Tutorial on basic mathematics
Q(1)(a)Evaluate (i)
5Q /( P  2)
1 3 5
2/7
22 / 7
13 11
, (iii)
,(iv)
,
  , (ii)
 , (iv )
3 5 7
33 / 49
66 39
9 / 11
1 /( P  2)
(v )
3/ x
.
x3
1 2
  5 , (iii) x 2  4 x  6  0 ,
x x
3
2  3Q
(iv) x 3  3x  0 , (v)

5
2Q  5
Q
(b) Solve (i) 2x  3  5x  8 , (ii)
Solution
1 3 5 5  7  3  3  7  5  3  5 35  63  75 173


(a)(i)   
3 5 7
3 5 7
105
105
1
(ii)
2 / 7 2 11 22
13 11 1 1 1
  

  
 0.349206 (iii)
 0.055556
66 39 6 3 18
9 / 11 7 9 63
(iv )
(b) (i)
(ii)
68
 1.647619 to 6 decimal places.
105
5Q /( P  2)
3
5Q
P2
3/ x
3
1


 
 5Q and
.

P2
1
x  3 x x  3 x( x  3)
1 /( P  2)
2 x  4  5x  8  2x  5x  8  4  3x  12  x  4 .
1 2
1 2
3
3
 5
 5   5  3  5x  x   0.6
x x
x
x
5
Solution of Quadratics.
Remember important formula for solving a quadratic equation, that is that gives roots of a
general quadratic equation ax 2  bx  c  0 .
 b  b 2  4ac
.
2a
Solve (i) x 2  x  6  0 ,(ii) x 2  8 x  16  0 and (iii) x 2  8 x  22  0 .
ax 2  bx  c  0  x 
(i) By formula above x 
1  12  4(1)( 6) 1  25 1  5


 3 or  2 .
2
2(1)
2
 8  8 2  4(1)(16)  8  0
 4 .
(ii) By formula above x 

2(1)
2
Thus we get two roots which are the same, namely x  4 .
 8  8 2  4(1)( 22)  8   20  8  20  1


2
2(1)
2
Thus the roots are complex.
(iii) By formula above x 
Calculating Percentages
Q(1) (i)A compact disc player is normally priced at €256 and is reduced in a sale by
20%.Calculate the sale price.
(ii) A bank deposit earns 7.5% interest in one year. Calculate the interest earned on
a deposit of €15000.
20
Solution (i) 20% of €256=256 
 €51.2.
100
Hence sale price is =€256-€51.2=€204.8.
80
Alternative sale price is 80% of original price, that is €256 
=€204.8.
100
(ii) Interest of 7.5% on €15000=€15000 
7 .5
= €1125.0.
100
Q(2) (i)A deposit of €750 increases by 9%.Calculate the resulting deposit.
(ii)A television set is advertised at €315.The retailer offers a 10% discount. How
much do you pay for the television.
Solution (i) Increase =€750 
9
 €67.5
100
Resulting deposit = €750+€67.5=€817.5.
(ii)Discount = €315 
10
 €31.5.
100
Sale price=€315-€31.5=€283.5.
Laws of Indices
Example 1
Simplify each of the following expressions, giving your answer with positive
indices only.
 2 L1 
2 4 2 3
(4)35 3 0.8
3 2 5 5


(a)
,
(b)
(c)
(d)
22
3 4 3  2. 2
2 3 5 2 3 3
 4L 
Solution
2 4 2 3 2 43 21
1
1
(a)
= 2 = 2 = 21 = .
2
2
2
2
2
2
2
(4)35 0.8
(4)35.8
(4)3530.8
(b) 4 2.2 =
= (b)
= 4  35.81.8 = 4  3 4 =324.
4  2.2
1.8
33
3
3
3 2 5 5
3 2 33
3 23
35
243
243
(c) 3 2 3 = 3 2 5 = 3 25 = 3 7 =
=
.
25
253
255
2 5 8  78125 625000
2
2
 2 L1 
 L1 
1
 1   1 


 = 
(d) 
= 
 = 2  = 2 4

 2 LL   2 L  2 L
 2L 
 4L 
Example 2
Simplify each of the following expressions, giving your answer with positive
indices only.
1
1
3(2 n )  4(2 n  2 )
a3
b3 
(a) ( 2 ) 2  ( 2 ) 2 , and (b)
.
2 n  2 n 1
b
a
Solution
2
2
1
1
3
1
1
 a3  2  a2  2
a2
a
a3
b3 
(a) ( 2 ) 2  ( 2 ) 2   2    3  
 3
b
a
b
b  b 
b2
3
2
3
2
a b
 a 2  b 2  ab

ba
1
1
 


   
3(2 n )  4(2 n  2 ) 3 2 n  2 2 2 n2
3 2n  2n
2 n 1
2 n 3  1


 22  4 .

=
n 1
n
n 1
n
n 1
n 1
n
n 1
2 2  1 2
2 2
2 2
2 2
http://www.mathcentre.ac.uk
and then click on Algebra
and then click on Powers or indices.
(b)
Changing the subject of an equation or Transposition of formulae
Example 1 Transpose d  2 h(2r  h) to make r the subject.
Solution
2
2
1
d
d 


 h(2r  h)     h2r  h  2   h(2r  h)
2


2
d2
2
h

2
2
d2
d2
4  r  4h  d
 2rh  h 2 
 2rh  h 2 
 r
4
4
8h
2h
Example 2
(a)
(i) If
1
1
1

 express R in terms of R1 and R2 .
R1 R2 R
ax 2
(ii) If y 
express x in terms of y , a and r .
1 r
3
(b) In an electrical alternating current circuit the impedance Z is given by
2

1  

Z   R 2   L 
.


C  



Transpose this formula making C the subject and show that
C

1
 L  Z  R
2
2

.
Hence evaluate C if Z  150 , R  100 ,   314 rad / sec and
L  0.5 henrys .
Example 2
Solution
(a)
(i)
R  R1 1
RR
1
1
1

  2
 R 1 2
R1 R2 R
R1 R2
R
R2  R1
1 r
ax 2
ax 2
y 3 1  r 
(ii) y 
 y3 
 ax 2  y 3 1  r   x 2 
 x  y2
1 r
1 r
a
a
3
3
2

1 

2

(b) Z  R   L 


C 



2
2
2

  Z 2  R 2   L  1     LC  1   Z 2  R 2 .
 C 

C 




 2 LC  1
 Z 2  R 2   2 LC  1  C Z 2  R 2   2 LC  C Z 2  R 2  1
C


C  2L   Z 2  R2  1  C 
C

 L  Z 2  R 2
1
314 314  0.5  150  100
2

1
2



1
 7.046  10 5 farads.
314157  111.8
For further material and examples click on
http://www.mathcentre.ac.uk
and then click on
Transposition of formulae
Solution of Simultaneous Equations and their Graphs
Q(1)
(a) Solve the following system of equations graphically and algebraically
x  y 1
6 x  5 y  15
(b) Using Gaussian Elimination solve the following system of equations
x1  2 x 2  3x3  14
2 x1  2 x 2  x3  1
7 x1  5 x 2  6 x3  1
Verify your answer.
Solution Q(1)
(a)
x  y 1
6 x  5 y  15
6x  6 y  6
6 x  5 y  15
__________
y  9
x  y  1  x  1  (9)  10 .
Verification

10  9  1
6(10)  5(9)  60  45  15
.
(b)
x1  2 x 2  3x3  14
x1  2 x 2  3x3  14

2 x1  2 x 2  x3  1
2 x1  2 x 2  x3
7 x1  5 x 2  6 x3  1
1
__________________
3x1
 4 x3  15 .
2 x1  2 x2  x3  1
5
7 x1  5 x 2  6 x3  1  2
Solve
3x1  4 x3  15  8
24 x1  7 x3  3  1


10 x1  10 x 2  5 x3  5
14 x1  10 x 2  12 x3  2
____________________
24 x1
 7 x3  3
24 x1  32 x3  120
24 x1  7 x3  3
_______________
39 x3  117  x3  3
3x1  4(3)  15  3x1  15  12  3  x1  1 .
x1  2 x2  3x3  1  2 x2  3(3)  14  2 x2  14  10  4  x2  2 .
x1  2 x 2  3x3  14  1  2(2)  3(3)  14
Verification 2 x1  2 x 2  x3  1  2(1)  2(2)  3  1
7 x1  5 x 2  6 x3  1  7(1)  5(2)  6(3)  17  18  1
For further material and examples click on
http://www.mathcentre.ac.uk
and then click on
Simultaneous Equations.