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Unformatted text preview: MA 265 HOMEWORK ASSIGNMENT #1 SOLUTIONS #1. Page 8; Exercise 1. In Exercises 1 through 14, solve each given linear system by the method of elimination. x + 2 y = 8 3 x 4 y = 4 Solution: First we eliminate x . Multiply the first equation by ( 3), then add to the second: 3 x 6 y = 24 3 x 4 y = 4 10 y = 20 = y = 2 . Substituting this value into the equation x + 2 y = 8 yields x + 4 = 8, so that x = 4. Hence x = 4, y = 2. #2. Page 8; Exercise 2. In Exercises 1 through 14, solve each given linear system by the method of elimination. 2 x 3 y + 4 z = 12 x 2 y + z = 5 3 x + y + 2 z = 1 Solution: First we eliminate x . Multiply the second equation by ( 2), then add to the first; and multiply the second equation by ( 3) then add to the third: 2 x 3 y + 4 z = 12 2 x + 4 y 2 z = 10 y + 2 z = 2 3 x + 6 y 3 z = 15 3 x + y + 2 z = 1 7 y z = 16 This gives the system of equations y + 2 z = 2 7 y z = 16 Now we eliminate y . Multiple the first equation by ( 7) and add to the second:...
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This note was uploaded on 03/11/2010 for the course MATH 261A taught by Professor ... during the Spring '10 term at Purdue University Calumet.
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