How do you write 4y-5x=3(4x-2y+1) in standard form?

Apr 30, 2017

See the entire solution process below:

Explanation:

First, expand the terms in parenthesis on the right side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:

$4 y - 5 x = \textcolor{red}{3} \left(4 x - 2 y + 1\right)$

$4 y - 5 x = \left(\textcolor{red}{3} \cdot 4 x\right) - \left(\textcolor{red}{3} \cdot 2 y\right) + \left(\textcolor{red}{3} \cdot 1\right)$

$4 y - 5 x = 12 x - 6 y + 3$

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.

To convert to the Standard Form of a linear equation we need to first subtract $\textcolor{red}{12 x}$ and add $\textcolor{b l u e}{6 y}$ to each side of the equation to get the $x$ and $y$ terms on left side of the equation while keeping the equation balanced:

$- \textcolor{red}{12 x} + \textcolor{b l u e}{6 y} + 4 y - 5 x = - \textcolor{red}{12 x} + \textcolor{b l u e}{6 y} + 12 x - 6 y + 3$

$- \textcolor{red}{12 x} - 5 x + \textcolor{b l u e}{6 y} + 4 y = - \textcolor{red}{12 x} + 12 x + \textcolor{b l u e}{6 y} - 6 y + 3$

$\left(- \textcolor{red}{12} - 5\right) x + \left(\textcolor{b l u e}{6} + 4\right) y = 0 - 0 + 3$

$- 17 x + 10 y = 3$

Now, multiply each side of the equation by $\textcolor{red}{- 1}$ to ensure the coefficient of the $x$ is a positive integer:

$\textcolor{red}{- 1} \left(- 17 x + 10 y\right) = \textcolor{red}{- 1} \cdot 3$

$\left(\textcolor{red}{- 1} \cdot - 17 x\right) + \left(\textcolor{red}{- 1} \cdot 10 y\right) = - 3$

$17 x + \left(- 10 y\right) = - 3$

$\textcolor{red}{17} x - \textcolor{b l u e}{10} y = \textcolor{g r e e n}{- 3}$