# How do you solve the system of equations 2x+2y=4 and 12-3x=3y?

Nov 16, 2017

No solutions

#### Explanation:

$2 x + 2 y = 4$
$- 3 x + 12 = 3 y$

We need to solve $2 x + 2 y = 4$ for $x$

Let's start by adding $\textcolor{red}{- 2 y}$ to both sides

$2 x + 2 y \textcolor{red}{- 2 y} = 4 \textcolor{red}{- 2 y}$

$2 x = - 2 y + 4$

$x = \frac{- 2 y + 4}{2}$

$x = - y + 2$

Then, substitute $- y + 2$ for $x$ in $- 3 x + 12 = 3 y$

$- 3 x + 12 = 3 y$

$- 3 \left(- y + 2\right) + 12 = 3 y$

$3 y - 6 + 12 = 3 y$

$3 y + 6 = 3 y$

Then add $- 3 y$ to both sides

$3 y - 3 y + 6 = 3 y - 3 y$

$6 = 0$

Finally, add $- 6$ to both sides

$6 - 6 = 0 - 6$

$0 = - 6$

Thus,

There is no solutions.

Nov 16, 2017

No solution

#### Explanation:

Rewriting the second equation gives $3 x + 3 y = 12$. Dividing each term by $3$ then yields $x + y = 4$.

For the first equation $2 x + 2 y = 4$, we can divide each term by $2$, yielding $x + y = 2$.

Then, we have that $x + y = 2$ and $x + y = 4$. But how could this ever be the case? $x + y$ should always give the same value for a fixed $x$ and fixed $y$. Thus there are no solutions.