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Unformatted text preview: CHAPTER 1 Review of Algebra Much of the material in this chapter is revision from GCSE maths (al though some of the exercises are harder). Some of it – particularly the work on logarithms – may be new if you have not done Alevel maths. If you have done Alevel, and are confident, you can skip most of the ex ercises and just do the worksheet, using the chapter for reference where necessary. — ./ — 1. Algebraic Expressions 1.1. Evaluating Algebraic Expressions Examples 1.1 : (i) A firm that manufactures widgets has m machines and employs n workers. The number of widgets it produces each day is given by the expression m 2 ( n 3). How many widgets does it produce when m = 5 and n = 6? Number of widgets = 5 2 × (6 3) = 25 × 3 = 75 (ii) In another firm, the cost of producing x widgets is given by 3 x 2 + 5 x + 4. What is the cost of producing (a) 10 widgets (b) 1 widget? When x = 10, cost = (3 × 10 2 ) + (5 × 10) + 4 = 300 + 50 + 4 = 354 When x = 1, cost = 3 × 1 2 + 5 × 1 + 4 = 3 + 5 + 4 = 12 It might be clearer to use brackets here, but they are not essential: the rule is that × and ÷ are evaluated before + and . (iii) Evaluate the expression 8 y 4 12 6 y when y = 2. (Remember that y 4 means y × y × y × y .) 8 y 4 12 6 y = 8 × ( 2) 4 12 6 ( 2) = 8 × 16 12 8 = 128 1 . 5 = 126 . 5 (If you are uncertain about using negative numbers, work through Jacques pp.7–9.) Exercises 1.1 : Evaluate the following expressions when x = 1, y = 3, z = 2 and t = 0: (a) 3 y 2 z (b) xt + z 3 (c) ( x + 3 z ) y (d) y z + 2 x (e) ( x + y ) 3 (f) 5 x +3 2 t z 1 2 1. REVIEW OF ALGEBRA 1.2. Manipulating and Simplifying Algebraic Expressions Examples 1.2 : (i) Simplify 1 + 3 x 4 y + 3 xy + 5 y 2 + y y 2 + 4 xy 8. This is done by collecting like terms , and adding them together: 1 + 3 x 4 y + 3 xy + 5 y 2 + y y 2 + 4 xy 8 = 5 y 2 y 2 + 3 xy + 4 xy + 3 x 4 y + y + 1 8 = 4 y 2 + 7 xy + 3 x 3 y 7 The order of the terms in the answer doesn’t matter, but we often put a positive term first, and/or write “higherorder” terms such as y 2 before “lowerorder” ones such as y or a number. (ii) Simplify 5( x 3) 2 x ( x + y 1). Here we need to multiply out the brackets first, and then collect terms: 5( x 3) 2 x ( x + y 1 = 5 x 15 2 x 2 2 xy + 2 x = 7 x 2 x 2 2 xy + 5 (iii) Multiply x 3 by x 2 . x 3 × x 2 = x × x × x × x × x = x 5 (iv) Divide x 3 by x 2 . We can write this as a fraction, and cancel: x 3 ÷ x 2 = x × x × x x × x = x 1 = x (v) Multiply 5 x 2 y 4 by 4 yx 6 . 5 x 2 y 4 × 4 yx 6 = 5 × x 2 × y 4 × 4 × y × x 6 = 20 × x 8 × y 5 = 20 x 8 y 5 Note that you can always change the order of multiplication. (vi) Divide 6 x 2 y 3 by 2 yx 5 ....
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This note was uploaded on 04/12/2010 for the course ECON DEAM taught by Professor Vines during the Spring '10 term at Oxford University.
 Spring '10
 Vines
 Economics

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