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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

⇒ x – y = 5       ……….(i)

and  

x2 – y2 = 325        ………….(ii)

⇒  x = y + 5     ………..(iii)

Substitute for x in equation (ii)

(y + 5)2 – y2 = 325 

y2 + 10y + 25 – y2 = 325

10y + 25 = 325

10y = 300

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

⇒ y = 30

Substitute for y in equation (iii)

x = y + 5     ………..(iii)

x = 30 + 5

⇒  x = 35

Check:

352 – 302 = 325 orx – y = 5
1,225 – 900 = 325or35 – 30 = 5
325 = 325or 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. x2 + y2 = 369  ………..   (i)

Also,

x – y = 3    ……………(ii)

i.e. x = y + 3       ……….(iii)

Substitute for x in equation (i)

⇒ (y + 3)2 + y2 = 369 

y2 + 6y + 9 + y2 =369

2y2 + 6y + 9 – 369 = 0

⇒  2y2 + 6y – 360 = 0

Divide both sides by 2

⇒   y2 + 3y -180 = 0

Factorise

⇒   (y + 15)(y – 12) = 0

⇒     y + 15 = 0      or     y – 12 = 0

y = -15       or    y = 12

Substitute for y = 12 in equation (iii)

⇒ x = 12 + 3

⇒   x = 15

Substitute for y = -15 in equation (3)

⇒    x = -15 + 3

x = -12

Check:     

x2 + y2 = 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

⇒     x + y = 35        ………..(i)

Also,

xy = 250      ………(ii)

Re-write equation (i) 

⇒      x = 35 – y    ………(iii)

substitute x in equ iii into equ ii

⇒      (35 – y)y = 250

⇒      35y – y2 = 250

⇒       y2 – 35y + 250 = 0

Factorise

⇒        (y – 25)(y – 10) = 0

⇒       y – 25 = 0      or      y – 10 = 0

⇒       y = 25        or     y = 10

Substitute for y in equation (3)

⇒    x = 35 – 25       or x = 35 – 10

x = 10          or     x = 25

Check:

x + y = 35          or       xy = 250

10 + 25 = 35       or        10 × 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:

Screenshot 2022 02 05 at 20.58.31

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

Area = l  x b = x(x + 5) = x2 + 5x

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

∴ A = 3P

x2 + 5x  = 3[2(2x + 5)]

⇒ x2 + 5x = 3(4x + 10)

⇒   x2 + 5x = 12x + 30

⇒ x2 – 7x – 30 = 0

Factorise:   

x2 – 10x + 3x – 30 = 0  

x(x – 10) + 3(x – 10) = 0  

(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.

⇒  length = 10 + 5 = 15cm

Check:   A = 3P

         l × b = 3[2(l + b)]

        15 × 10 = 3[2(10 + 15)]

        150 = 6 × 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 = n2 + (n + 2)2

= n2 + n2 + 4n + 4

= 2n2 + 4n + 4

⇒ 2n2 + 4n + 4 = 164

Divide both sides by 2

⇒      n2 + 2n + 2 = 82

⇒      n2 + 2n – 80 = 0

Factorise:

n2 + 10n – 8n – 80 = 0

(n – 8)(n + 10) = 0

n – 8 = 0            or       n + 10 = 0

⇒    n = 8           or      -10

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

Check:  

n2 + (n + 2)2 = 164

82 + 102 = 164

64 + 100 = 164

Example 6

Consider the rectangle ABCD below where EF = 10cm, EH = 8cm, if the area ABCD is 288cm2, find:

(i) x

(ii) The length and breadth of ABCD

SS2 Second term

Solution:

Area = L x B

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

288 = 80 + 20x + 16x + 4x2

288 = 80 + 36x + 4x2

Collect the like terms

⇒     4x2 + 36x – 208 = 0

Divide both sides by 4

⇒     x2 + 9x – 52 = 0    

Factorise:

(x + 13)(x – 4) = 0

⇒  x + 13 = 0      or     x – 4 = 0

x = -13     or     x = 4

Since measurement is positive, therefore x = 4cm.

Length = 10 + 2(4) and   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 son’s age be x

∴  the mother’s age be 5x

Three years ago, the son’s age was (x – 3)

Three years ago, the mother’s age was (5x – 3)

The product of their ages three years ago will be 

(x – 3)(5x – 3) = 185

5x2 – 3x – 15x + 9 = 185

5x2 – 18x – 176 = 0

Factorise

(5x + 22)(x – 8x) = 0

⇒    5x + 22 = 0          or     x – 8 = 0

⇒     5x = -22             or      x = 8

⇒     x = \( – \frac{22}{5} \)              or          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 × 8 – 3) = 185

(5)(40 – 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:

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(i) At the set-out Journey:

D = Speed x Time

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

⇒   S = vt       …….. (i) 

⇒     120 = vt,        v =  \( \frac{120}{t} \)     …….(ii)

On the return Journey

Speed = (v + 5) km/h

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

⇒   \( \scriptsize s = (v + 5) \; \times \; \left ( \scriptsize t – \normalsize \frac{1}{3} \right) \scriptsize h \)

⇒     120 = \( \scriptsize vt \; – \; \normalsize \frac{v}{3}\scriptsize \; + \; 5t \; – \; \normalsize \frac{5}{3}  \)     …………(iii)

Substitute for vt in equation (iii)

⇒     120 = \( \scriptsize 120 \; – \; \normalsize \frac{v}{3}\scriptsize \; + \; 5t \; – \; \normalsize \frac{5}{3}  \)

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

Multiply both sides by 3

⇒ 0 = -v + 15t – 5

⇒    15t – v = 5    ………(iv)

Substitute for v in equation (iv)

⇒     15t – \(\frac{120}{t}\) =5

Multiply both sides by t

⇒   15t2 – 5t – 120 = 0

Divide both sides by 5

⇒    3t2 – t – 24 = 0

Factorise:

⇒     (3t + 8)(t – 3) = 0

⇒ 3t + 8 = 0         or     t – 3 = 0

⇒        3t = -8        or     t = 3

⇒       3t = -8       or    t = 3

⇒       t = \(– \frac{8}{3} \)          or      3

Since time is not given as negative value, then

t = 3 hours

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

⇒   Speed = \( \frac{120}{3} \scriptsize = 40 \; hours \)

(ii)   Total time taken for to and fro Journey is given as 

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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