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Unformatted text preview: ==+= +19 3 4 15 3 2 4 3 3 z y x z y x z y x Solving by Elimination Step 2: Now write 4 and 5 as a system. Solve for x and z. 3x+2z = 11 x=5, z = 2 == + 34 2 6 11 2 3 z x z x Solving by Elimination Step 3: Substitute values for x and z into one of the original equations (1,2, or 3) and solve for y. x=5, z=2 x – 3y + 3z = 4 2 You try! == +=+ 1 2 3 5 2 z y x z y x z y x Solving by Elimination You may have to multiply an equation by a nonzero number in order to eliminate. For example: =+ = + + =+ 12 2 6 15 16 2 4 5 2 z y x z y x z y x Homework p. 157 # 19...
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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 University.
 Fall '09
 Johnson
 Algebra

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