Example 1

The difference of the squares of two positive integers is 325 and their difference is 5. What are the numbers?

**Solution:**

**Let the two numbers be x and y**

â‡’ x – y = 5 Â Â Â â€¦â€¦â€¦.(i)

and

x^{2} – y^{2} = 325 â€¦â€¦â€¦â€¦.(ii)

â‡’ Â x = y + 5 Â Â â€¦â€¦â€¦..(iii)

**Substitute for x in equation (ii)**

(y + 5)^{2} – y^{2} = 325

y^{2} + 10y + 25 – y^{2} = 325

10y + 25 = 325

10y = 300

y = \( \frac{300}{10} \)

â‡’ y = 30

**Substitute for y in equation (iii)**

x = y + 5 Â Â â€¦â€¦â€¦..(iii)

x = 30 + 5

â‡’Â x = 35

**Check:**

35^{2} – 30^{2} = 325Â | or | x – y = 5 |

1,225 – 900 = 325 | or | 35 – 30 = 5 |

325 = 325 | or | 5 – 5 |

Example 2:

The sum of the squares of two integers is 369 and their difference is 3. What are the numbers?

**Solution:**

**Let the two numbers be x and y**

i.e. x^{2} + y^{2} = 369 â€¦â€¦â€¦.. (i)

**Also,**

x – y = 3 â€¦â€¦â€¦â€¦â€¦(ii)

i.e. x = y + 3 â€¦â€¦â€¦.(iii)

**Substitute for x in equation (i)**

â‡’ (y + 3)^{2} + y^{2} = 369Â

y^{2} + 6y + 9 + y^{2} =369

2y^{2} + 6y + 9 – 369 = 0

â‡’ Â 2y^{2} + 6y – 360 = 0

**Divide both sides by 2**

â‡’ Â y^{2} + 3y -180 = 0

**Factorise**

â‡’ Â (y + 15)(y – 12) = 0

â‡’ Â Â y + 15 = 0Â Â Â or Â Â y – 12 = 0

y = -15 or y = 12

**Substitute for y = 12 in equation (iii)**

â‡’ x = 12 + 3

â‡’ Â x = 15

**Substitute for y = -15 in equation (3)**

â‡’Â Â x = -15 + 3

x = -12

**Check:**Â Â Â

x^{2} + y^{2} = 369

152 + 122 = 369

225 + 144 = 369

369 = 369

Or

(-12)^{2} + (-15)^{2} = 369

144 + 225 = 369

369 = 369

**Also,**

x – y = 3

15 – 12 = 3

3 = 3

x – y = 3

– 12 -(-15) = 3

-12 + 15 = 3

3 = 3

Example 3

Two numbers have a sum of 35 and a product of 250. What are the numbers?

**Solution:**

Let the numbers be x and y

â‡’ Â Â x + y = 35Â Â Â Â â€¦â€¦â€¦..(i)

**Also,**

xy = 250 â€¦â€¦â€¦(ii)

Re-write equation (i)

â‡’Â Â Â x = 35 – yÂ Â â€¦â€¦â€¦(iii)

**substitute x in equ iii into equ ii**

â‡’Â Â Â (35 â€“ y)y = 250

â‡’Â Â Â 35y – y^{2} = 250

â‡’ Â Â Â y^{2} – 35y + 250 = 0

**Factorise**

â‡’Â Â Â Â (y – 25)(y – 10) = 0

â‡’ Â Â Â y – 25 = 0Â Â Â orÂ Â Â y – 10 = 0

â‡’ Â Â Â y = 25Â Â Â Â or Â Â y = 10

**Substitute for y in equation (3)**

â‡’Â Â x = 35 – 25 Â Â Â or x = 35 – 10

x = 10 or x = 25

**Check:**

x + y = 35 or xy = 250

10 + 25 = 35 or 10 Ã— 25 = 250

35 = 35 or 250 = 250

Example 4

The area of a rectangle is three times its perimeter. If the length of the rectangle is 5cm longer than its width, find the length and width of the rectangle.

**Solution:**

Perimeter = 2(l + b) = 2(x + x + 5) = 2(2x + 5)

Area = lÂ x b = x(x + 5) = x^{2} + 5x

The area is three times its perimeter, for example, if the area is 6 the perimeter will be 2.

âˆ´ A = 3P

x^{2} + 5x = 3[2(2x + 5)]

â‡’ x^{2} + 5x = 3(4x + 10)

â‡’ Â x^{2} + 5x = 12x + 30

â‡’ x^{2} – 7x – 30 = 0

**Factorise:**Â Â

x^{2} – 10x + 3x – 30 = 0 Â

x(x – 10) + 3(x – 10) = 0 Â

(x – 10)(x + 3) = 0

x – 10 = 0 or x + 3 = 0

x = 10 or x = -3

**Measurement cannot be negative, therefore the width is 10cm.**

â‡’Â length = 10 + 5 = 15cm

**Check:** Â A = 3P

l Ã— b = 3[2(l + b)]

Â Â Â Â Â Â Â Â 15 Ã— 10 = 3[2(10 + 15)]

150 = 6 Ã— 25

150 = 150

Example 5

Two consecutive positive even numbers are n and n + 2. Write down an algebraic expression for the sum of their squares. Find the numbers if the sum of the squares is 164.

**Solution:** Sum of their squares = n^{2} + (n + 2)^{2}

= n^{2} + n^{2} + 4n + 4

= 2n^{2} + 4n + 4

â‡’ 2n^{2} + 4n + 4 = 164

**Divide both sides by 2**

â‡’Â Â Â n^{2} + 2n + 2 = 82

â‡’Â Â Â n^{2} + 2n – 80 = 0

**Factorise: **

n^{2} + 10n – 8n – 80 = 0

(n – 8)(n + 10) = 0

n – 8 = 0 or n + 10 = 0

â‡’Â Â n = 8 Â Â Â Â Â orÂ Â Â -10

Since we are dealing with positive even integers, the numbers are 8 and 10.

**Check: **Â

n^{2} + (n + 2)^{2} = 164

8^{2} + 10^{2} = 164

64 + 100 = 164

Example 6

Consider the rectangle ABCD below where EF = 10cm, EH = 8cm, if the area ABCD is 288cm^{2}, find:

**(i)** x

**(ii)** The length and breadth of ABCD

**Solution:**

Area = L x B

288 = (10 + 2x)(8 + 2x)

288 = 80 + 20x + 16x + 4x^{2}

288 = 80 + 36x + 4x^{2}

**Collect the like terms**

â‡’ Â Â 4x^{2} + 36x – 208 = 0

**Divide both sides by 4**

â‡’ Â Â x^{2} + 9x – 52 = 0Â Â Â Â

**Factorise:**

(x + 13)(x – 4) = 0

â‡’Â x + 13 = 0Â Â Â or Â Â x – 4 = 0

x = -13 or x = 4

Since measurement is positive, therefore x = 4cm.

Length = 10 + 2(4) and Â Breadth = 8 + 2(4)

= 10+8 = 8+8

= 18cm = 16cm

**Check:**

Area = 288

L x B = 288

18 x 16 = 288

288 = 288

Example 7

Mrs Fawehinmi is 5 times older than her son. 3 years ago the product of their age was 185. How old are they now?

**Solution:**

Let the sonâ€™s age be x

âˆ´ the mother’s age be 5x

Three years ago, the sonâ€™s age was (x – 3)

Three years ago, the motherâ€™s age was (5x – 3)

The product of their ages three years ago will be

(x – 3)(5x – 3) = 185

5x^{2} – 3x – 15x + 9 = 185

5x^{2} – 18x – 176 = 0

**Factorise**

(5x + 22)(x – 8x) = 0

â‡’Â Â 5x + 22 = 0Â Â Â Â Â or Â Â x – 8 = 0

â‡’ Â Â 5x = -22 Â Â Â Â Â Â orÂ Â Â x = 8

â‡’ Â Â x = \( – \frac{22}{5} \)Â Â Â Â Â Â Â orÂ Â Â Â Â x = 8

x = – 4.4 or x = 8.

Since the age of the son cannot be -4.4, therefore, the age of the son is 8 years, while the age of the mother is 5 x 8 = 40 years.

**Check:**

3 years ago product of their ages is:

(x – 3)(5x – 3) = 185

i.e. (8 – 3)(5 Ã— 8 – 3) = 185

(5)(40 â€“ 3) = 185

(5)(37) = 185

185 = 185

Example 8

A man travelled a distance of 120km by car from Badagry to Ondo. On his return journey, his average speed was increased by 5km/h and he found out that he took 20 minutes less

**(i) **What was his average speed for the first part of the journey?

**(ii)** How long did he take for the double journey?

**Solution:**

**(i)** At the set-out Journey:

D = Speed x Time

Let distance = S (km), Speed = V (km/h) and Time t (h)

â‡’ Â S = vt Â Â Â …….. (i)Â

â‡’ Â Â 120 = vt,Â Â Â Â v = Â \( \frac{120}{t} \) Â Â â€¦â€¦.(ii)

On the return Journey

Speed = (v + 5) km/h

Time = \( \left ( \scriptsize t – \normalsize \frac{20}{6} \right) \scriptsize h = \left ( \scriptsize t – \normalsize \frac{1}{3} \right) \scriptsize h\)

â‡’Â Â \( \scriptsize s = (v + 5) \; \times \; \left ( \scriptsize t – \normalsize \frac{1}{3} \right) \scriptsize h \)

â‡’ Â Â 120 = \( \scriptsize vt \; – \; \normalsize \frac{v}{3}\scriptsize \; + \; 5t \; – \; \normalsize \frac{5}{3} Â \) Â Â â€¦â€¦â€¦â€¦(iii)

Substitute for vt in equation (iii)

â‡’ Â Â 120 = \( \scriptsize 120 \; – \; \normalsize \frac{v}{3}\scriptsize \; + \; 5t \; – \; \normalsize \frac{5}{3} Â \)

0 = \(– \frac{1}{3}\scriptsize \; + \; 5t \; – \; \normalsize \frac{5}{3} \)

**Multiply both sides by 3**

â‡’ 0 = -v + 15t – 5

â‡’Â Â 15t – v = 5Â Â â€¦â€¦â€¦(iv)

**Substitute for v in equation (iv)**

â‡’ Â Â 15t – \(\frac{120}{t}\) =5

**Multiply both sides by t**

â‡’ Â 15t^{2} – 5t – 120 = 0

**Divide both sides by 5**

â‡’Â Â 3t^{2} – t – 24 = 0

**Factorise:**

â‡’ Â Â (3t + 8)(t – 3) = 0

â‡’ 3t + 8 = 0 Â Â Â Â or Â Â t – 3 = 0

â‡’Â Â Â Â 3t = -8Â Â Â Â or Â Â t = 3

â‡’ Â Â Â 3t = -8 Â Â Â orÂ Â t = 3

â‡’ Â Â Â t = \(– \frac{8}{3} \)Â Â Â Â Â orÂ Â Â 3

Since time is not given as negative value, then

t = 3 hours

âˆ´ Average Speed for the 1st Journey = \( \frac{Distance}{Time} \)

â‡’ Â Speed = \( \frac{120}{3} \scriptsize = 40 \; hours \)

**(ii) **Â Total time taken for to and fro Journey is given asÂ

T = \( \scriptsize t + (t – \frac{1}{3}) \)

=\( \scriptsize 3 + 3\; – \; \frac{1}{3} \)

= \( \scriptsize 3 + 2 \normalsize \frac{2}{3} \)

= \( \scriptsize 5 \normalsize \frac{2}{3} \scriptsize hours \)

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