Section 3.1 Equations and the Rectangular Coordinate System 203 �� ������������������������ ������������������ A problem and an incorrect solution are given. Provide the correct solution and an explanation of the error. 6a. Is the ordered pair (4, -2) a solution of y = 2x - 10? �������������������������������������� ���������������������������������������������������������������� In the ordered pair (4, -2), the x-value is 4 and the y-value is -2. y = 2x - 10 y = 2x-10 4 4 = 2(-2) -10 4 = -4 - 10 4 = -14 -2 = 2(4) -10 -2 = 8 - 10 -2 = -2 Since the point makes the equation true, it is a solution of the equation. Since the point makes the equation false, it is not a solution. 6b. Plot the points (0, 3) and (-2, 0). �������������������������������������� ���������������������������������������������������������������� When the first coordinate is 0, there is no movement left or right from the origin. When the second coordinate is 0, there is no movement up or down from the x-axis. The correct plot is shown here. (0, 3) (–2, 0) 2 2 4 4 –2 –4 –2 –4 x y The points are shown in the graph. (0, 3) 2 y (–2, 0) 4 2 4 –2 –4 –2 –4 6c. Graph y = ux + 3u. x �������������������������������������� ���������������������������������������������������������������� While the points in the table are solutions of the equation, we cannot assume that the graph continues in this manner. It is always a good idea to input values for x that make the expression inside the absolute value negative. So, we find a couple more points to get x y = ux + 3u (x, y) -4 y = u-4 + 3u = u-1u = 1 (-4, 1) -3 y = u-3 + 3u = 0 (-3, 0) The table that corresponds to this graph is x y = ux + 3u (x, y) -2 y = u-= 1 (-2, 1) -1 = u-1 + 3u = 2 (- 1, 2) 0 y = u0 + 3u = 3 (0, 3) 1 y = u1 + 3u = 4 (1, 4) 2 y = u2 + 3u = 5 (2, 5) So, the graph is 6 2 (2, 5) (–1, 2) –6 –4 –2 x 2 4 4 –2 y (1, 4) (–2, 1) 1 makes he is s aph. u 2 + 3u y ,3 ) (2, 5) (1, 4) (–1, 2) (–4, 1) 6 2 –8 –6 –4 –2 2 x 4 –2 y
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