MAC1105HW3 - Solutions to Section 2.2 and 2.3...

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Unformatted text preview: Solutions to Section 2.2 and 2.3 harringt@math.ufl.edu June 29, 2005 Section 2.2 30.) a(a + 4) − 6(a + 4) a2 + 4a − 6a − 24 −2a − 24 −4a a = = = = = a(a − 4) + 6(a − 4) a2 − 4a + 6a − 24 2a − 24 0 0 32.) 3x − [5 − 2(x + 1)] 3x − [5 − 2x − 2] 3x − [3 − 2x] 3x − 3 + 2 x 5x − 3 11x x = = = = = = = x − 7(x + 2) x − 7x − 14 −6x − 14 −6x − 14 −6x − 14 −11 −1 34.) 5d − [6d + 4(5 − d)] = 3 − [2d − (2d − 1)] 5d − [6d + 20 − 4d] = 3 − [2d − 2d + 1] 5d − [6d + 20 − 4d] = 3 − [1] 5d − [2d + 20] = 3 − 1 1 5d − 2d − 20 = 2 3d − 20 = 2 3d = 22 22 d= 3 36.) 2a 1 − 3 2 4a 3 − 6 6 4a − 3 6 4a − 3 −2a − 3 −2a =a =a =a = 6a =0 =3 −3 a= 2 38.) 5a −a 2 5a 2a − 2 2 5a − 2a 2 3a −a a = = = = = = 1 2 4a 1 + 2 2 4a + 1 2 4a + 1 1 −1 2a + 40.) t−5 t − 3 2 2t 3 ∗ (t − 5) − 6 6 2 = 6 = 6 2t 3t − 15 − 6 6 2t − 3t + 15 6 −t + 15 6 −t + 15 −t t = 6 = 6 = 6 = 36 = 21 = −21 Section 2.3 44.) 5(x − 2) − 3(x + 4) ≥ 2x − 22 5x − 10 − 3x − 12 ≥ 2x − 22 2x − 22 ≥ 2x − 22 This is an identity. So the solution in interval notation is: (−∞, ∞) 46.) −3 < r − 3 ≤ 1 0<r ≤ 4 The solution in interval notation is: (0, 4] 54.) 13 < 7 − 2(x − 3) 13 < 7 − 2x + 6 13 < 13 − 2x 0 < −2x 0>x −9 ≤ x ≤ ≤ ≤ ≤ ≥ < The solution in interval notation is: [−9, 0) 3 31 31 31 18 −9 0 ...
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