Questions
 Five years ago, a mother’s age was four times that of her daughter. In four years to come, she will be 2 ½ times the age of her daughter. Calculate the sum of their present ages
 Simplify;
 6a – 2b + 7a – 4b + 2

 Simplify
 Given that x + y = 8 and x^{2} + y^{2} = 34
Find; the value of x^{2} + 2xy + y^{2}
 Find the value of ; 2xy
 x^{2} – 2xy + y^{2}
 Value of x and y
 Simplify the expression.
 Simplify the expression
^{2}/_{3}(3x 2) – ^{3}/_{4}(2x 2)  Simplify by factorizing completely:
 Simplify as far as possible.
 Simplify:

 hence solve:

 Factorize completely the expression
75x^{2} – 27y^{2}  Simplify
 Simplify the expression:
 Given that (x3) (Ax^{2}+bx+c) = x^{3}7x6, find the value of A, B and C

 solve for y in 8x(2^{2})^{y}=6x2^{y}1
 Simplify completely
 Simplify the expression.:
 Simplify
 The sum of two numbers is 15. The difference between five times the first number and three times the second number is 19. Find the two numbers.
 Simplify the following expressions by reducing it to a single fraction
 Simplify the expression:
Answers
 Let the daughter’s age 5yrs ago be x
Mother 4x
come;
Daughter = x + 9
Mother = 4x+ 9
4x + 9 = ^{5}/_{2}(x +9)
4x + 9 = 2.5x + 22.5
1.5x = 13.5
x = 9
Mother = 41yrs
14 + 41 =55 
 6a + 7a – 2b – 4b + 2
= 13a – 6b + 2 
 6a + 7a – 2b – 4b + 2


 From x + y and x^{2} + y^{2} = 34
X = 8 – y
Substituting for x in x^{2} – y^{2} = 34
(8 – y) (8 – y) + y^{2} = 34
64 – 8y – 8y + y^{2} + y^{2} = 34
64 – 16y + 2y^{2} = 34
2y^{2} – 16y + 64 – 34 = 0
2y^{2} – 16y + 30 = 0
y^{2} = 8y + 15 = 0
y (y – 3) – 5 (y3) = 0 (y5) (y – 3)
y is either 5 or 3
but x – y = 8
x is either 5 0r 3
x^{2} + 2xy + y^{2} = 32 + 2 x 3 x 5 + 25
= 9 + 30 + 25 = 64  2xy = 2 x 3 x 5 = 30
 x^{2} – 2xy + y^{2} = 9 – 2 x 3 x 5 + 25 = 4
 x  y = 8 and x^{2} + y^{2} = 34
x = 8 – y
(8 – y)^{2} + y^{2} = 34
y^{2} – 8y + 15 = 0
y^{2} – 3y – 5y + 15 = 0
y(y 3) – 5(y – 3)
(y3) = 0 y = 3
(y5) = 0 y = 5
x + 3 = 8, x = 5 or x + 5 = 8
x = 3
x is either 3 or 5
y is either 3 or 5
 From x + y and x^{2} + y^{2} = 34





 3( 25x^{2} – 9y^{2})
3(5x + 3y)(5x – 3y)  Factorizing the numerator
= p(p^{2} – q^{2}) + q(p^{2}q)
= (p+q) (p^{2}q^{2})
= (p+q) (p+q) n(pq)
Factorising the denominator
(p+q) (p+q)
Numerator = p  q
Denominator 
( 3x + 2y ) ( 3x  2y )
( 3x + 2y ) ( 3x  2y )
3x + 2y
4x + 3y  (x – 3) (AX^{2} +BX + C) = x^{3} – 7x – 6
AX^{3} + BX^{2} +CX – 3AX^{2} – 3BX – 3c = x^{3} – 7x – 6
A = 1
B – 3A = 0
B – 3 x 1 = 0
B = 3
3c = 6
c = 2 
 8(2^{2})^{y} = 6 x 2^{y} – 1
let t = 2^{y}
8t^{2} = 6t – 1
8t^{2} – 4t – 2t + 1 = 0
(4t – 1) (2t – 1) = 0
t = ¼ or ½
t = 2^{y} = ¼ = 2^{2}
y = 2
or t = 2y = ½ = 2^{1}
y = 1
y = 2 or 1 
 8(2^{2})^{y} = 6 x 2^{y} – 1


 Let the numbers be a and b
a + b = 15 x3
5a – 3b = 19 x 1
3a + 3b = 45
5a – 3b = 19
8a = 64
a = 8
b = 7 
