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Solving Systems of Equations by Elimination Day 2

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Presentation on theme: "Solving Systems of Equations by Elimination Day 2"— Presentation transcript:

1 Solving Systems of Equations by Elimination Day 2
Teacher Twins©2014 Systems

2 Warm Up (-2, 5) (6, -2) (0, 3) Solve each system. –x + y = 7 x + y = 3

3 Solving Systems of Equations by Elimination Day 2

4 Elimination Using Multiplication

5 Solve the system by elimination. 2x + y = 23 3x + 2y = 37
Multiply first equation by -2. To do elimination you must have opposites. In this problem you must make opposites. Look for the easiest way to make opposites. -2( 2x + y = 23) -4x + -2y = -46 -4x + -2y = -46 3x + 2y = 37 Add the equations. Using the new equation for the first equation. -x = -9 x = 9 Find the y value by using one of the equations. 2(9) + y = 23 18 + y = 23 y = 5 Answer (9, 5)

6 Try on your own 2x – y = 4 7x + 3y = 27 (3, 2)

7 4(-1) + 3y = 8 -4 + 3y = 8 Answer (-1, 4) y = 4
Solve the system by elimination. 4x + 3y = 8 3x – 5y = -23 You must multiply both equations to get opposites. Think of a common denominator. I will make the 3y and -5y , 15y and -15y. 5(4x + 3y = 8) 3(3x – 5y = -23) 20x + 15y = 40 9x – 15y = -69 Add the new equations. 29x + 0 = -29 x = - 1 4(-1) + 3y = 8 -4 + 3y = 8 y = 4 Use one of the original equations to find y. Answer (-1, 4)

8 Try on your own 9x – 2y = -8 -7x + 3y = 12 (0, 4)

9 Closure 2x + 4y = 5 3x – 5y = 6 Which variable is the easiest to turn in to opposites? Why? The y variable would be the easiest because it already has opposite signs.


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