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If you'd like best grades and thorough figuring out of precalculus, this robust examine instrument is the simplest teach you could have! It takes you step by step throughout the topic and offers you greater than six hundred accompanying similar issues of absolutely labored strategies. you furthermore mght get lots of perform difficulties to do by yourself, operating at your individual velocity. (Answers supplied to teach you the way you are doing.) recognized for his or her readability, wealth of illustrations and examples, and shortage of dreary trivialities, Schaum's Outlines have offered greater than 30 million copies worldwideand this consultant will exhibit you why!

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**Sample text**

Symmetric with respect to the line y = x if (b,a) is on the graph whenever (a,@ is on the graph. TESTS FOR SYMMETRY (Fig. 8-3): 1. If substituting -x for x leads to the same equation, the graph has symmetry with respect to the y-axis. 2. If substituting -y for y leads to the same equation, the graph has symmetry with respect to the x-axis. 3. If simultaneously substituting -x for x and -y for y leads to the same equation, the graph has symmetry with respect to the origin. Note: a graph may have none of these three symmetries, one, or all three.

4s LINEAR AND NONLINEAR INEQUALITIES CHAP. 61 + + + - Sign o f x - 10 Signofx+2 Sign of result + + l l l l l l l l l l l 1 1 1 1 l 1 1 1 1 1 1 1 1 1 1 1 1 0 -2 1 l 1 ) I 1 10 Fig. 6-3 The critical points divide the real number line into the intervals (-=,-2), (-2,10), and ( 1 0 , ~ )In. (--,-2), x - 10 and x + 2 are negative, hence the product is positive. In (-2,lO). r - 10 is negative and x + 2 is positive; hence the product is negative. In ( l o , ~ ) both , factors are positive; hence the product is positive.

In ( 1,=), both factors are positive; hence the product is positive. + + + Signofx- 1 Signofx+ 2 - Sign of product+ I I I -3 I -2 I I I -1 I I I 0 I I I 1 I I I I I I ~ 2 Fig. 6-2 The inequality holds when vals: (-=,-2) U (1 ,=). ( x - l)(x + 2) is positive. 1. Solve: 3(y - 5 ) - 4(y + 6) 5 7 Eliminate parentheses, combine terms, and isolate the variable: 3(y - 5) - 4(y + 6) I 7 3y- 15-4y-24I7 -y-3917 -y I 4 6 y 2 -46 The solution set is [-46, -). 2. Solve: -- 3 6 3x 8 LINEAR AND NONLINEAR INEQUALITIES 44 [CHAP.