## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 2 - 2.1 - Linear Equations in Two Variables - 2.1 Exercises - Page 170: 84

#### Answer

0 = $\frac{-3}{2}$x + $\frac{1}{2}$y + 1

#### Work Step by Step

The intercept form of an equation is: $\frac{x}{a}$ + $\frac{y}{b}$ = 1 where a is (a,0), the x-intercept, and b is (0,b), the y-intercept. In this problem the x-intercept is ($\frac{2}{3}$,0), so a = $\frac{2}{3}$ In this problem the y-intercept is (0, -2), so b = -2 Using the equation above for the intercept form, we have: $\frac{x}{\frac{2}{3}}$ + $\frac{y}{-2}$ = 1 Which can be rewritten as: $\frac{3x}{2}$ + $\frac{y}{-2}$ = 1 We want to manipulate the equation to be in the general form which is: 0 = ax + by + c Note: a, b, and c are constants. First, we want to get the left hand side to be 0. We can do this by subtracting $\frac{3x}{2}$ from each side, and subtracting $\frac{y}{-2}$ from each side. $\frac{3x}{2}$ + $\frac{y}{-2}$ - $\frac{3x}{2}$ - $\frac{y}{-2}$ = 1 - $\frac{3x}{2}$ - $\frac{y}{-2}$ This gets: 0 = 1 - $\frac{3x}{2}$ - $\frac{y}{-2}$ We can rearrange the variables to get the general form: 0 = $\frac{-3}{2}$x + $\frac{1}{2}$y + 1

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