Section 3.1 The Coordinate System, Graphing Equations, and the Midpoint Formula 151 Solutions 2a. Replace the variables with their corresponding values. (1, -3) (0, 6) (2, 0) 3x - y =6 3x - y =6 3x - y = 6 3(1) - (-3) =6 3(0) - (6) =6 3(2) - (0) = 6 3 + 3 =6 0- 6 =6 6- 0 = 6 6 = 6 -6 =6 6= 6 True False True The ordered pairs (1, -3) and (2, 0) are solutions of 3x - y = 6 since they make the equation a true statement. The ordered pair (0, 6) is not a solution of 3x - y = 6 since it doesn’t make the equation true. 2b. Replace the variables with their corresponding values. a - 5 4 , 0b a3 4 , 8b a1 4 , 4b y = -4x + 5 y = -4x + 5 y = -4x + 5 0=-4 a- 5 4 b + 5 8=-4 a3 4 b + 5 4=-4 a1 4 b + 5 0 = 5 + 5 8 = -3 + 5 4 = -1 + 5 0 = 10 8 = 2 4 = 4 False False True The ordered pair a1 4 , 4b is a solution of y = -4x + 5 since it makes the equation a true statement. The ordered pairs a- 5 4 , 0b and a3 4 , 8b are not solutions of y = -4x + 5 since they do not make the equation true. 2c. Replace the variables with their corresponding values. (-5, 7) (-3, 5) (4, 2) y = |x - 2| y = |x - 2| y = |x - 2| 7 = | -5 - 2| 5 = | -3 - 2| 2 = |4 - 2| 7 = | -7| 5 = | -5| 2 = | 2| 7 = 7 5 = 5 2 = 2 True True True All three ordered pairs make the equation true, so all are solutions of y = |x - 2|. 2d. Replace the variables with their corresponding values. (2, 3) (-1, 3) (1, 2) y = x2 - x + 1 y = x2 - x + 1 y = x2 - x + 1 3 = (2)2 - (2) + 1 3 = (-1)2 - (-1) + 1 2 = (1)2 - (1) + 1 3 = 4 - 2 + 1 3 = 1 + 1 + 1 2 = 1 - 1 + 1 3 = 3 3 = 3 2 = 1 True True False The ordered pairs (2, 3) and (-1, 3) are solutions of the equation y = x2 - x + 1 since they make the equation true. The ordered pair (1, 2) is not a solution of y = x2 - x + 1 since it doesn’t make the equation true.
hendricks_intermediate_algebra_1e_ch1_3
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