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198 Chapter 3 Linear Equations in Two Variables ���������������������� Graphing Solutions of an Equation Step 1: Determine at least three solutions of the equation. Use a table to show the solutions. a. Substitute any value for x. b. Solve the resulting equation or simplify the resulting expression to find the value of y. c. The ordered pair (x, y) is a solution of the equation. Step 2: Plot the solutions on a coordinate system. Step 3: Connect the points and use their pattern to sketch the graph. ������������ We will study graphing special types of equations in later sections and chapters. This method is a basic introduction to graphing. �� ������������������������ ������������������ Graph each equation by completing a table of solutions. 3a. y = x + 3 3b. y = x2 - 1 3c. y = ux + 2u Solutions 3a. x y = x + 3 (x, y) -2 y = -2 + 3 = 1 (-2, 1) -1 y = -1 + 3 = 2 (-1, 2) 0 y = 0 + 3 = 3 (0, 3) 1 y = 1 + 3 = 4 (1, 4) 2 y = 2 + 3 = 5 (2, 5) Now plot the solutions and connect them to form the graph. x y 6 4 2 (–1, 2) (1, 4) –6 –4 2 –2 –2 –4 (–2, 1) (0, 3) (2, 5) x y 6 4 2 (–1, 2) (1, 4) –6 –4 2 –2 –2 –4 (–2, 1) (0, 3) (2, 5) Notice the graph of y = x + 3 is a line. 3b. x y = x2 - 1 (x, y) -2 y = (-2)2 -1 = 4 - 1 = 3 (-2, 3) -1 y = (-1)2 - 1 = 1 - 1 = 0 (-1, 0) 0 y = (0)2 - 1 = 0 - 1 = - 1 (0, -1) 1 y = (1)2 - 1 = 1 - 1 = 0 (1, 0) 2 y = (2)2 - 1= 4 - 1 = 3 (2, 3) Now plot the solutions and connect them to form the graph. (–2, 3) (2, 3) (–1, 0) (1, 0) (0, –1) 2 2 4 4 –2 –4 –2 –4 x y (–2, 3) (2, 3) (–1, 0) (1, 0) (0, –1) 2 2 4 4 –2 –4 –2 –4 x y Notice the graph of y = x2 - 1 has a “U-shape” pattern.


hendricks_beginning_algebra_1e_ch1_3
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