# How do you write the equation y+4=-1/3(x-12) in standard form?

Apr 16, 2017

x+3y=0#

#### Explanation:

The equation of a line in $\textcolor{b l u e}{\text{standard form}}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{A x + B y = C} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where A is a positive integer and B, C are integers.

$\text{Rearrange " y+4=-1/3(x-12)" into this form}$

$\text{multiply ALL terms by 3 to eliminate the fraction}$

$3 y + 12 = \left({\cancel{3}}^{1} \times - \frac{1}{\cancel{3}} ^ 1 \left(x - 12\right)\right)$

$\Rightarrow 3 y + 12 = - x + 12$

$\text{add x to both sides}$

$x + 3 y + 12 = \cancel{- x} \cancel{+ x} + 12$

$\Rightarrow x + 3 y + 12 = 12$

$\text{subtract 12 from both sides}$

$x + 3 y \cancel{+ 12} \cancel{- 12} = 12 - 12$

$\Rightarrow x + 3 y = 0 \leftarrow \textcolor{red}{\text{ in standard form}}$

Apr 16, 2017

See the entire solution process below:

#### Explanation:

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

First, multiply each side of the equation by $\textcolor{red}{3}$ to eliminate the fractions while keeping the equation balanced:

$\textcolor{red}{3} \left(y + 4\right) = \textcolor{red}{3} \times - \frac{1}{3} \left(x - 12\right)$

$\left(\textcolor{red}{3} \times y\right) + \left(\textcolor{red}{3} \times 4\right) = \cancel{\textcolor{red}{3}} \times - \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} \left(x - 12\right)$

$3 y + 12 = - 1 \left(x - 12\right)$

$3 y + 12 = - 1 x + 12$

Now, subtract $\textcolor{red}{12}$ and add $\textcolor{red}{1 x}$ to each side of the equation to put in standard form while keeping the equation balanced:

$\textcolor{red}{1 x} + 3 y + 12 - \textcolor{red}{12} = \textcolor{red}{1 x} - 1 x + 12 - \textcolor{red}{12}$

$1 x + 3 y + 0 = 0 + 0$

$\textcolor{red}{1} x + \textcolor{b l u e}{3} y = \textcolor{g r e e n}{0}$