3.6 part 2 notes c

# 3.6 part 2 notes c - Put them into a system and solve for y...

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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.
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Unformatted text preview: Put them into a system 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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