both equations.

  9. Check Substitute (–1, –1) into the system. y = –2x– 3 y = x (–1) (–1) (–1)–2(–1)–3  –12– 3 –1 –1  –1 – 1 Example 2A: Solving a System Equations by Graphing Solve the system by graphing. Check your answer. y = x Graph the system. y = –2x – 3 The solution appears to be at (–1, –1). y = x • (–1, –1) y = –2x – 3 (–1, –1) is the solution of the system.

  10. y + x = –1 y + x = –1 y = x –6 −x−x y = Example 2B: Solving a System Equations by Graphing Solve the system by graphing. Check your answer. y = x –6 Graph using a calculator and then use the intercept command. Rewrite the second equation in slope-intercept form.

  11. Check Substitute into the system. y = x–6 y = x –6 + – 1 – 6 –1  –1  –1 – 1 The solution is . Example 2B Continued Solve the system by graphing. Check your answer.

  12. Example 3:Problem-Solving Application Wren and Jenni are reading the same book. Wren is on page 14 and reads 2 pages every night. Jenni is on page 6 and reads 3 pages every night. After how many nights will they have read the same number of pages? How many pages will that be?

  13. 1 Understand the Problem Example 3 Continued The answer will be the number of nights it takes for the number of pages read to be the same for both girls. List the important information: Wren on page 14 Reads 2 pages a night Jenni on page 6 Reads 3 pages a night

  14. Make a Plan Total pages every night already read. number read is plus Wren y = 2 x 14 + 2 x y 3 + Jenni = 6 Example 3 Continued Write a system of equations, one equation to represent the number of pages read by each girl. Let x be the number of nights and y be the total pages read.

  15. 3 Solve  (8, 30) Nights Example 3 Continued Graph y = 2x + 14 and y = 3x + 6. The lines appear to intersect at (8, 30). So, the number of pages read will be the same at 8 nights with a total of 30 pages.

  16. 2(8) + 14 = 16 + 14 = 30  3(8) + 6 = 24 + 6 = 30 4 Look Back Example 3 Continued Check (8, 30) using both equations. Number of days for Wren to read 30 pages. Number of days for Jenni to read 30 pages.

  17. Warm-Up Tell whether the ordered pair is a solution of the given system. 1. (–3, 1) 2. (2, –4) 3. Joy has 5 collectable stamps and will buy 2 more each month. Ronald has 25 collectable stamps and will sell 3 each month. After how many months will they have the same number of stamps? How many will that be?

  18. Objective Solve linear equations in two variables by substitution.

  19. Sometimes it is difficult to identify the exact solution to a system by graphing. In this case, you can use a method called substitution. The goal when using substitution is to reduce the system to one equation that has only one variable. Then you can solve this equation by the methods taught in Chapter 2.

  20. Solve for one variable in at least one equation, if necessary. Step 1 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. Write the values from steps 3 and 4 as an ordered pair, (x, y), and check.

  21. Step 2 y = x – 2 3x = x – 2 Step 3 –x –x 2x = –2 2x = –2 2 2 x = –1 Example 1A: Solving a System of Linear Equations by Substitution Solve the system by substitution. y = 3x y =x – 2 Step 1y = 3x Both equations are solved for y. y = x – 2 Substitute 3x for y in the second equation. Solve for x. Subtract x from both sides and then divide by 2.

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

  23. Step 2 4x + y = 6 4x+(x + 1) = 6 Step 3 –1 –1 5x = 5 5x = 5 5 5 x = 1 Example 1B: Solving a System of Linear Equations by Substitution Solve the system by substitution. y =x + 1 4x + y = 6 The first equation is solved for y. Step 1y = x + 1 Substitute x + 1 for y in the second equation. 5x + 1 = 6 Simplify. Solve for x. Subtract 1 from both sides. Divide both sides by 5.

  24. Step 4 y = x + 1 y = 1 + 1 y = 2 Step 5 (1, 2) y = x + 1 4x + y = 6 2 1 + 1 4(1)+ 2 6 2 2  6 6  Example1B Continued Solve the system by substitution. Write one of the original equations. Substitute 1 for x. Write the solution as an ordered pair. Check Substitute (1, 2) into both equations in the system.

  25. Sometimes you substitute an expression for a variable that has a coefficient. When solving for the second variable in this situation, you can use the Distributive Property.

  26. Caution When you solve one equation for a variable, you must substitute the value or expression into the other originalequation, not the one that had just been solved.

  27. Step 1 y + 6x = 11 –6x – 6x y = –6x + 11 Step 2 3x + 2y = –5 3x+ 2(–6x + 11) = –5 Example 2: Using the Distributive Property y + 6x = 11 Solve by substitution. 3x + 2y = –5 Solve the first equation for y by subtracting 6x from each side. Substitute –6x + 11 for y in the second equation. 3x + 2(–6x + 11) = –5 Distribute 2 to the expression in parenthesis.

  28. – 22 –22 –9x = –27 –9x = –27 –9–9 Example 2 Continued y + 6x = 11 Solve by substitution. 3x + 2y = –5 Step 3 Simplify. Solve for x. 3x + 2(–6x) +2(11) = –5 3x –12x + 22 = –5 –9x + 22 = –5 Subtract 22 from both sides. Divide both sides by –9. x = 3

  29. Step 4 y + 6x = 11 –18 –18 y = –7 Example 2 Continued y + 6x = 11 Solve by substitution. 3x + 2y = –5 Write one of the original equations. y + 6(3) = 11 Substitute 3 for x. Simplify. y + 18 = 11 Subtract 18 from each side. Step 5 (3, –7) Write the solution as an ordered pair.

  30. Example 2: Consumer Economics Application Jenna is deciding between two cell-phone plans. The first plan has a $50 sign-up fee and costs $20 per month. The second plan has a $30 sign-up fee and costs $25 per month. After how many months will the total costs be the same? What will the costs be? If Jenna has to sign a one-year contract, which plan will be cheaper? Explain. Write an equation for each option. Let t represent the total amount paid and m represent the number of months.

  31. + Option 1 t = $50 m $20 $30 $25 t + Option 2 = m Step 1 t = 50 + 20m t = 30 + 25m Step 2 50 + 20m = 30 + 25m Example 2 Continued Total paid sign-up fee for each month. payment amount is plus Both equations are solved for t. Substitute 50 + 20m for t in the second equation.

  32. Step 3 50 + 20m = 30 + 25m –20m – 20m –30–30 20 = 5m 5 5 m =4 Step 4 t = 30 + 25m t = 30 + 100 t = 130 Example 2 Continued Solve for m. Subtract 20m from both sides. 50 = 30 + 5m Subtract 30 from both sides. 20 = 5m Divide both sides by 5. Write one of the original equations. t = 30 + 25(4) Substitute 4 for m. Simplify.

  33. Example 2 Continued Write the solution as an ordered pair. Step 5 (4, 130) In 4 months, the total cost for each option would be the same $130. If Jenna has to sign a one-year contract, which plan will be cheaper? Explain. Option 1: t = 50 + 20(12) = 290 Option 2: t = 30 + 25(12) = 330 Jenna should choose the first plan because it costs $290 for the year and the second plan costs $330.

  34. Warm-Up

  35. Objectives Solve systems of linear equations in two variables by elimination. Compare and choose an appropriate method for solving systems of linear equations.

  36. Another method for solving systems of equations is elimination. Like substitution, the goal of elimination is to get one equation that has only one variable. To do this by elimination, you add the two equations in the system together.

  37. Write the system so that like terms are aligned. Step 1 Eliminate one of the variables and solve for the other variable. Step 2 Substitute the value of the variable into one of the original equations and solve for the other variable. Step 3 Solving Systems of Equations by Elimination Write the answers from Steps 2 and 3 as an ordered pair, (x, y), and check. Step 4

  38. Step 1 3x– 4y = 10 4x = 8 4 4 x = 2 Example 1: Elimination Using Addition 3x –4y = 10 Solve by elimination. x + 4y = –2 Write the system so that like terms are aligned. x + 4y = –2 Add the equations to eliminate the y-terms. Step 2 4x + 0 = 8 4x = 8 Simplify and solve for x. Divide both sides by 4.

  39. Step 3 x + 4y = –2 –2 –2 4y = –4 4y –4 44 y = –1 Example 1 Continued Write one of the original equations. 2 + 4y = –2 Substitute 2 for x. Subtract 2 from both sides. Divide both sides by 4. Step 4 (2, –1) Write the solution as an ordered pair.

  40. When two equations each contain the same term, you can subtract one equation from the other to solve the system. To subtract an equation add the oppositeof each term.

  41. 2x + y = –5 –2x+ 5y = –13 Step 2 0 + 6y = –18 6y = –18 y = –3 Example 2: Elimination Using Subtraction 2x + y = –5 Solve by elimination. 2x – 5y = 13 2x + y = –5 Step 1 Add the opposite of each term in the second equation. –(2x – 5y = 13) Eliminate the x term. Simplify and solve for y.

  42. Step 3 2x + y = –5 +3 +3 Step 4 (–1, –3) Example 2 Continued Write one of the original equations. 2x + (–3) = –5 Substitute –3 for y. 2x – 3 = –5 Add 3 to both sides. 2x = –2 Simplify and solve for x. x = –1 Write the solution as an ordered pair.

  43. Remember! Remember to check by substituting your answer into both original equations.

  44. In some cases, you will first need to multiply one or both of the equations by a number so that one variable has opposite coefficients. This will be the new Step 1.

  45. x + 2y = 11 Step 1 –2(–3x + y = –5) x + 2y = 11 +(6x –2y = +10) Step 2 7x = 21 x = 3 Example 3A: Elimination Using Multiplication First Solve the system by elimination. x + 2y = 11 –3x + y = –5 Multiply each term in the second equation by –2 to get opposite y-coefficients. Add the new equation to the first equation. 7x + 0 = 21 Simplify and solve for x.

  46. x + 2y = 11 Step 3 –3 –3 2y = 8 Step 4 (3, 4) Example 3A Continued Write one of the original equations. 3 + 2y = 11 Substitute 3 for x. Subtract 3 from each side. Simplify and solve for y. y = 4 Write the solution as an ordered pair.

  47. Step 1 2(–5x + 2y =32) 5(2x + 3y =10) –10x + 4y = 64 +(10x + 15y =50) Example 3B: Elimination Using Multiplication First Solve the system by elimination. –5x + 2y = 32 2x + 3y = 10 Multiply the first equation by 2 and the second equation by 5 to get opposite x-coefficients Add the new equations. Step 2 19y =114 Simplify and solve for y. y =6

  48. 2x + 3y = 10 Step 3 –18 –18 (–4, 6) Step 4 Write the solution as an ordered pair. 2x =–8 Example 3B Continued Write one of the original equations. 2x + 3(6) = 10 Substitute 6 for y. 2x + 18 = 10 Subtract 18 from both sides. x =–4 Simplify and solve for x.

  49. Step 1 3(2x + 5y = 26) Step 2 0 + 7y = 28 Check It Out! Example 3b Solve the system by elimination. 2x + 5y = 26 –3x – 4y = –25 Multiply the first equation by 3 and the second equation by 2 to get opposite x-coefficients +(2)(–3x – 4y = –25) 6x + 15y = 78 +(–6x –8y = –50) Add the new equations. y = 4 Simplify and solve for y.

  50. 2x + 5y = 26 Step 3 2x + 20 = 26 Write the solution as an ordered pair. Step 4 (3, 4) –20 –20 2X = 6 Check It Out! Example 3b Continued Write one of the original equations. 2x + 5(4) = 26 Substitute 4 for y. Subtract 20 from both sides. Simplify and solve for x. x = 3

More Related