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5-Minute Check on Activity 3-2

5-Minute Check on Activity 3-2. Solve the systems of equations: y = 2x + 1 y = 3x – 2. One None Infinite. Only a). 3x – 2 = y = 2x + 1 3x – 2 = 2x + 1 x – 2 = 1 x = 3 and y = 3(3) – 2 = 7.

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5-Minute Check on Activity 3-2

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  1. 5-Minute Check on Activity 3-2 Solve the systems of equations: y = 2x + 1y = 3x – 2 One None Infinite Only a) 3x – 2 = y = 2x + 1 3x – 2 = 2x + 1 x – 2 = 1 x = 3 and y = 3(3) – 2 = 7 Click the mouse button or press the Space Bar to display the answers.

  2. Activity 3 - 3 Healthy Lifestyle

  3. Objectives • Solve a 2x2 linear system algebraically using substitution method and the addition method • Solve equations containing parentheses

  4. Vocabulary • Addition method – combining multiples of equations to eliminate a variable

  5. Healthy Lifestyle You are trying to maintain a healthy lifestyle. You eat a well-balanced diet and exercise regularly. One of your favorite exercise activities is a combination of walking and jogging in your nearby park. One day, it takes you 1.3 hours to walk and jog a total of 5.5 miles in the park. You are curious about the amount of time you spent walking and the amount of time you spent jogging during the workout.

  6. Activity Continued One day, it takes you 1.3 hours to walk and jog a total of 5.5 miles in the park. Write an equation using x and y, for the total time of your walk/jog workout in the park. TT = x + y

  7. Activity Continued One day, it takes you 1.3 hours to walk and jog a total of 5.5 miles in the park. 2. If you walk at 3 miles per hour, write an expression (not an equation) that represents the distance you walked. • If you jog at 5 miles per hour, write an expression that represents the distance you jogged. • Write an equation for the total distance you walked/jogged in the park.  3x 5y D = 3x + 5y

  8. Activity Continued One day, it takes you 1.3 hours to walk and jog a total of 5.5 miles in the park. 5. Solve each of the equations (from 1 and 4) for y 6. Solve the system of equations from part 5 using the substitution method for last lesson TT = x + y D = 3x + 5y 5.5 – 3x = 5y 1/5(5.5 – 3x) = y 1.3 – x = y 1/5(5.5 – 3x) = y = 1.3 – x 5.5 – 3x = 6.5 – 5x 2x = 1 x = 0.5 y = 1.3 – 0.5 = 0.8

  9. y x Activity Continued Graph your equations and solve graphically (Window: Xmin= -2.5, Xmax= 2.5, Ymin= -2.5 and Ymax=2.5)

  10. Addition Method • Step 1:Line up like terms in each equation vertically. If necessary, multiply one or both equations by constants so that the coefficients of one of the variables are opposites • Step 2:Add the corresponding sides of the two equations (to eliminate a variable) • Step 3: Solve the resulting equation for the remaining variable • Step 4: Substitute the value from step 3 into one of the original equations and solve for the other variable

  11. Addition Example Solving systems of equations using addition method 3x + 4y = 12 and 2x + 5y = 20 Step 1: 3x + 4y = 12 2x + 5y = 20 Step 2: Setting up to get rid of x variables -2  (3x + 4y = 12)  -6x - 8y = -24 3  (2x + 5y = 20)  6x + 15y = 60 7y = 36 Step 3: 7y = 36 y = 36/7 = 5 1/7 Step 4: 3x + 4(36/7) = 12 3x + 144/7 = 12 3x = -60/7 x = -20/7

  12. Problem 4 Solving systems of equations using addition method x + y = 1.3 and 3x + 5y = 5.5 Step 1: x + y = 1.3 3x + 5y = 5.5 Step 2: Setting up to get rid of x variables -3  (x + y = 1.3)  -3x - 3y = -3.9 (3x + 5y = 5.5)  3x + 5y = 5.5 2y = 1.6 Step 3: 2y = 1.6 y = 0.8 Step 4: 3x + 5(0.8) = 5.5 3x + 4 = 5.5 3x = 1.5 x = 0.5

  13. Problem 5 Solving systems of equations using addition method -2x + 5y = -16 and 3x + 2y = 5 Step 1: -2x + 5y = -16 3x + 2y = 5 Step 2: Setting up to get rid of x variables 3  (-2x + 5y = -16)  -6x + 15y = -48 2  ( 3x + 2y = 5)  6x + 4y = 10 19y = -38 Step 3: 19y = -38 y = -38/19 = -2 Step 4: -2x + 5(-2) = -16 -2x + -10 = -16 -2x = -6 x = 3

  14. Problem 6 1.3 – x = y Solving walk/jog problem using addition method 5.5 – 3x = 5y Step 1: 3x + 4y = 12 2x + 5y = 20 Step 2: Setting up to get rid of x variables -2  (3x + 4y = 12)  -6x - 8y = -24 3  (2x + 5y = 20)  6x + 15y = 60 7y = 36 Step 3: 7y = 36 y = 36/7 = 5 1/7 Step 4: 3x + 4(36/7) = 12 3x + 144/7 = 12 3x = -60/7 x = -20/7

  15. Summary and Homework • Summary • The two methods for solving a 2 x 2 system of linear equations algebraically: • Substitution method • Addition method • Homework • pg 319 problems 1 – 3

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