 4.3.1: If c 7 0, u = c is equivalent to u 5 ______________ or u 5 ________...
 4.3.2: u = v is equivalent to u 5 ______________ or u 5 ______________.
 4.3.3: If c 7 0, u 6 c is equivalent to ______________ 6 u 6 ______________.
 4.3.4: If c 7 0, u 7 c is equivalent to u , ______________ or u 7 ________...
 4.3.5: u 6 c has no solution if c ______________ 0.
 4.3.6: u 7 c is true for all real numbers if c ______________ 0.
 4.3.7: 3x  1 = 5
 4.3.8: 3x  1 5
 4.3.9: 3x  1 5
 4.3.10: 3x  1 = x
 4.3.11: 3x  1 6 5
 4.3.12: 3x  1 7 5
 4.3.13: 2 y + 6 = 10
 4.3.14: 3 y + 5 = 12
 4.3.15: 3 2x  1 = 21
 4.3.16: 2 3x  2 = 14
 4.3.17: 6y  2 + 4 = 32
 4.3.18: 3y  1 + 10 = 25
 4.3.19: 7 5x + 2 = 16
 4.3.20: 7 3x + 2 = 16
 4.3.21: x + 1 + 5 = 3
 4.3.22: x + 1 + 6 = 2
 4.3.23: 4y + 1 + 10 = 4
 4.3.24: 3y  2 + 8 = 1
 4.3.25: 2x  1 + 3 = 3
 4.3.26: 3x  2 + 4 = 4
 4.3.27: 5x  8 = 3x + 2
 4.3.28: 4x  9 = 2x + 1
 4.3.29: 2x  4 = x  1
 4.3.30: 6x = 3x  9
 4.3.31: 2x  5 = 2x + 5
 4.3.32: 3x  5 = 3x + 5
 4.3.33: x  3 = 5  x
 4.3.34: x  3 = 6  x
 4.3.35: 2y  6 = 10  2y
 4.3.36: 4y + 3 = 4y + 5
 4.3.37: ` 2x 3  2 ` = ` x 3 + 3 `
 4.3.38: x 2  2 ` = ` x  1 2 `
 4.3.39: x 6 3
 4.3.40: x 6 5
 4.3.41: x  2 6 1
 4.3.42: x  1 6 5
 4.3.43: x + 2 1
 4.3.44: x + 1 5
 4.3.45: 2x  6 6 8
 4.3.46: 3x + 5 6 17
 4.3.47: x 7 3
 4.3.48: x 7 5
 4.3.49: x + 3 7 1
 4.3.50: x  2 7 5
 4.3.51: x  4 2
 4.3.52: x  3 4
 4.3.53: 3x  8 7 7
 4.3.54: 5x  2 7 13
 4.3.55: 2(x  1) + 4 8
 4.3.56: 3(x  1) + 2 20
 4.3.57: ` 2x + 6 3 ` 6 2
 4.3.58: ` 3x  3 4 ` 6 6
 4.3.59: ` 2x + 2 4 ` 2
 4.3.60: ` 3x  3 9 ` 1
 4.3.61: 3  2x 3 ` 7 5
 4.3.62: ` 3  3x 4 ` 7 9
 4.3.63: x  2 6 1
 4.3.64: x  3 6 2
 4.3.65: x + 6 7 10
 4.3.66: x + 4 7 12
 4.3.67: x + 2 + 9 16
 4.3.68: x  2 + 4 5
 4.3.69: 2 2x  3 + 10 7 12
 4.3.70: 3 2x  1 + 2 7 8
 4.3.71: 4 1  x 6 16
 4.3.72: 2 5  x 6 6
 4.3.73: 3 2x  1
 4.3.74: 9 4x + 7
 4.3.75: Let f(x) = 5  4x . Find all values of x for which f(x) = 11.
 4.3.76: Let f(x) = 2  3x . Find all values of x for which f(x) = 13.
 4.3.77: Let f(x) = 3  x and g(x) = 3x + 11 . Find all values of x for whic...
 4.3.78: Let f(x) = 3x + 1 and g(x) = 6x  2 . Find all values of x for whic...
 4.3.79: Let g(x) = 1 + 3(x + 1) . Find all values of x for which g(x) 5.
 4.3.80: Let g(x) = 3 + 4(x + 1) . Find all values of x for which g(x) 3.
 4.3.81: Let h(x) = 2x  3 + 1. Find all values of x for which h(x) 7 6.
 4.3.82: Let h(x) = 2x  4  6. Find all values of x for which h(x) 7 18.
 4.3.83: When 3 times a number is subtracted from 4, the absolute value of t...
 4.3.84: When 4 times a number is subtracted from 5, the absolute value of t...
 4.3.85: ax + b 6 c
 4.3.86: ax + b c
 4.3.87: 4  x = 1
 4.3.88: 4  x 6 5
 4.3.89: 2x + 1 3
 4.3.90: 2x + 1 3
 4.3.91: Solve the inequality: x  21 3. Interpret the solution in terms of ...
 4.3.92: Solve the inequality: x  15 3. Interpret the solution in terms of ...
 4.3.93: The inequality T  57 7 describes the range of monthly average temp...
 4.3.94: The inequality T  50 22 describes the range of monthly average tem...
 4.3.95: Solve: x  8.6 0.01. If the length of the machine part is supposed ...
 4.3.96: Solve: x  9.4 0.01. If the length of the machine part is supposed ...
 4.3.97: If a coin is tossed 100 times, we would expect approximately 50 of ...
 4.3.98: Explain how to solve an equation containing one absolute value expr...
 4.3.99: Explain why the procedure that you described in Exercise 98 does no...
 4.3.100: Describe how to solve an absolute value equation with two absolute ...
 4.3.101: Describe how to solve an absolute value inequality of the form u 6 ...
 4.3.102: Explain why the procedure that you described in Exercise 101 does n...
 4.3.103: Describe how to solve an absolute value inequality of the form u 7 ...
 4.3.104: Explain why the procedure that you described in Exercise 103 does n...
 4.3.105: x + 1 = 5
 4.3.106: 3(x + 4) = 12
 4.3.107: 2x  3 = 9  4x
 4.3.108: 2x + 3 6 5
 4.3.109: 2x  1 3 ` 6 5 3
 4.3.110: x + 4 6 1
 4.3.111: 2x  1 7 7
 4.3.112: 0.1x  0.4 + 0.4 7 0.6
 4.3.113: x + 4 7 1
 4.3.114: Use a graphing utility to verify the solution sets for any five equ...
 4.3.115: I solved x  2 = 5 by rewriting the equation as x  2 = 5 or x + 2 ...
 4.3.116: I solved x  2 7 5 by rewriting the inequality as 5 6 x  2 7 5.
 4.3.117: Because the absolute value of any expression is never less than a n...
 4.3.118: Ill win the contest if I can complete the crossword puzzle in 20 mi...
 4.3.119: All absolute value equations have two solutions.
 4.3.120: The equation x = 6 is equivalent to x = 6 or x = 6.
 4.3.121: Values of 5 and 5 satisfy x = 5, x 5, and x 5.
 4.3.122: The absolute value of any linear expression is greater than 0 for a...
 4.3.123: Write an absolute value inequality for which the interval shown is ...
 4.3.124: The percentage, p, of defective products manufactured by a company ...
 4.3.125: Solve: 2x + 5 = 3x + 4.
 4.3.126: 3x  5y = 15 (Section 2.4, Example 1)
 4.3.127: f(x) =  2 3 x (Section 2.4, Example 5)
 4.3.128: f(x) = 2 (Section 2.4, Example 6)
Solutions for Chapter 4.3: Equations and Inequalities Involving Absolute Value
Full solutions for Intermediate Algebra for College Students  6th Edition
ISBN: 9780321758934
Solutions for Chapter 4.3: Equations and Inequalities Involving Absolute Value
Get Full SolutionsChapter 4.3: Equations and Inequalities Involving Absolute Value includes 128 full stepbystep solutions. Intermediate Algebra for College Students was written by and is associated to the ISBN: 9780321758934. This textbook survival guide was created for the textbook: Intermediate Algebra for College Students, edition: 6. This expansive textbook survival guide covers the following chapters and their solutions. Since 128 problems in chapter 4.3: Equations and Inequalities Involving Absolute Value have been answered, more than 9510 students have viewed full stepbystep solutions from this chapter.

Cholesky factorization
A = CTC = (L.J]))(L.J]))T for positive definite A.

Commuting matrices AB = BA.
If diagonalizable, they share n eigenvectors.

Diagonal matrix D.
dij = 0 if i # j. Blockdiagonal: zero outside square blocks Du.

Diagonalization
A = S1 AS. A = eigenvalue matrix and S = eigenvector matrix of A. A must have n independent eigenvectors to make S invertible. All Ak = SA k SI.

GramSchmidt orthogonalization A = QR.
Independent columns in A, orthonormal columns in Q. Each column q j of Q is a combination of the first j columns of A (and conversely, so R is upper triangular). Convention: diag(R) > o.

Hessenberg matrix H.
Triangular matrix with one extra nonzero adjacent diagonal.

Identity matrix I (or In).
Diagonal entries = 1, offdiagonal entries = 0.

Inverse matrix AI.
Square matrix with AI A = I and AAl = I. No inverse if det A = 0 and rank(A) < n and Ax = 0 for a nonzero vector x. The inverses of AB and AT are B1 AI and (AI)T. Cofactor formula (Al)ij = Cji! detA.

Kronecker product (tensor product) A ® B.
Blocks aij B, eigenvalues Ap(A)Aq(B).

Least squares solution X.
The vector x that minimizes the error lie 112 solves AT Ax = ATb. Then e = b  Ax is orthogonal to all columns of A.

Linear transformation T.
Each vector V in the input space transforms to T (v) in the output space, and linearity requires T(cv + dw) = c T(v) + d T(w). Examples: Matrix multiplication A v, differentiation and integration in function space.

Markov matrix M.
All mij > 0 and each column sum is 1. Largest eigenvalue A = 1. If mij > 0, the columns of Mk approach the steady state eigenvector M s = s > O.

Multiplier eij.
The pivot row j is multiplied by eij and subtracted from row i to eliminate the i, j entry: eij = (entry to eliminate) / (jth pivot).

Normal equation AT Ax = ATb.
Gives the least squares solution to Ax = b if A has full rank n (independent columns). The equation says that (columns of A)·(b  Ax) = o.

Outer product uv T
= column times row = rank one matrix.

Positive definite matrix A.
Symmetric matrix with positive eigenvalues and positive pivots. Definition: x T Ax > 0 unless x = O. Then A = LDLT with diag(D» O.

Projection matrix P onto subspace S.
Projection p = P b is the closest point to b in S, error e = b  Pb is perpendicularto S. p 2 = P = pT, eigenvalues are 1 or 0, eigenvectors are in S or S...L. If columns of A = basis for S then P = A (AT A) 1 AT.

Singular Value Decomposition
(SVD) A = U:E VT = (orthogonal) ( diag)( orthogonal) First r columns of U and V are orthonormal bases of C (A) and C (AT), AVi = O'iUi with singular value O'i > O. Last columns are orthonormal bases of nullspaces.

Trace of A
= sum of diagonal entries = sum of eigenvalues of A. Tr AB = Tr BA.

Tridiagonal matrix T: tij = 0 if Ii  j I > 1.
T 1 has rank 1 above and below diagonal.
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