HOMEWORK 5
RICKY NG
Write the solution in interval notation:
Question 5.1. x ≥ 5
Solution. x is on the right of 5 on the number line, including 5. So, [5, ∞).
Question 5.2. x < 7
Solution. x is on the left of 7 on the number line, excluding 7. So, (−∞, 7).
Question 5.3. −7 ≤ x ≤ 2
Solution. x is in between −7 and 2, including both endpoints. So, [−7, 2].
Question 5.4. 3x ≥ 24
Solution. Treat as equation and divide by 3 ( positive number !):
3x ≥ 24
3x
24
≥
3
3
x≥8,
so [8, ∞).
Question 5.5. −4x < 40
Solution. Divide by −4 ( negative number ! ):
−4x < 40
−4x 40
>
−4 −4
x> − 10,
so (−10, ∞).
Question 5.6. 3x + 4 ≤ −17
Solution. Solve like an equation. Note that 3 is positive so inequality sign is unchanged.
3x + 4−4 ≤ −17−4
3x ≤ −21
−21
3x
≤
3
3
x ≤ −7,
so (−∞, −7].
1
Question 5.7. 5 − 9x ≤ 3x − 7
Solution. Solve like an equation: separate x’s and constants:
5−5 − 9x−3x ≤ 3x−3x − 7−5
−12x ≤ −12
Divide by −12, which is negative:
−12x −12
≥
−12 −12
x ≥ 1,
so [1, ∞).
Question 5.8. −4(3 − 2x) ≤ −(x + 20)
Solution. First get grid of the parentheses by multiplying inside:
−12 + 8x ≤ −x − 20.
Then separate unknown and constants:
−12+12 + 8x+x ≤ −x+x − 20+12
9x ≤ 8
9x
8
≤
9
9
8
x≤ ,
9
so (−∞, 89 ].
Question 5.9. 52 (x + 12 ) > − 31 (10 − x)
Solution. This is in lecture 13: Recap. Solution is (−53, ∞).
Question 5.10. −19 < 5 − 4x ≤ −3
Solution. Keep the balancing rule in mind, even though we have three expressions. First
single out the x
−19−5 <
−24 <
5−5 − 4x ≤
−4x ≤
Divide by −4 and flip the inequality signs:
−24
−8
>x≥ ,
−4
−4
so 6 > x ≥ 2, and solution is [2, 6).
2
−3−5
−8