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By: Jeffrey Bivin Lake Zurich High School jeff.bivin@lz95.org Last Updated: October 9, 2007

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What is the intersection of 3 planes? We will be looking at the case where they intersect in a point. Jeff Bivin -- LZHS

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I think I’ll do some substitution. 3z = 12 Jeff Bivin -- LZHS 2x + 3y - 5z = 13 - 5y + z = -1

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3z = 12 z = 4 - 5y + 4 = -1 2x + 3(1) – 5(4) = 13 2x + 3 – 20 = 13 2x - 17 = 13 x = 15 - 5y + 4 = -1 - 5y = -5 y = 1 2x = 30 Jeff Bivin -- LZHS 2x + 3y - 5z = 13 - 5y + z = -1 3z = 12

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Decide which variable you want to eliminate. I think I’ll choose to eliminate the y variable. 2x + 5y + 8z = 8 3x - 2y + 4z = 26 2x + 4y + 3z = -3 Jeff Bivin -- LZHS

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2x + 5y + 8z = 8 3x - 2y + 4z = 26 2x + 4y + 3z = -3 6x - 4y + 8z = 52 2x + 4y + 3z = -3 8x + 11z = 49 4x + 10y + 16z = 16 15x - 10y + 20z = 130 19x + 36z = 146 -152x - 288z = -1168 152x + 209z = 931 -79z = -237 z = 3 19x + 36(3) = 146 19x = 38 x = 2 2(2) + 5y + 8(3) = 8 4 + 5y + 24 = 8 5y = -20 y = -4 Jeff Bivin -- LZHS

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Decide which variable you want to eliminate. I think I’ll choose to eliminate the x variable. 5x + 2y + 7z = 19 3x - 3y + 5z = 34 x + 3y - 5z = -22 Jeff Bivin -- LZHS

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3x - 3y + 5z = 34 -3x - 9y + 15z = 66 -12y + 20z = 100 5x + 2y + 7z = 19 -5x - 15y + 25z = 110 -13y + 32z = 129 154y - 384z = -1548 -154y + 260z = 1300 -124z = -248 z = 2 -13y + 32(2) = 129 -13y = 65 y = -5 5x + 2(-5) + 7(2) = 19 5x - 10 + 14 = 19 5x = 15 x = 3 5x + 2y + 7z = 19 3x - 3y + 5z = 34 x + 3y - 5z = -22 Jeff Bivin -- LZHS

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Decide which variable you want to eliminate. I think I’ll choose to eliminate the z variable. 2x + 3y - 5z = -12 4x - 5y + 3z = 12 2x - 3y = -7 Jeff Bivin -- LZHS

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6x + 9y - 15z = -36 20x - 25y + 15z = 60 26x - 16y = 24 -26x + 39y = 91 23y = 115 y = 5 26x – 80 = 24 26x = 104 x = 4 2(4) + 3(5) – 5z = -12 8 + 15 – 5z = -12 -5z = -35 z = 7 2x + 3y - 5z = -12 4x - 5y + 3z = 12 2x - 3y = -7 26x – 16(5) = 24 Jeff Bivin -- LZHS

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