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Solving Systems of Equations by Elimination

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Presentation on theme: "Solving Systems of Equations by Elimination"— Presentation transcript:

1 Solving Systems of Equations by Elimination
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2 Using the Addition Method
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3 Using the Addition Method
Finish problem by solving for c using substitution. Multiply by -2 BACK

4 Addition Method - Seven Steps
Step 1 Write both equations of the system in standard form Ax + By = C Step 2 If necessary multiply one or both equations by appropriate numbers so that the coefficients of x or y are negatives of each other. BACK

5 Addition Method - Seven Steps
Step 3 Add the two equations to get one equation with only one variable. Step 4 Solve the equation from Step 3. Step 5 Substitute the solution from Step 4 into either of the original equations. BACK

6 Addition Method - Seven Steps
Step 6 Solve the resulting equation from Step 5 for the remaining variable. Step 7 Check the answer. BACK

7 Solve. Multiply by 2 Multiply by 3 BACK

8 Your turn, Example Step 1 Write both equations of the system in standard form Ax + By = C BACK

9 Standard Form

10 Multiply by -2 Multiply by 3

11 Example Step 6 Solve the resulting equation from Step 5 for the remaining variable. BACK

12 Example Step 7 Check the answer. BACK

13 Example BACK

14 Using the Addition Method
Multiply by -2 What does it mean? BACK

15 A false statement would indicate the lines of the two equation would not intersect and therefore has no solution. BACK

16 Addition Method Multiply by 4 Multiply by 3 What does it mean? BACK

17 This is a true statement, therefore the equations are the same and would make two lines that coincide with each other and produce an infinite amount of solutions BACK

18 Once upon a time… …a handsome math professor was in a barnyard that was full of pigs and chickens. BACK

19 Once upon a time… The handsome math professor counted all of the heads of the pigs and chickens. The result was 30. The handsome math professor counted all of the feet of the pigs and chickens. The result was 84. How many pigs were there? How many chickens were there?

20 Once upon a time… 4p + 2c = 84 Let p = # of pigs Let c = # chickens
Counting Equation p + c = 30 Value Equation 4p + 2c = 84 BACK

21 Pigs and Chickens p + c = 30 4p + 2c = 84 -2p – 2c = -60 2p = 24
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22 Systems of Equations Elimination Using Addition, Subtraction, & Multiplication
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