Algebra: Basic Operation, Law of Indices

Transcription

1 Algera: Basi Operation, Law of Indies MD MARUFUR RAHMAN Ms Sustainale Energy Systems Beng (Hons) Mehanial Engineering Bs (Hons) Computer Siene & Engineering

2 05-06 Inreasingly, diffiulty in understanding algera is proving a prolem for many students as they ommene studying GCSE, A level and engineering ourses. Inevitaly there are a lot of formulae and alulations involved with engineering studies that require a sound lenh of algera. This doument offers a quik revision of the main areas of algera essential for further study, i.e. asi algera, simple equations, transposition of formulae, simultaneous equations and quadrati equations. Blok : Basi Operations Algera is that part of mathematis in whih the relations and properties of numers are investigated y means of general symols. For example, the area of a retangle is found y multiplying the length y the readth; this is expressed algeraially as A L, where A represents the area, L the length and the readth. Let a,, and d represent any four numers. (i) (ii) (iii) (iv) (v) a + ( + ) (a + ) + a() (a) a + + a a a a( + ) a + a (vi) (vii) (a + )( + d) a + ad + + d Let a 6,, and d 5 in eah of the aove, and hek that the left hand side of eah equation equals the right hand side. Prolem. Evaluate: a - + a when a, and 5 Replaing a, and with their numerial values gives: a - + a Prolem. Find the value of p q r, given that p, q and r Replaing p, q and r with their numerial values gives: P a g e

3 05-06 p q r ( ) 7 Prolem. Find the sum of a,,, -a, -5 and 6 Eah symol must e dealt with individually. For the a terms: +a - a a For the terms: For the terms: Thus a (-a) + (-5) + 6 a a a Prolem. Find the sum of: 5a -, a +, - 5d and - a + d - The algerai expressions may e taulated as shown elow, forming olumns for the a's, 's, 's and d's. Thus: Adding gives: +5a - +a d - a d 6a d Prolem 5. Sutrat x + y - z from x - y + 5z x - y + 5z x + y - z Sutrating gives: -x - 5y + 9z (Note that +5z - - z +5z + z 9z) An alternative method of sutrating algerai expressions is to hange the signs of the ottom line and add. Hene: Adding gives: x - y + 5z -x - y + z -x - 5y + 9z P a g e

4 05-06 Prolem 6. Multiply a + y a + Eah term in the first expression is multiplied y a, then eah term in the first expression is multiplied y, and the two results are added. The usual layout is shown elow. a + a + Multiplying y a a + a Multiplying y + a + Adding gives: a + 5a + Prolem 7. Multiply x - y + xy y x - 5y x - y + xy x - 5y Multiplying y x 6x - xy + 8x y Multiplying y -5y - 0xy - 5xy + 0y Adding gives: 6x - xy + 8x y - 5xy + 0y Prolem 8. Simplify: p 8pq p 8pq means p. This an e redued y anelling as in arithmeti. 8 p q Thus: p p 8pq 8 p q q Exerises: Find the sum of a, -a, -6a, 5a and a [ a ] Evaluate pq r when p, q - and r - [ -8 ] Add together d + e, -e + f, d - f, d - e + f - e [ 9d - e ] Sutrat a - + from - a - [ -5 a + - ] Multiply x + y y x - y [ ] Multiply a y a + [ ] Simplify (i) a 9a (ii) a a [(i) (ii) a ] P a g e

5 05-06 Blok : Laws of indies There are numer of rules that enale us to manipulate expressions involving indies. These rules are known as laws of indies. In eah ase note that the ase must e the same throughout. The laws of indies are: (i) a m a n (ii) (iii) (a m ) n a mn (iv) (v) (vi) (vii) Prolem 9. Simplify: a a 5 Grouping like terms gives: a a 5 Using the first law of indies gives: a i.e. a 5 6 a 5 6 Prolem 0. Simplify: 6 Using the first law of indies, Prolem. Simplify: Using the seond law of indies, 5 or 5 a and evaluate when a, a a a, a and and - 6 Thus a 6 ; When a, and -, a 6 () (-)6 (9) (6) Prolem. Simplify: p q r 6 p q r Using the seond law of indies gives: and evaluate when p 6, q 9 and r, taking positive roots only. 6 p q r p q r 5 P a g e

6 05-06 When p 6, q 9 and r, p q r ( ) ( ) ( ) ()( )() 08 Prolem. Simplify: + xy xy Algerai expressions of the form a + an e split into a + + xy Thus + x - y - + x - y - xy + y xy (sine x 0, from the sixth law of indies) Prolem. Simplify: The highest ommon fator (H.C.F.) of eah of the three terms omprising the numerator and denominator is xy. Dividing eah term y xy gives: x y Prolem 5. Simplify: ( p ) ( ) q Using the third law of indies gives: p q 8 p q or 8 p q ( m n ) Prolem 6. Simplify: m n The rakets indiate that eah letter in the raket must e raised to the power outside. Using the third law of indies gives: ( m n ) m n Using the seond law of indies gives: m n m n m n m n 6 m n 6 Prolem 7. Simplify: ( 5 )( ) m n m- n 6- m n 5 and evaluate when a, 6 and. 6 P a g e

7 05-06 Exerises: Using the fourth law of indies, the expression an e written as: Using the first law of indies gives: It is usual to express the answer in the same form as the question. Hene, When a, 6 and, a ( )( ) ( ) ( ) Simplify (x y z)(x yz ) and evaluate when x, y and z [, 6 ] Simplify and evaluate when a, and [, ± 9 ] Simplify: and evaluate when a, and [, 9 ] In Prolems to 0, simplify the given expressions: [ ] 7 P a g e