AB
YL
D
U
ST
, AB
YL
D
U
ST
, AB
YL
D
U
ST
, 1.1 Linear, Rational, and Absolute Value Equations 35
28. Subtract 5a from both sides of 5a = 6a to 40. Multiply by 60x.
get 0 = a. The latter equation is conditional
whose solution set is {0}. 12 − 15 + 20 = −17x
17 = −17x
29. Note, 5x = 4x is equivalent to x = 0. The
latter equation is conditional whose solution A conditional equation with solution set {−1}.
set is {0}.
41. Multiply by 3(z − 3).
30. Note, 4(y − 1) = 4y − 4 is true by the
3(z + 2) = −5(z − 3)
distributive law. The equation is an identity
and the solution set is R. 3z + 6 = −5z + 15
8z = 9
31. Equivalently, we get 2x + 6 = 3x − 3 or 9 =
ST
x. The latter equation is conditional whose 9
A conditional equation with solution set .
solution set is {9}. 8
32. Equivalently, we obtain 2x + 2 = 3x + 2 or 42. Multiply by (x − 4).
0 = x. The latter equation is conditional 2x − 3 = 5
whose solution set is {0}.
2x = 8
U
33. Using the distributive property, we find x = 4
3x − 18 = 3x + 18 Since division by zero is not allowed, x = 4
D
−18 = 18. does not satisfy the original equation. We have
an inconsistent equation and so the solution
The equation is inconsistent and the solution
set is ∅.
set is ∅.
YL
43. Multiplying by (x − 3)(x + 3).
34. Since 5x = 5x + 1 or 0 = 1, the equation is
inconsistent and the solution set is ∅. (x + 3) − (x − 3) = 6
35. An identity and the solution set is {x|x 6= 0}. 6 = 6
36. An identity and the solution set is {x|x 6= −2}. An identity with solution set
{x|x 6= 3, x 6= −3}.
AB
37. Multiplying by 2(w − 1), we get
44. Multiply by (x + 1)(x − 1).
1 1 1
− =
w − 1 2w − 2 2w − 2 4(x + 1) − 9(x − 1) = 3
2 − 1 = 1. 4x + 4 − 9x + 9 = 3
An identity and the solution set is {w|w 6= 1} −5x = −10
38. Multiply by x(x − 3). A conditional equation with solution set {2}.
(x − 3) + x = 9 45. Multiply by (y − 3).
2x = 12
4(y − 3) + 6 = 2y
A conditional equation with solution set {6}. 4y − 6 = 2y
39. Multiply by 6x. y =3
6−2 = 3+1 Since division by zero is not allowed, y = 3
does not satisfy the original equation. We have
4 = 4
an inconsistent equation and so the solution
An identity with solution set {x|x 6= 0}. set is ∅.
Copyright 2015 Pearson Education, Inc.
YL
D
U
ST
, AB
YL
D
U
ST
, AB
YL
D
U
ST
, 1.1 Linear, Rational, and Absolute Value Equations 35
28. Subtract 5a from both sides of 5a = 6a to 40. Multiply by 60x.
get 0 = a. The latter equation is conditional
whose solution set is {0}. 12 − 15 + 20 = −17x
17 = −17x
29. Note, 5x = 4x is equivalent to x = 0. The
latter equation is conditional whose solution A conditional equation with solution set {−1}.
set is {0}.
41. Multiply by 3(z − 3).
30. Note, 4(y − 1) = 4y − 4 is true by the
3(z + 2) = −5(z − 3)
distributive law. The equation is an identity
and the solution set is R. 3z + 6 = −5z + 15
8z = 9
31. Equivalently, we get 2x + 6 = 3x − 3 or 9 =
ST
x. The latter equation is conditional whose 9
A conditional equation with solution set .
solution set is {9}. 8
32. Equivalently, we obtain 2x + 2 = 3x + 2 or 42. Multiply by (x − 4).
0 = x. The latter equation is conditional 2x − 3 = 5
whose solution set is {0}.
2x = 8
U
33. Using the distributive property, we find x = 4
3x − 18 = 3x + 18 Since division by zero is not allowed, x = 4
D
−18 = 18. does not satisfy the original equation. We have
an inconsistent equation and so the solution
The equation is inconsistent and the solution
set is ∅.
set is ∅.
YL
43. Multiplying by (x − 3)(x + 3).
34. Since 5x = 5x + 1 or 0 = 1, the equation is
inconsistent and the solution set is ∅. (x + 3) − (x − 3) = 6
35. An identity and the solution set is {x|x 6= 0}. 6 = 6
36. An identity and the solution set is {x|x 6= −2}. An identity with solution set
{x|x 6= 3, x 6= −3}.
AB
37. Multiplying by 2(w − 1), we get
44. Multiply by (x + 1)(x − 1).
1 1 1
− =
w − 1 2w − 2 2w − 2 4(x + 1) − 9(x − 1) = 3
2 − 1 = 1. 4x + 4 − 9x + 9 = 3
An identity and the solution set is {w|w 6= 1} −5x = −10
38. Multiply by x(x − 3). A conditional equation with solution set {2}.
(x − 3) + x = 9 45. Multiply by (y − 3).
2x = 12
4(y − 3) + 6 = 2y
A conditional equation with solution set {6}. 4y − 6 = 2y
39. Multiply by 6x. y =3
6−2 = 3+1 Since division by zero is not allowed, y = 3
does not satisfy the original equation. We have
4 = 4
an inconsistent equation and so the solution
An identity with solution set {x|x 6= 0}. set is ∅.
Copyright 2015 Pearson Education, Inc.
