Deborah Hughes-Hallett, William G. McCallum, Eric Connally's Algebra: Form and Function PDF

By Deborah Hughes-Hallett, William G. McCallum, Eric Connally

ISBN-10: 0470226668

ISBN-13: 9780470226667

Shape is expounded to operate. An aircraft wing has the shape it does due to its lifting functionality. The pillars of the Parthenon and the girders of a skyscraper are formed to the aim of assisting their giant constructions. equally, the shape of an algebraic expression or equation displays its functionality. Algebra: shape and serve as initial variation introduces each one function--linear, energy, quadratic, exponential, polynomial--and offers a learn of the fundamental kind of expressions for that functionality. Readers are inspired to check the fundamental types, see how they're developed, and view the function of every part. through the textual content, there are instruments sections positioned on the ends of chapters to assist readers gather the talents they should practice simple algebraic manipulations.

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Additional resources for Algebra: Form and Function

Example text

For example, we can rewrite 12/6 as 12 · 61 . In general we have a 1 = a · . rewrite division as multiplication b b If b = 0, the number 1/b is called the reciprocal of b. 1 1 a = 2· a. 1. At the beginning of Example 1 we regrouped the multiplication to see that 2 Using Evaluation to See When Expressions Are Not Equivalent Many common errors in algebra result from thinking that expressions are equivalent when in fact they are not. It is usually easy to find out when two expressions are not equivalent by evaluating them.

39. A number is doubled and then added to itself. The result is 99. 26 Chapter 1 THE KEY CONCEPTS OF ALGEBRA For Exercises 40–47, which of the given values of the variable are solutions? 40. x2 + 2 = 3x for x = 0, 1, 2 41. 2x2 + 3x3 = 5x5 for x = −1, 0, 1 42. t + 1/(t2 + 1) = (t + 1)/(t2 + 1) for t = −1, 0, 1 43. 5 44. 2(r − 6) = 5r + 12, for r = 8, −8 45. n2 − 3n = 2n + 24, for n = 8, −3 In Exercises 58–62, explain what operation can be used to transform the first equation into the second equation.

Give as many solutions of the equations in Problems 47–49 as you can find from the table. 57. 2 x −1 0 1 2 Expression 1 1 2 −1 0 Expression 2 1 0 −1 0 Expression 3 0 2 −1 −1 13 59. 1 = −3 4z 2 58. 3x2 =1 3x2 − 1 60. a+1 1 = 2a 2 61. Given that x = 4 is a solution to 47. Expression 1 = Expression 2 48. Expression 1 = Expression 3 49. Expression 2 = Expression 3 2jx + z = 3, evaluate the expression 16j + 2z. In Problems 50–56, does the equation have a solution? Explain how you know without solving it.

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Algebra: Form and Function by Deborah Hughes-Hallett, William G. McCallum, Eric Connally

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