Page 20 - Math Course 3 (Book 1)

Solving Linear Inequalities

Empty Set and All Reals
MO. 1 - L3b

Solving Linear Inequalities: Examples
Distributive Property
Solve –7(s + 4) + 11s ≥ 8s – 2(2s + 1).

–7(s + 4) + 11s ≥ 8s – 2(2s + 1) Original inequality
Let’s Begin –7s – 28 + 11s ≥ 8s – 4s – 2 Distributive Property

4s – 28 ≥ 4s – 2 Combine like terms.
Distributive Property 4s – 28 – 4s ≥ 4s – 2 – 4s Subtract 4s from

Examples each side.

– 28 ≥ – 2 Simplify.

Solve 3(–2x + 4) > –12. Since the inequality results in a
Answer false statement, the solution set
3(–2x + 4) > –12 Original inequality is the empty set Ø

–6x + 12 > –12 Distributive Property

–6x + 12 – 12 > –12 – 12 Subtract 12 from Solve 4d – 3 > 2d + 3(d + 3) – (d + 12).
each side.
4d – 3 > 2d + 3(d + 3) – (d + 12)Original inequality
–6x > –24 Simplify.
4d – 3 > 2d + 3d + 9 – d – 12 Distributive Property
–6x –24 Divide each side by
–6 < –6 –6 and change > to <. 4d – 3 > 4d + –3 Combine like terms.

x < 4 Simplify. 4d > 4d Add 3 to both sides.
d > d Divide each side
Answer The solution set is {x | x < 4} . by 4.

Since the inequality is always
Solve 6c + 3(2 – c) ≥ –2c + 1. Answer true, the solution set is
{d | d is a real number}.
6c + 3(2 – c) ≥ –2c + 1 Original inequality

6c + 6 – 3c ≥ –2c + 1 Distributive Property
Your Turn!
3c + 6 ≥ –2c + 1 Combine like terms.

3c + 6 + 2c ≥ –2c + 1 + 2c Add 2c to each side. Distributive Property

5c + 6 ≥ 1 Simplify. Solve 2(3x – 4) > – 20.

5c + 6 – 6 ≥ 1 – 6 Subtract 6 from each A. x > –2
side. B. x < –2
C. x > 2
5c ≥ –5 Simplify. D. x < 2

c ≥ –1 Divide each side by 5. Answer

Answer The solution set is {c | c ≥ –1}.

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