In Exercises 1–6, solve the equation Ax = b by using the LU factorization given for A. In Exercises 1 and 2, also solve Ax = b by ordinary row reduction.
2.
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- In Exercises 49–54, solve for y in terms of t or x, as appropriate.arrow_forwardFor Exercises 5–10, a. Simplify the expression. b. Substitute 0 for h in the simplified expression. 2(x + h)? + 3(x + h) · 5. (2x + 3x) 3(x + h - 4(x + h) – (3x - 4x) 6. h 1 1 1 1 (x + h) – 2 7. x - 2 2(x + h) + 5 8. 2x + 5 h (x + h) – x 9. (x + h) 10. - X h harrow_forwardExercises 7–12: Determine whether the equation is linear or nonlinear by trying to write it in the form ax + b = 0. 7. 3x – 1.5 = 7arrow_forward
- For Exercises 33–42, solve the equation on the interval [0, 2π). (See Examples 5–7)arrow_forwardFor Exercises 101–104, verify by substitution that the given values of x are solutions to the given equation. 101. x + 25 = 0 102. x + 49 = 0 a. x = 5i a. x = 7i b. x = -5i b. x = -7i 103. x - 4x + 7 = 0 104. x - 6x + 11 = 0 a. x = 2 + iV3 b. x = 2 – iV3 a. x = 3 + iVā b. x = 3 – iV2arrow_forwardYou may find it helpful to review the information in the Reasonable Answers box from this section before answering Exercises 13–16. Verify that the solutions you found to Exercise 9 are indeed homogeneous solutions.arrow_forward
- In Exercises 59–64, solve and check each linear equation. 59. 2x – 5 = 7 60. 5x + 20 = 3x 61. 7(x – 4) = x + 2 62. 1 - 2(6 – x) = 3x + 2 63. 2(x – 4) + 3(x + 5) = 2x – 2 64. 2x 4(5x + 1) = 3x + 17arrow_forward2. Expand and simplify to express each equation in standard form. [2, 2, 2, 3, 3, 2] a) y = (x + 5)(x + 2) b) y = (x- 4)(x+ 3) c) y = (2x- 3)(x-6) d) y = (3x + 4)2 e) y = 3(x – 4)2 +1 f) y = (4x + 3)(4x- 3)arrow_forwardFind a and b if a + (5 – 3b) i = ai + 2 (b + 5)arrow_forward
- z=a+ib a-arrow_forwardIn Exercises 43–54, solve each absolute value equation or indicate the equation has no solution. 43. |x – 2| = 7 45. |2x – 1| = 5 47. 2|3x – 2| = 14 44. |x + 1| = 5 46. |2r – 3| = 11 48. 3|2x – 1| = 21 %3D %3D 5 24 - + 6 = 18 50. 4 1 x + 7 = 10 51. |x + 1| + 5 = 3 53. |2x – 1| + 3 = 3 52. |x + 1| + 6 = 2 54. |3x – 2| + 4 = 4arrow_forwardExercises 43–52: Complete the following. (a) Solve the equation symbolically. (b) Classify the equation as a contradiction, an identity, or a conditional equation. 43. 5x - 1 = 5x + 4 44. 7- 9: = 2(3 – 42) – z 45. 3(x - 1) = 5 46. 22 = -2(2x + 1.4) 47. 0.5(x – 2) + 5 = 0.5x + 4 48. 눈x-2(x-1)3-x + 2 2x + 1 2x 49. 50. x – 1.5 2- 3r - 1.5 51. -6 52. 0.5 (3x - 1) + 0.5x = 2x – 0.5arrow_forward
- Algebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:Cengage