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By: Jeffrey Bivin Lake Zurich High School Last Updated: October 11, 2005.

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Presentation on theme: "By: Jeffrey Bivin Lake Zurich High School Last Updated: October 11, 2005."— Presentation transcript:

1 By: Jeffrey Bivin Lake Zurich High School Last Updated: October 11, 2005

2 y = 3x + 5 2x + 4y = 34 2x + 4(3x + 5) = 34 2x + 12x + 20 = 34 14x + 20 = 34 14x = 14 x = 1 y = 3(1) + 5 y = y = 8 Jeff Bivin -- LZHS

3 Decide which variable you want to eliminate. I think I’ll choose to eliminate the y variable. Jeff Bivin -- LZHS 3x – 5y = 14 2x + 4y = -20

4 3x – 5y = 14 2x + 4y = x - 20y = 56 10x + 20y = x = -44 x = -2 3(-2) – 5y = y = 14 -5y = 20 y = -4 Jeff Bivin -- LZHS

5 Decide which variable you want to eliminate. I think I’ll choose to eliminate the x variable. Jeff Bivin -- LZHS 2x + 7y = 48 3x + 5y = 28

6 2x + 7y = 48 3x + 5y = 28 6x + 21y = x - 10y = y = 88 y = 8 3x + 5(8) = 28 3x + 40 = 28 3x = -12 x = -4 Jeff Bivin -- LZHS

7 Decide which variable you want to eliminate. I think I’ll choose to eliminate the x variable. Jeff Bivin -- LZHS 4x + 3y = -19 6x + 5y = -32

8 4x + 3y = -19 6x + 5y = x + 9y = x - 10y = 64 -y = 7 y = -7 6x + 5(-7) = -32 6x - 35 = -32 6x = 3 Jeff Bivin -- LZHS

9 y = -2x - 6 6x + 3y = 11 6x + 3(-2x - 6) = 11 6x - 6x - 18 = = 11 Jeff Bivin -- LZHS

10 x = 5y + 1 2x - 10y = 2 2(5y + 1) - 10y = 2 10y y = 2 2 = 2 Jeff Bivin -- LZHS


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