# How do you solve for y in 2x + 3y = 3?

Mar 27, 2018

$y = - \frac{2}{3} x + 1$

#### Explanation:

Move the $x$ term to the right side, then divide everything by $3$:

$2 x + 3 y = 3$

$3 y + 2 x = 3$

$3 y + 2 x \textcolor{b l u e}{-} \textcolor{b l u e}{2 x} = 3 \textcolor{b l u e}{-} \textcolor{b l u e}{2 x}$

$3 y \textcolor{red}{\cancel{\textcolor{b l a c k}{\textcolor{b l a c k}{+} 2 x \textcolor{b l u e}{-} \textcolor{b l u e}{2 x}}}} = 3 \textcolor{b l u e}{-} \textcolor{b l u e}{2 x}$

$3 y = 3 \textcolor{b l u e}{-} \textcolor{b l u e}{2 x}$

$3 y = - 2 x + 3$

$\textcolor{b l u e}{\frac{\textcolor{b l a c k}{3 y}}{3}} = \textcolor{b l u e}{\frac{\textcolor{b l a c k}{- 2 x + 3}}{3}}$

$\textcolor{b l u e}{\frac{\textcolor{b l a c k}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} y}}{\textcolor{red}{\cancel{\textcolor{b l u e}{3}}}}} = \textcolor{b l u e}{\frac{\textcolor{b l a c k}{- 2 x + 3}}{3}}$

$y = \textcolor{b l u e}{\frac{\textcolor{b l a c k}{- 2 x + 3}}{3}}$

$y = \textcolor{b l u e}{\frac{\textcolor{b l a c k}{- 2 x}}{3}} + \textcolor{b l u e}{\frac{\textcolor{b l a c k}{3}}{3}}$

$y = \textcolor{b l u e}{\frac{\textcolor{b l a c k}{- 2 x}}{3}} + 1$

$y = - \frac{2}{3} x + 1$

That is solving for $y$. Hope this helped!