# How do you write the equation y+3=-5(x+1) in standard form?

Apr 23, 2017

$5 x + y = - 8$

#### 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.

$\text{Rearrange " y+3=-5(x+1)" into this form}$

$\text{distribute bracket}$

$y + 3 = - 5 x - 5$

$\text{add 5x to both sides}$

$5 x + y + 3 = \cancel{- 5 x} \cancel{+ 5 x} - 5$

$\text{subtract 3 from both sides}$

$5 x + y \cancel{+ 3} \cancel{- 3} = - 5 - 3$

$\Rightarrow 5 x + y = - 8 \leftarrow \textcolor{red}{\text{ in standard form}}$

Apr 23, 2017

See the 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, expand the terms on the right side of the equation:

$y + 3 = \textcolor{red}{- 5} \left(x + 1\right)$

$y + 3 = \left(\textcolor{red}{- 5} \cdot x\right) + \left(\textcolor{red}{- 5} \cdot 1\right)$

$y + 3 = - 5 x + - 5$

$y + 3 = - 5 x - 5$

Now, add $\textcolor{red}{5 x}$ and subtract $\textcolor{b l u e}{3}$ from each side of the equation to isolate the $x$ and $y$ terms on the left side of the equation and the constant on the right side of the equation while keeping the equation balanced:

$\textcolor{red}{5 x} + y + 3 - \textcolor{b l u e}{3} = \textcolor{red}{5 x} - 5 x - 5 - \textcolor{b l u e}{3}$

$5 x + y + 0 = 0 - 8$

$\textcolor{red}{5} x + \textcolor{b l u e}{1} y = \textcolor{g r e e n}{- 8}$