3.6 part 2 notes

# 3.6 part 2 notes - and solve for y and z(elimination 4-7y...

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1 3.6 Notes (part 2) Solving Three-Variable Systems by Substitution Solving by Substitution = - + = - + - - = + - 10 2 2 1 2 4 4 2 z y x z y x z y x Step 1: Choose an equation and solve for one of its variables 1) x – 2y + z = -4 Step 2: Substitute the expression for x into each of the other two equations. 2) -4x + y – 2z = 1 3) 2x + 2y – z = 10 Solving by Substitution Step 3: Call those new equations 4 and 5. Put them into a system
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Unformatted text preview: and solve for y and z (elimination). 4)-7y + 2z = -15 5) 6y – 3z = 18 Solving by Substitution Step 4: Substitute values for y and z into one of the original equations. 1) x – 2y + z = -4 You try! = + +--=--= +-7 2 2 5 2 6 3 z y x z y x z y x Homework p. 153 # 10-15...
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## This note was uploaded on 05/13/2011 for the course MTH 98 taught by Professor Johnson during the Fall '09 term at Grand Valley State.

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