## Algebra and Trigonometry 10th Edition

0 = 6x + $\frac{3}{2}$y + 1
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 ($\frac{-1}{6}$,0), so a = $\frac{-1}{6}$ In this problem the y-intercept is (0, $\frac{-2}{3}$), so b = $\frac{-2}{3}$ Using the equation above for the intercept form, we have: $\frac{x}{\frac{-1}{6}}$ + $\frac{y}{\frac{-2}{3}}$ = 1 Which can be rewritten as: $\frac{6x}{-1}$ + $\frac{3y}{-2}$ = 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 $\frac{6x}{-1}$ from each side, and subtracting $\frac{3y}{-2}$ from each side. $\frac{6x}{-1}$ + $\frac{3y}{-2}$ - $\frac{6x}{-1}$ - $\frac{3y}{-2}$ = 1 - $\frac{6x}{-1}$ - $\frac{3y}{-2}$ This gets: 0 = 1 - $\frac{6x}{-1}$ - $\frac{3y}{-2}$ We can rearrange the variables to get the general form: 0 = 6x + $\frac{3}{2}$y + 1