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51. 3x – 7> 2 52. 2x + 5 >1 53. 1 — 2х < 3 54. 2 – 3x < 5 55. 3x – 12 3 + x 56. 2х - 2 2 3 +х 57. – 2(x + 3) < 8 58. — 3(1 — х) < 12 59. 4 — 3(1 — х) < 3 60. 8 – 4(2 – x) < -2r 61. – (x – 4) > x + 8 62. 3x + 4 >(x – 2) 63. 0 - 2х — 6- 4 64. 4 < 2x + 2 < 10 65. -5 < 4 – 3x s 2 2х — 1 <0 4 Зх + 2 66. – 3 < 3 – 2x < 9 67. – 3 < 68. 0 <= < 4 1 69. 1 <1- 1 70. 0 < 1 Ex <1 71. (x + 2) (x – 3) > (x – 1) (x + 1) 72. (x – 1) (x + 1) > (x – 3) (x + 4) 73. |2x| < 8 74. 3x| < 15 2.

51. 3x – 7> 2 52. 2x + 5 >1 53. 1 — 2х < 3 54. 2 – 3x < 5 55. 3x – 12 3 + x 56. 2х - 2 2 3 +х 57. – 2(x + 3) < 8 58. — 3(1 — х) < 12 59. 4 — 3(1 — х) < 3 60. 8 – 4(2 – x) < -2r 61. – (x – 4) > x + 8 62. 3x + 4 >(x – 2) 63. 0 - 2х — 6- 4 64. 4 < 2x + 2 < 10 65. -5 < 4 – 3x s 2 2х — 1 <0 4 Зх + 2 66. – 3 < 3 – 2x < 9 67. – 3 < 68. 0 <= < 4 1 69. 1 <1- 1 70. 0 < 1 Ex <1 71. (x + 2) (x – 3) > (x – 1) (x + 1) 72. (x – 1) (x + 1) > (x – 3) (x + 4) 73. |2x| < 8 74. 3x| < 15 2.

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.4: Applications
Problem 2EQ: 2. Suppose that in Example 2.27, 400 units of food A, 500 units of B, and 600 units of C are placed...
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Solve each inequality algebraically. Express your answer using set notation or interval notation. Graph the solution set. Verify your results using a graphing utility.

51. 3x – 7> 2 52. 2x + 5 >1 53. 1 — 2х < 3 54. 2 – 3x < 5 55. 3x – 12 3 + x 56. 2х - 2 2 3 +х 57. – 2(x + 3) < 8 58. — 3(1 — х) < 12 59. 4 — 3(1 — х) < 3 60. 8 – 4(2 – x) < -2r 61. – (x – 4) > x + 8 62. 3x + 4 >(x – 2) 63. 0 - 2х — 6- 4 64. 4 < 2x + 2 < 10 65. -5 < 4 – 3x s 2 2х — 1 <0 4 Зх + 2 66. – 3 < 3 – 2x < 9 67. – 3 < 68. 0 <= < 4 1 69. 1 <1- 1 70. 0 < 1 Ex <1 71. (x + 2) (x – 3) > (x – 1) (x + 1) 72. (x – 1) (x + 1) > (x – 3) (x + 4) 73. |2x| < 8 74. 3x| < 15 2.
Transcribed Image Text:51. 3x – 7> 2 52. 2x + 5 >1 53. 1 — 2х < 3 54. 2 – 3x < 5 55. 3x – 12 3 + x 56. 2х - 2 2 3 +х 57. – 2(x + 3) < 8 58. — 3(1 — х) < 12 59. 4 — 3(1 — х) < 3 60. 8 – 4(2 – x) < -2r 61. – (x – 4) > x + 8 62. 3x + 4 >(x – 2) 63. 0 - 2х — 6- 4 64. 4 < 2x + 2 < 10 65. -5 < 4 – 3x s 2 2х — 1 <0 4 Зх + 2 66. – 3 < 3 – 2x < 9 67. – 3 < 68. 0 <= < 4 1 69. 1 <1- 1 70. 0 < 1 Ex <1 71. (x + 2) (x – 3) > (x – 1) (x + 1) 72. (x – 1) (x + 1) > (x – 3) (x + 4) 73. |2x| < 8 74. 3x| < 15 2.
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