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Quiz #6b Answers July 15, 2014 Math 141 1. solve the system of equations by any method shown in class. x + 3y + z = 10 7y − z + 2x = 21 41 = z + 13y + 4x x 1 2 4 y 3 7 13 z 1 −1 1 10 21 41 rref → x 1 0 0 y 0 1 0 z 10 −3 0 7 1 0 Answer: x = 7 − 10z y = 1 + 3z z = any number 2. Parametric solution: x = −8 + 4z y = 33 − 5z z = any number We can not buy a part of an animal. So all of the variables must be an integer. We also know that all of the variables must be greater than or equal to zero. x≥0 −8 + 4z ≥ 0 4z ≥ 8 z≥2 y≥0 33 − 5z ≥ 0 33 ≥ 5z 33/5 ≥ z z ≤ 33 5 ≈ 6.6 z≥0 In addition we know that the variables can not be any larger than 25 x ≤ 25 −8 + 4z ≤ 24 4z ≤ 32 z≤8 y ≤ 25 33 − 5z ≤ 24 7 ≤ 5z 7/5 ≤ z z ≥ 75 ≈ 1.4 z ≤ 25 Taken all together, we find that z = 2, 3, 4, 5, 6.