Solve $2a + 5b = 20$ Is this equation solvable? It seems like you should be able to get a right number!
If this is solvable can you tell me step by step on how you solved it.
$$\begin{align}
{2a + 5b} & = {20}
\end{align}$$
My thinking process: 
$$\begin{align}
{2a + 5b} & = {20} &   {2a + 5b} & = {20} \\
{0a + 5b} & = {20} &   {a + 0b} & = {20} \\
{0a + b} & = {4} &     {a + 0b} & = {10} \\
{0a + b} & = {4/2} &   {a + 0b} & = {10/2} \\
{0a + b} & = {2} &     {a + 0b} & = {5} \\
\end{align}$$
The problem comes out to equal:
$$\begin{align}
{2(5) + 5(2)} & = {20} \\
{10 + 10} & = {20} \\
{20} & = {20}
\end{align}$$
since the there are two different variables could it not be solved with the right answer , but only "a answer?"
What do you guys think?

Generally one can use the Extended Euclidean algorithm, but that's overkill here. First note that since $\rm\,2a+5b = 20\:$ we see $\rm\,b\,$ is even, say $\rm\:b = 2n,\:$ hence dividing by $\,2\,$ yields $\rm\:a = 10-5n.$
Remark $\ $ The solution  $\rm\:(a,b) = (10-5n,2n) = (10,0) + (-5,2)\,n\:$ is the  (obvious) particular solution $(10,0)\,$ summed with the general solution $\rm\,(-5,2)\,n\,$ of the associated homogeneous equation $\rm\,2a+5b = 0,\:$ i.e. the general form of a solution of a nonhomogeneous linear equation.
