Download presentation

Presentation is loading. Please wait.

Published byMaia Patt Modified over 2 years ago

1
Do Now 1/13/10 Take out HW from last night. Text p. 447, #6-28 evens Text p. 447, #6-28 evens Copy HW in your planner Text p. 454, #10-26 evens, #37 & #39 Text p. 454, #10-26 evens, #37 & #39 Quiz Sections Friday Quiz Sections Friday In your journal, list the possible ways to solve a linear system. Then solve the following systems. 5x + 6y = 50 -x + 6y = 26 -8y + 6x = 36 6x – y = 15 (4, 5) (2, -3)

2
Homework Text p. 447, #6-28 evens 6) (12,6) 8) (-1,2) 10) (-2,2) 12) (0,-10) 14) (-3,1) 16) (-15,-62) 18) (-6,-1) 20) (5,3) 22) B 24) Did not change the sign for -3x when rearranging the equations. y = 9 26) (4,-9) 28) (5,8)

3
Objective SWBAT to solve linear systems using elimination (adding, subtracting, and multiplying).

4
“How Do You Solve a Linear System???” (1) Solve Linear Systems by Graphing (7.1) (2) Solve Linear Systems by Substitution (7.2) (3) Solve Linear Systems by ELIMINATION!!! (7.3) Adding or Subtracting (4) Solve Linear Systems by Multiplying First (7.4) Then eliminate.

5
Section 7.3 “Solve Linear Systems by Adding or Subtracting” ELIMINATION- adding or subtracting equations to obtain a new equation in one variable. Solving Linear Systems Using Elimination (1) Add or Subtract the equations to eliminate one variable. (2) Solve the resulting equation for the other variable. (3) Substitute in either original equation to find the value of the eliminated variable.

6
Equation 1 Equation 1 -2x + 5y = 13 -2x + 5y = 13 Equation 2 Equation 2 2x + 3y = 11 2x + 3y = 11 “Solve Linear Systems by Elimination” Equation 1 Equation 1 2x + 3y = 11 2x + 3y = 11 Substitute value for y into either of the original equations 2x + 3(3) = 11 2x + 3(3) = 11 2x + 9 = 11 2x + 9 = 11 The solution is the point (1,3). Substitute (1,3) into both equations to check. -2(1) + 5(3) = 13 -2(1) + 5(3) = = 13 2(1) + 3(3) = 11 2(1) + 3(3) = = 11 ADDITION + 8y = 24 8y = 24 y = 3 y = 3 x = 1 x = 1 Eliminated

7
Section 7.4 “Solve Linear Systems by Multiplying First” ELIMINATION- adding or subtracting equations to obtain a new equation in one variable. Solving Linear Systems Using Elimination (1) Multiply the whole equation by a constant in order to be able to eliminate a variable. (2) Add or Subtract the equations to eliminate one variable. (3) Solve the resulting equation for the other variable. (4) Substitute in either original equation to find the value of the eliminated variable.

8
Equation 1 Equation 1 2x + 3y = 5 2x + 3y = 5 Equation 2 Equation 2 6x + 5y = 19 6x + 5y = 19 “Solve Linear Systems by Elimination Multiplying First!!” Equation 2 Equation 2 2x + 3y = 5 2x + 3y = 5 Substitute value for y into either of the original equations 2x + 3(-1) = 5 2x + 3(-1) = 5 2x - 3 = 5 2x - 3 = 5 The solution is the point (4,-1). Substitute (4,-1) into both equations to check. 6(4) + 5(-1) = 19 6(4) + 5(-1) = = 19 2(4) + 3(-1) = 5 2(4) + 3(-1) = 5 5 = 5 MultiplyFirst + -4y = 4 -4y = 4 y = -1 y = -1 x = 4 x = 4 Eliminated x (-3) -6x – 9y = x – 9y = -15 6x + 5y = 19 6x + 5y = 19

9
Equation 1 Equation 1 3x + 10y = -3 3x + 10y = -3 Equation 2 Equation 2 2x + 5y = 3 2x + 5y = 3 “Solve Linear Systems by Elimination Multiplying First!!” Equation 1 Equation 1 2x + 5y = 3 2x + 5y = 3 Substitute value for x into either of the original equations 2(9) + 5y = 3 2(9) + 5y = y = y = 3 The solution is the point (9,-3). Substitute (9,-3) into both equations to check. 2(9) + 5(-3) = 3 2(9) + 5(-3) = 3 3 = 3 3(9) + 10(-3) = -3 3(9) + 10(-3) = = -3 MultiplyFirst + -x = -9 -x = -9 x = 9 x = 9 y = -3 y = -3 Eliminated x (-2) 3x + 10y = -3 3x + 10y = -3 -4x - 10y = -6 -4x - 10y = -6

10
Equation 1 Equation 1 -3x + 2y = -9 -3x + 2y = -9 Equation 2 Equation 2 4x + 5y = 35 4x + 5y = 35 “Solve Linear Systems by Elimination Multiplying First!!” Equation 1 Equation 1 4x + 5y = 35 4x + 5y = 35 Substitute value for x into either of the original equations 4(5) + 5y = 35 4(5) + 5y = y = y = 35 The solution is the point (5,3). Substitute (5,3) into both equations to check. 4(5) + 5(3) = 35 4(5) + 5(3) = = 35 -3(5) + 2(3) = -9 -3(5) + 2(3) = = -9 MultiplyFirst + 23x = x = 115 x = 5 x = 5 y = 3 y = 3 Eliminated x (2) 15x - 10y = 45 8x + 10y = 70 8x + 10y = 70 x (-5)

11
Equation 1 Equation 1 6x + 13y = -9 6x + 13y = -9 Equation 2 Equation 2 9x + 2y = 39 9x + 2y = 39 “Solve Linear Systems by Elimination Multiplying First!!” Equation 1 Equation 1 9x + 2y = 39 9x + 2y = 39 Substitute value for y into either of the original equations 9x + 2(-3) = 39 9x + 2(-3) = 39 9x - 6 = 39 9x - 6 = 39 The solution is the point (5,-3). Substitute (5,-3) into both equations to check. 9(5) + 2(-3) = 39 9(5) + 2(-3) = = 39 6(5) + 13(-3) = -9 6(5) + 13(-3) = = -9 MultiplyFirst + -35y = y = 105 y = -3 y = -3 x = 5 x = 5 Eliminated x (2) -18x - 39y = 27 18x + 4y = 78 18x + 4y = 78 x (-3)

12
Guided Practice x + y = 2 x + y = 2 2x + 7y = 9 6x – 2y = 1 -2x + 3y = -5 (1,1) (-0.5, -2) 3x - 7y = 5 9y = 5x + 5 (-10,-5)

13
Communicators Solve the linear systems using elimination on your communicators. When you are finished raise your board!

14
Homework Text p. 454, #10-26 evens, #37 & #39 Text p. 454, #10-26 evens, #37 & #39 Quiz Sections Friday. Quiz Sections Friday. NJASK7 prep

Similar presentations

© 2016 SlidePlayer.com Inc.

All rights reserved.

Ads by Google