Solve the inequality.

$\frac{1}{3}$(2x - 9) ≥ 2(2x + 1)

Get rid of th fraction by multiplying both sides of the equation by the denominator, 3

3 x $\frac{1}{3}$(2x - 9) ≥ 3 x 2(2x + 1)

1 x (2x - 9) ≥ 6(2x + 1)

2x - 9 ≥ 12x + 6

2x - 12x ≥ 6 + 9

-10x ≥ 15

Divide both sides by -10. When dividing by a negative, the inequality sign changes.

x ≤ $\frac{15}{-10}$

x ≤ -1.5

Notes:

1. -1.5 is between -1 and - 2

2. For ≥ and ≤, the solid is shaded

3. For ≤, the arrow is to the left just as the direction of the sign and for ≥, the arrow is to the right

More than 18 km means the distance d is greater than 18 km and hence 18 km is less than the distance (18 < d)

Not more than 23 km means it can be 23 or less hence the less than or equal to sign is used (d ≤ 23)

When you combine the two, the inequality will be 18 < d ≤ 23 which is read as 18 is less than d and d is also less than or equal to 23.

$\frac{1}{3}$(5 - 3x) < $\frac{2}{5}$(3 - 7x)

Get rid of the fractions by multiplying both sides by the L.C.M of the denominators (3 and 5).

The L.C.M of 3 and 5 is 15

15 x $\frac{1}{3}$(5 - 3x) < 15 x $\frac{2}{5}$(3 - 7x)

5(5 - 3x) < 3 x 2(3 - 7x)

25 - 15x < 6(3 - 7x)

25 - 15x < 18 - 42x

- 15x + 42x < 18 - 25

27x < -7

Divide both sides by 27

x < $\frac{-7}{27}$

The amount of money(f) which Dede has was not enough to buy the car
Which implies that Dede's amount is less than the cost of the car.

f(Dede's amount) < (less than) the cost of car (₦3,000,000.00)
but Kofi had enough money(k) to buy the car

This implies Kofi has either the exact amount or more than the cost of the car

⇒k ≥ ₦3,000,000.00 and the cost of the car is less than or equal (≤) to Kofi's amount
f < ₦3,000,000.00 ≤ k
Read as Dede's amount(f) is less than ₦3,000,000.00 and ₦3,000,000.00 is less than or equal to Kofi's amount(k)