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homework_1_solutions

# homework_1_solutions - MA 265 HOMEWORK ASSIGNMENT#1...

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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: - 7 y - 14 z = 14 7 y - z = 16 - 15 z = 30 = z = - 2 . Substituting this value into the equation y + 2 z = - 2 yields y - 4 = - 2, so that y = 2. Finally, sub-

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homework_1_solutions - MA 265 HOMEWORK ASSIGNMENT#1...

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