College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.3 - Lines - 1.3 Exercises - Page 112: 4

Answer

$6$; $4$; $y=-\dfrac{2}{3}x+4$; $m=-\dfrac{2}{3}$

Work Step by Step

RECALL: (1) The slope-intercept form of a line's equation is $y=mx + b$ where m = slope and b = y-intercept. (2) The x-intercept is a point where the graph touches or crosses the x-axis. The x-intercept can be found by setting $y=0$ then solving for $x$. (3) The y-intercept is a point where the graph touches or crosses the y-axis. The y-intercept can be found by setting $x=0$ then solving for $y$. Convert $2x+3y-12=0$ to slope-intercept form by solving for $y$ to obtain: $2x + 3y -12 -2x+12=0-2x+12 \\3y=-2x+12 \\\dfrac{3y}{3} = \dfrac{-2x+12}{3} \\y = \dfrac{-2x}{3} + \dfrac{12}{3} \\y = -\dfrac{2}{3}x + 4$ The equation above has a slope of $,=-\dfrac{2}{3}$ and a y-intercept of $4$. To find the x-intercept, set $y=0$ then solve for $x$ to obtain: $y=-\dfrac{2}{3}x+4 \\0 = -\dfrac{2}{3}x + 4 \\0+\dfrac{2}{3}x = -\dfrac{2}{3}x+4 + \dfrac{2}{3}x \\\dfrac{2}{3}x = 4$ Multiply $\dfrac{3}{2}$ to both sides of the equation to obtain: $\\\dfrac{3}{2} \cdot \dfrac{2}{3}x = 4 \cdot \dfrac{3}{2} \\x = 6$ The x-intercept is $6$.
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