Lesson 4, Topic 4
In Progress

# Problems Leading to Linear Equations

Lesson Progress
0% Complete

Example 4

(i) Tunde is four times as old as his son. In four years’ time, he will be three times as old. How old are they now?

Solution:

⇒ Let the son’s age = x

⇒ Thus, Tunde’s age = 4x

⇒ In four year’s time Son’s age = (x + 4)

⇒ In four year’s time the son will be three times as old = 3(x + 4)

⇒ In four year’s time Tunde’s age = 4x + 4

⇒ Therefore, (4x + 4) = 3(x + 4)

Expand the brackets

⇒     4x + 4 = 3x + 12

Collect like terms

⇒    4x – 3x = 12 – 4

⇒     x = 8

∴         Son’s age = 8years  and Tunde’s age = 4 × 8 = 32 years

(ii). The sum of two numbers is 40. When $$\scriptsize 3\frac{1}{4}$$ times the larger number is subtracted from $$\scriptsize 5 \frac{1}{2}$$ times the smaller, the difference is -25. Find the two numbers.

Solution:

Let the two numbers be x and y

x + y = 40……………..(i)

Let “x” be the larger number and “y” be the smaller number

⇒     $$\frac{11y}{2} \: – \: \frac{13x}{4} = \: \scriptsize -25$$

Clear the fractions by multiplying both sides by 4

⇒ $$\scriptsize 4 \: \times \: \normalsize \frac{11y}{2} \: – \: \scriptsize 4 \: \times \: \normalsize \frac{13x}{4} \scriptsize = 4 \: \times \: – 25$$

22y – 13x = -100……………..(ii)

Recall  x + y = 40

∴     x = 40 – y    ……(iii)

Substitute for x in equation (ii)

⇒     22y – 13(40 – y) = -100

⇒     22y – 520 + 13y = -100

Collect like terms

⇒ 22y + 13y = 520 – 100

⇒     35y = 420

Divide both sides by 35

⇒ $$\frac {35y}{35} = \frac {420}{35}$$

∴ y = 12

Substitute for y in equation (iii)

⇒ x = 40 – 12

⇒ x = 28

Therefore, the two numbers are 12 and 28.

(iii) The distance between two villages P and Q is xkm. A train running between P and Q arrives 3 minutes behind the scheduled time when it travels at an average speed of 45km/h, but arrives 9 minutes ahead of the scheduled time  when it travels at an average speed of 60km/h, calculate:

a. The scheduled time for the journey

b. The distance x (WAEC)

Solution:

a. Recall D = Speed x Time

1st movement:

Time = (t + 3)

Speed = 45km/h,

Let xkm distance = 45t + 3      …………(i)

2nd movement:

Time = (t – 9)

Speed = 60km/h

Let xkm distance = 60t – 9    ………..(ii)

Equating equations (i) and (ii), we have

45(t + 3) = 60(t – 9)

Divide both sides by 15

⇒ 3(t + 3) = 4(t- 9)

Expand the brackets

⇒      3t + 9 = 4t – 36

Collect like terms

⇒    4t – 3t = 36 + 9

t = 45 minutes

⇒    $$\frac {45}{60} \scriptsize hr = \normalsize \frac {3}{4}\scriptsize hr$$

b. Recall: Distance = Speed x Time

Using the 1st movement

Speed = 45km/h

Time = t + 3 = 45 + 3

= 48 minutes

= $$\frac {48}{60} \scriptsize hr = \normalsize \frac {4}{5} \scriptsize hr$$

= $$\scriptsize 45 \: \times \: \normalsize \frac{4}{5}$$

= 9 x 4 = 36km

Distance = 36km

Hint: To solve a word problem leading to an equation follow these steps:

1. Decide the unknown quantity and represent it by a letter such as x
2. State clearly the units used when necessary and convert to the same unit.
3. Form an equation to represent the facts provided by the problems
4. Solve the equation using a suitable method and use the solution to answer the questions in words.
5. Check the solution (optional but useful)

error: