Section 3.1 Equations and the Rectangular Coordinate System 195 There are, in fact, infinitely many solutions of an equation in two variables. Just as there are infinitely many solutions of this type of equation, there are infinitely many ordered pairs that do not satisfy the equation x + y = 3. Any ordered pair whose values do not add to 3 is not a solution of x + y = 3. Some ordered pairs that are not solutions are (1, 5), (3, -3), and (-1, -2). (x, y) x + y = 3 Solution? 1 (1, 5) + 5 = 3 6 ≠ 3 No (3, -3) 3 + (-3) = 3 0 ≠ 3 No (-1, -2) -1 + (-2) = 3 -3 ≠ 3 No ���������������������� Determining if an Ordered Pair Is a Solution of an Equation in Two Variables Step 1: Replace the values of x and y with the numbers given in the ordered pair. Step 2: Simplify each side of the equation. Step 3: If the resulting equation is true, then the ordered pair is a solution of the equation. If it is not true, then the ordered pair is not a solution. �� ������������������������ ������������������ Determine if the ordered pair is a solution of the equation. 1a. (3, -1); 2x - y = 7 1b. a3 2 , 0b ; y = 2x + 3 1c. (0, 5); y = x2 - 3x + 5 Solutions 1a. 2x - y = 7 2(3) - (-1) = 7 Let x = 3 and y = -1. 6 + 1 = 7 Simplify. 7 = 7 True Since (3, -1) makes the equation a true statement, it is a solution of 2x - y = 7. 1b. y = 2x + 3 0 = 2a3 2 b + 3 Let x = 3 2 and y = 0. 0 = 3 + 3 Simplify. 0 = 6 False Since a3 2 , 0b makes the equation a false statement, it is not a solution of y = 2x + 3. 1c. y = x2 - 3x + 5 5 = (0)2 - 3(0) + 5 Let x = 0 and y = 5. 5 = 0 + 5 Simplify. 5 = 5 True Since (0, 5) makes the equation a true statement, it is a solution of y = x2 - 3x + 5. Student Check 1 Determine if the ordered pair is a solution of the equation. a. (-2, 6); 3x - y = 0 b. a1 4 , 0b ; y = 4x - 1 c. (-3, 5); y = ux - 2u
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