Check: Show that (4, -3) and (-2, 9) are solutions of the equation y = -2x + 5. (x, y) y = -2x + 5 Solution? (4, -3) -3 = -2(4) + 5 -3 = -8 + 5 Yes -3 = -3 (-2, 9) 9 = -2(-2) + 5 9 = 4 + 5 Yes 9 = 9 3b. Find the slope of the line. m = y2 - y1 x2 - x1 Begin with the slope formula. m = 12 - 0 4 - (-4) Let (x1, y1) = (-4, 0) and (x2, y2) = (4, 12). m = 12 8 Simplify the numerator and denominator. m = 3 2 Simplify the result. Use the slope-intercept form of a line to write the equation. Either of the given points can be used in the slope-intercept form to find b. m = 3 2 ; (x, y) = (-4, 0) y = mx + b 0 = 3 2 (-4) + b 0 = -6 + b 0 + 6=-6 + b + 6 6 = b Equation: y = 3 2 x + 6 m = 3 2 ; (x, y) = (4, 12) y = mx + b 12 = 3 2 (4) + b 12 = 6 + b 12 - 6 = 6 + b - 6 6 = b Equation: y = 3 2 x + 6 Check: Show that (-4, 0) and (4, 12) are solutions of the equation y = 3 2 x + 6. (x, y) y = 3 2 x + 6 Solution? (-4, 0) 0 = 3 2 (-4) + 6 0=-6 + 6 0 = 0 Yes (4, 12) 12 = 3 2 (4) + 6 12 = 6 + 6 12 = 12 Yes 3c. Find the slope of the line. m = y2 - y1 x2 - x1 m = 11 - 4 -2 - (-2) m = 7 0 m = undefined Begin with the slope formula. Let (x1, y1) = (-2, 4) and (x2, y2) = (-2, 11). Simplify the numerator and denominator. Simplify the result. 254 Chapter 3 Linear Equations in Two Variables
hendricks_beginning_intermediate_algebra_1e_ch1_3
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