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CHAPTER3.7T
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à 3.7ïSolving Inequalities.
äïPlease solve the following inequalities.
â
êêêê2x - 3 < x + 8
êêêê2x - x - 3 < x - x + 8
êêêêx - 3 < 8
êêêêx - 3 + 3 < 8 + 3
êêêêx < 11
éS
The addition property for inequalities simply states that you can add
the same number to each side of an existing inequality.ïThis is the
same as the addition property for equations. The multiplication property
for inequalities has two cases.ïThe first case says that you can multi-
ply both sides of an existing inequality with the same positive number.
The second case states that if you multiply both sides of an existing
inequality with a negative number, then you must reverse the direction
of the inequality symbol.ïIn the problem, -2x < 4, it is necessary to
êêêè 1
multiply both sides by - ─ .ïThe inequality symbol must be reversed.
êêêè 2
êêê ┌è1 ┐êè ┌è1 ┐
êï-2x < 4ë │ - ─ │∙[-2x]ï>ï│ - ─ │∙[4]ë xï> -2
êêê └è2 ┘êè └è2 ┘
1
êêêè Solveè -9z > -72
èA)ï-8 < zê B) 9 < zê C) z < 8ê D) å of ç
üêêë-9z > -72
êêêè ┌ï1 ┐êè ┌ï1 ┐
êêêè │- ─ │∙[-9z]ï<ï│- ─ │∙[-72]
êêêè └ï9 ┘êè └ï9 ┘
êêêè z < 8
Ç C
2
êêêïSolveè24 - 6t < 8 + 2t
èA)ït > 2êïB) 6 < tê C) -8 > têD) å of ç
ü 24 - 6t < 8 + 2têêè ┌ï1 ┐êï┌ï1 ┐
ê24 - 6t - 2t < 8 + 2t -2tè ┌─>ï│- ─ │∙(-8t) > │- ─ │∙(-16)
ê24 - 8t < 8 + 0êê│è └ï8 ┘êï└ï8 ┘
ê24 - 8t < 8êêè │
ê24 - 24 - 8t < 8 - 24ê │è t > 2
ê0 - 8t < -16êêè│
ê-8t < -16 ───────────────────┘
Ç A
3
êêêëSolveè4k ≥ 2k - 10
èA)ï10 ≥ kê B) k ≥ -5êC) -4 ≤ kêD) å of ç
ü
êï4k ≥ 2k - 10êêê 1êë1
êï4k - 2k ≥ 2k - 2k - 10ë ┌─>è─ ∙ (2k)ï≥ï─ ∙ (-10)
êï2k ≥ 0 - 10êêè│ë2êë2
êï2k ≥ -10 ───────────────────┘
êêêêêêk ≥ -5
Ç B
4
êêêïSolveè3ï≤ïx + 4ï≤ï9
èA)ï2 ≤ x ≤ 8ëB) -1 ≤ x ≤ 5èC) -4 ≤ x ≤ 2èD) å of ç
ü
êêêè 3ï≤ïx + 4ï≤ï9
êêêè 3 + (-4)ï≤ïx + 4 + (-4)ï≤ï9 + (-4)
êêêè -1ï≤ïx + 0ï≤ï5
êêêè -1ï≤ïxï≤ï5
Ç B
5
êêê Solveè-9ï≤ï2x - 3ï≤ï5
èA)ï2 ≤ x ≤ 8ëB) -3 ≤ x ≤ 4èC) -4 ≤ x ≤ 2èD) å of ç
üêêï-9ï≤ï2x - 3ï≤ï5
êêê -9 + 3ï≤ï2x - 3 + 3ï≤ï5 + 3
êêê -6ï≤ï2xï≤ï8
êêê 1êë1êè1
êêê ─ ∙ (-6)ï≤ï─ ∙ 2xï≤ï─ ∙ 8
êêê 2êë2êè2
êêê -3ï≤ïxï≤ï4
Ç B
6èTranslate and solve. If twice a number is added to six, the
êëresult is greater than nine.ïFind the solution.
êêêêêêï3
èA)ïx > -4ê B) x < -3êC) x > ─ê D) å of ç
êêêêêêï2
üêè2x + 6 > 9êêêêï3
êêï2x + 6 + (-6)ï>ï9 + (-6)è ┌─> xï>ï─
êêï2x > 3êêêè│êï2
êêï1êë1êêï│
êêï─ ∙ (2x)ï>ï─ ∙ (3) ─────────┘
êêï2êë2
Ç C
7
êêêSolveè 6ï<ïx - 7ï<ï8
A) 13 < x < 15ëB) -1 < x < 1ëC) -1 < x < 8ëD) å of ç
üêêê6 < x - 7 < 8
êêêë 6 + 7 < x - 7 + 7 < 8 + 7
êêêë 13ï<ïxï<ï15
Ç A
8
êêêïSolveè 4m - 8 ≥ 7m + 4
ëA)ïm ≥ 7êB)ïm ≥ -4ë C)ïm ≤ -4ë D) å of ç
üêêë4m - 8 ≥ 7m + 4
êêêè 4m - 4m - 8 ≥ 7m - 4m + 4
êêêè -8 - 4 ≥ï3m + 4 - 4
êêêè -12 ≥ 3m
êêêè -12è3m
êêêë── ≥ ──
êêêë 3è 3
êêêë-4 ≥ mïorïm ≤ -4
Ç C
9
êêêSolveè 8ï<ï2m + 6 ≤ï15
êê9
èA) 1 < m ≤ ─ë B) 2 < m < 5è C) 2 < m ≤ 16èD) å of ç
êê2
üêêê8 < 2m + 6 ≤ 15
êêêë 8 - 6 < 2m + 6 - 6 ≤ 15 - 6
êêêë 2 < 2m ≤ 9
êêêë 2è2mè9
êêêë ─ < ── ≤ ─
êêêë 2è 2è2
êêêêê9
êêêë 1 < m ≤ ─
êêêêê2
Ç A
10
êêêSolveè 7ï<ï3m + 4 ≤ï15
è 7êêêè 15êêï19
A) ─ < m ≤ 12ëB) 1 < m ≤ ──ëC) 1 < m ≤ ──ëD) å of ç
è 3êêêë3êêè3
üêêê7 < 3m + 4 ≤ 15
êêêë 7 - 4 < 3m + 4 - 4 ≤ 15 - 4
êêêë 3 < 3m ≤ 11
êêêë 3è3mè11
êêêë ─ < ── ≤ ──
êêêë 3è 3è 3
êêêêê11
êêêë 1 < m ≤ ──
êêêêê 3
Ç D