�� ������������������������ ������������������ A problem and an incorrect solution are given. Provide the correct solution and an explanation of the error. 6a. Graph the equation 5x + y = 0. �������������������������������������� ���������������������������������������������������������������� We must have two points to draw the graph of this line. To find another point, make a substitution for x. If x = 1, then 5x + y = 0 5(1) + y = 0 5 + y = 0 y = -5 We can graph the equation using the points (0, 0) and (1, -5). 2 (1, –5) 4 –2 –4 –2 –4 –6 x y (0, 0) 6b. Find the x- and y-intercepts of 8x - y = -16. �������������������������������������� ���������������������������������������������������������������� The first error is in solving the first equation. It should be 8(0) - y = -16 -y = -16 y = 16 The other error was in assigning the points. The point (0, 16) is the y-intercept and the point (-2, 0) is the x-intercept. The x-intercept is (0, 0) and the y-intercept is (0, 2 2 4 4 –2 –4 –2 –4 x y 8(0) - y = -16 y = -16 8x 8 - 0 0 = -16 8x = - 16 x = -2 x-intercept is The (0, -16)1 and the y-intercept rcept is (-2, 0 0). y intercep 0). So, the graph is ANSWERS TO STUDENT CHECKS Student Check 1 a. yes; x - 2y = -4; A = 1, B = -2, C = -4 b. yes; x + 0y = 4; A = 1, B = 0, C = 4 c. no d. yes; 2x - 3y = -12; A = 2, B = -3, C = -12 Student Check 2 a. y = x - 2 b. y=- 1 3 x + 6 c. x + 3y = 9 (–1, –3) (0, –2) (3, 1) 2 2 4 –4 –2 4 6 –4 –6 x y 8 y 17 1, 3 (0, 6) (3, 5) 2 2 4 –2 4 6 x 6 (–3, 4) (0, 3) (3, 2) 2 –4 –2 x 2 4 4 –2 y 220 Chapter 3 Linear Equations in Two Variables
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