How do you write the equation y-1=5/6(x-4) in standard form?

Nov 20, 2017

See a 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 $\textcolor{red}{6}$ to eliminate the fractions while keeping the equation balanced:

$\textcolor{red}{6} \left(y - 1\right) = \textcolor{red}{6} \times \frac{5}{6} \left(x - 4\right)$

$\textcolor{red}{6} \left(y - 1\right) = \cancel{\textcolor{red}{6}} \times \frac{5}{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}}} \left(x - 4\right)$

$\left(\textcolor{red}{6} \times y\right) - \left(\textcolor{red}{6} \times 1\right) = 5 \left(x - 4\right)$

$6 y - 6 = \left(5 \times x\right) - \left(5 \times 4\right)$

$6 y - 6 = 5 x - 20$

Next, add $\textcolor{red}{6}$ and subtract $\textcolor{b l u e}{5 x}$ from each side of the equation to isolate the constant on the right side of the equation while keeping the equation balanced:

$- \textcolor{b l u e}{5 x} + 6 y - 6 + \textcolor{red}{6} = - \textcolor{b l u e}{5 x} + 5 x - 20 + \textcolor{red}{6}$

$- 5 x + 6 y - 0 = 0 - 14$

$- 5 x + 6 y = - 14$

Now multiply each side of the equation by $\textcolor{red}{- 1}$ to make the $x$ coefficient positive while keeping the equation balanced:

$\textcolor{red}{- 1} \left(- 5 x + 6 y\right) = \textcolor{red}{- 1} \times - 14$

$\left(\textcolor{red}{- 1} \times - 5 x\right) + \left(\textcolor{red}{- 1} \times 6 y\right) = 14$

$5 x + \left(- 6 y\right) = 14$

$\textcolor{red}{5} x - \textcolor{b l u e}{6} y = \textcolor{g r e e n}{14}$