Problem:
The slope of a line is [tex] - \frac{1}{3} [/tex] and the y-intercept is [tex] \frac{10}{3} [/tex]. What is the equation of the line written in general form?
Choices:
○ 10x + 3y - 1 = 0
○ x + 3y + 10 = 0
○ x + 3y - 10 = 0
Remember to use this formula:
[tex] \quad \quad\quad\quad\boxed{\tt{y = mx + b}}[/tex]
Given that:
[tex]\quad \quad\quad\quad\boxed{\tt{Slope (m) = \frac{1}{3} }}[/tex]
[tex]\quad \quad\quad\quad\boxed{\tt{y \: intercept(b) = \frac{10}{3} }}[/tex]
Lets try!
[tex] \quad \quad\quad\quad\boxed{\tt{y = - \frac{ 1}{3}x + \frac{10}{3} }}[/tex]
Then convert it in general form using this formula:
[tex] \quad \quad\quad\quad\boxed{\tt{ax + by + c = 0}}[/tex]
Each side will be multiply by 3
[tex] \quad \quad\quad\quad\boxed{\tt{ (3)y = 3(- \frac{ 1}{3}x + \frac{10}{3} )}}[/tex]
[tex] \quad \quad\quad\quad\boxed{\tt{ 3y = \cancel{ \color{red}3}(- \frac{ 1}{ \cancel{ \color{red}3}}x + \frac{10}{ \cancel{ \color{red}3}} )}}[/tex]
[tex] \quad \quad\quad\quad\boxed{\tt{3y = - x + 10}}[/tex]
Let's convert "x" like this.
[tex] \quad\quad\quad\boxed{\tt{3y = - x + 10}} \: ➡ \: \boxed{\tt{x + 3y = 10}}[/tex]
Convert "10" like this.
[tex] \quad\quad\quad \boxed{\tt{x + 3y = 10}} \: ➡ \: \boxed{ \tt{x + 3y - 10 = 0}}[/tex]
Hence, the answer is:
[tex]\quad \quad\quad\quad \boxed{ \color{green}{ \tt{x + 3y - 10 = 0}}}[/tex]
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