# How do you find a standard form equation for the line with slope of a line is -1/3, and the y-intercept is 10/3?

##### 1 Answer
Jan 16, 2017

See the entire solution process below:

#### Explanation:

First, we can use the slope-intercept formula to find an equation for this line.

The slope-intercept form of a linear equation is:

$y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$

Where $\textcolor{red}{m}$ is the slope and color(blue)(b is the y-intercept value.

Substituting the slope and y-intercept from the problem gives:

$y = \textcolor{red}{- \frac{1}{3}} x + \textcolor{b l u e}{\frac{10}{3}}$

Next, we need to transform to the standard form.

The standard form of a linear equation is:

$\textcolor{red}{A} x + \textcolor{b l u e}{B} y = \textcolor{g r e e n}{C}$

where, if at all possible, $\textcolor{red}{A}$, $\textcolor{b l u e}{B}$, and $\textcolor{g r e e n}{C}$are integers, and A is non-negative, and, A, B, and C have no common factors other than 1

$y = - \frac{1}{3} x + \frac{10}{3}$

$\textcolor{red}{3} \times y = \textcolor{red}{3} \left(- \frac{1}{3} x + \frac{10}{3}\right)$

$3 y = \left(\textcolor{red}{3} \times - \frac{1}{3} x\right) + \left(\textcolor{red}{3} \times \frac{10}{3}\right)$

3y = (cancel(color(red)(3)) xx -1/color(red)(cancel(color(black)(3)))x) + (cancel(color(red)(3)) xx 10/cancel(color(black)(3))))

$3 y = - 1 x + 10$

$\textcolor{red}{1 x} + 3 y = \textcolor{red}{1 x} - 1 x + 10$

$1 x + 3 y = 0 + 10$

$\textcolor{red}{1} x + \textcolor{b l u e}{3} y = \textcolor{g r e e n}{10}$