Let’s solve each inequality one by one. We’ll isolate the variable on one side, just like solving equations — but remember: if you multiply or divide by a negative number, flip the inequality sign!
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1) 3x – 5 < 10
Add 5 to both sides:
3x < 15
Divide by 3:
x < 5
✔ Final Answer for #1:
x < 5
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2) –2 + 2c ≥ –16
Add 2 to both sides:
2c ≥ –14
Divide by 2:
c ≥ –7
✔ Final Answer for #2:
c ≥ –7
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3) 8 ≥ –3h – 1
Add 1 to both sides:
9 ≥ –3h
Divide by –3 → FLIP the sign!
–3 ≤ h (or h ≥ –3)
✔ Final Answer for #3:
h ≥ –3
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4) –4 < –16 + 4d
Add 16 to both sides:
12 < 4d
Divide by 4:
3 < d (or d > 3)
✔ Final Answer for #4:
d > 3
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5) 13 + 4z ≥ 9
Subtract 13 from both sides:
4z ≥ –4
Divide by 4:
z ≥ –1
✔ Final Answer for #5:
z ≥ –1
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6) –12 < 3(–p + 2)
First, distribute the 3:
–12 < –3p + 6
Subtract 6 from both sides:
–18 < –3p
Divide by –3 → FLIP the sign!
6 > p (or p < 6)
✔ Final Answer for #6:
p < 6
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7) –45 ≥ –5(c + 3)
Distribute the –5:
–45 ≥ –5c – 15
Add 15 to both sides:
–30 ≥ –5c
Divide by –5 → FLIP the sign!
6 ≤ c (or c ≥ 6)
✔ Final Answer for #7:
c ≥ 6
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8) 2(3x – 6) < 12
Distribute the 2:
6x – 12 < 12
Add 12 to both sides:
6x < 24
Divide by 6:
x < 4
✔ Final Answer for #8:
x < 4
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9) 7(v – 8) < 14
Divide both sides by 7 first (easier than distributing):
v – 8 < 2
Add 8 to both sides:
v < 10
✔ Final Answer for #9:
v < 10
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10) 3(²⁄₃ – s) ≥ –4
Wait — this looks like it might be a typo? Probably meant:
3(⅔s – s) ≥ –4? Or maybe
3(²⁄₃ – s) ≥ –4?
Looking again: It says “3(²⁄₃ – s) ≥ –4” — so let’s assume it’s written correctly:
3 times (two-thirds minus s) is greater than or equal to –4.
So:
Distribute the 3:
3*(²⁄₃) – 3s ≥ –4
→ 2 – 3s ≥ –4
Subtract 2 from both sides:
–3s ≥ –6
Divide by –3 → FLIP the sign!
s ≤ 2
✔ Final Answer for #10:
s ≤ 2
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Final Answers:
1) x < 5
2) c ≥ –7
3) h ≥ –3
4) d > 3
5) z ≥ –1
6) p < 6
7) c ≥ 6
8) x < 4
9) v < 10
10) s ≤ 2
Parent Tip: Review the logic above to help your child master the concept of solving inequalities practice worksheet.