5 - ELEMENTARY MATHEMATICS W W L CHEN and X T DUONG c W W L...

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ELEMENTARY MATHEMATICS W W L CHEN and X T DUONG c ° W W L Chen, X T Duong and Macquarie University, 1999. This work is available free, in the hope that it will be useful. Any part of this work may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or any information storage and retrieval system, with or without permission from the authors. Chapter 5 POLYNOMIAL EQUATIONS 5.1. Linear Equations Consider an equation of the type ax + b =0 , (1) where a, b R are constants and a 6 = 0. To solve such an equation, we Frst subtract b from both sides of the equation to obtain ax = b, (2) and then divide both sides of this latter equation by a to obtain x = b a . Occasionally a given linear equation may be a little more complicated than (1) or (2). However, with the help of some simple algebra, one can reduce the given equation to one of type (1) or type (2). Example 5.1.1. Suppose that 4 x x +2 = 18 5 . Multiplying both sides by 5( x + 2), the product of the two denominators, we obtain 20 x = 18( x +2)=18 x +36 . Subtracting 18 x from both sides, we obtain 2 x = 36, and so x =36 / 2 = 18. This chapter was written at Macquarie University in 1999.
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5–2 W W L Chen and X T Duong : Elementary Mathematics Example 5.1.2. Suppose that x 6 2 + 3 x 4 = x +1 . Multiplying both sides by 4, we obtain 2( x 6) + 3 x =4( x + 1). Now 2( x 6) + 3 x =5 x 12 and 4( x +1)=4 x + 4. It follows that 5 x 12 = 4 x + 4, so that x 16 = 0, giving x = 16. Example 5.1.3. Suppose that 6 x 1 4 x +3 = 3 x 7 2 x 5 . Multiplying both sides by (4 x + 3)(2 x 5), the product of the two denominators, we obtain (6 x 1)(2 x 5) = (3 x 7)(4 x +3) . Now (6 x 1)(2 x 5) = 12 x 2 32 x + 5 and (3 x 7)(4 x +3)=12 x 2 19 x 21. It follows that 12 x 2 32 x +5=12 x 2 19 x 21, so that 13 x + 26 = 0, whence x =2. Example 5.1.4. Suppose that 3 x 2 x = 3 x 3 2 x +5 . Multiplying both sides by (2 x + 3)(2 x + 5), the product of the two denominators, we obtain (3 x + 1)(2 x +5)=(3 x 3)(2 x . Now (3 x + 1)(2 x +5)=6 x 2 +17 x + 5 and (3 x 3)(2 x +3)=6 x 2 x 9. Check that the original equation is equivalent to 14 x + 14 = 0, so that x = 1. We next consider a pair of simultaneous linear equations in two variables, of the type a 1 x + b 1 y = c 1 , a 2 x + b 2 y = c 2 , (3) where a 1 ,a 2 ,b 1 2 ,c 1 2 R . Multiplying the Frst equation in (3) by b 2 and multiplying the second equation in (3) by b 1 , we obtain a 1 b 2 x + b 1 b 2 y = c 1 b 2 , a 2 b 1 x + b 1 b 2 y = c 2 b 1 . (4) Subtracting the second equation in (4) from the Frst equation, we obtain ( a 1 b 2 x + b 1 b 2 y ) ( a 2 b 1 x + b 1 b 2 y )= c 1 b 2 c 2 b 1 . Some simple algebra leads to ( a 1 b 2 a 2 b 1 ) x = c 1 b 2 c 2 b 1 . (5) On the other hand, multiplying the Frst equation in (3) by a 2 and multiplying the second equation in (3) by a 1 , we obtain a 1 a 2 x + b 1 a 2 y = c 1 a 2 , a 1 a 2 x + b 2 a 1 y = c 2 a 1 .
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5 - ELEMENTARY MATHEMATICS W W L CHEN and X T DUONG c W W L...

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