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hendricks_beginning_algebra_1e_ch1_3

252 Chapter 3 Linear Equations in Two Variables 2b. Method 1 m=- 3 4 and (x, y) = (1, 5) y = mx + b 5 = - 3 4 (1) + b 5 = - 3 4 + b 4(5) = 4a- 3 4 + bb 20=-3 + 4b 20 + 3=-3 + 4b + 3 23 = 4b 23 4 = b Equation in slope-intercept form: y = - 3 4 x + 23 4 Equation in standard form: y=- 3 4 x + 23 4 Method 2 m=- 3 4 and (x1, y1) = (1, 5) y - y1 = m(x - x1) y - 5 = - 3 4 (x - 1) y - 5 = - 3 4 x + 3 4 4(y - 5) = 4a- 3 4 x + 3 4 4y - 20=-3x + 3 4y - 20 + 20=-3x + 3 + 20 4y=-3x + 23 y=- 3 4 x + 23 4 Equation in slope-intercept form: y=- 3 4 x + 23 4 Begin with the slope-intercept form. 4(y) = 4a- 3 4 x + 23 4 b Multiply each side by 4. 4y=-3x + 23 Simplify. 4y + 3x=-3x + 23 + 3x Add 3x to each side. 3x + 4y = 23 Simplify. Check: Show that (1, 5) is a solution of y=- 3 4 x + 23 4 b or 3x + 4y = 23. 3x + 4y = 23 Begin with the standard form. 3(1) + 4(5) = 23 Let x = 1 and x = 5. 3 + 20 = 23 Simplify. 23 = 23 Simplify. Since the resulting equation is true, our work is correct. 2c. Method 1 m = 0 and (x, y) = (-2, 1) y = mx + b 1 = 0(-2) + b 1 = 0 + b 1 = b Method 2 m = 0 and (x1, y1) = (-2, 1) y - y1 = m(x - x1) y - 1 = 0x - (-2) y - 1 = 0 y - 1 + 1 = 0 + 1 y = 1 Equation: y = 0x + 1 or y = 1 Equation: y = 1 Recall that a line with a slope of zero is horizontal and that horizontal lines are of the form y = k, where k is the y-value of a point on the line. 2d. A line with undefined slope is a vertical line. A vertical line is written as x = h, where h is the x-value of a point on the line. Since the given point is (-3, 9), h = -3. So, the equation of the line is x = -3.


hendricks_beginning_algebra_1e_ch1_3
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