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?
 A: Also, this equation has solutions in the integers in the greatest common divisor of 2 and 5 is 1, which divides 20.
To get an explicit solution, use the Euclidean algorithm. (BTW these are called Diophantine equations if you want to do further reading on them)
A: You have what is known as a linear diophantine equation. An equation of the form
$$ax + by = c$$
is solvable in $x$ and $y$ if and only if $\gcd(a,\ b)\mid c$. In your particular case the equation is solvable. 
You've generated one solution already, the pair $(x,\ y)=(5,\ 2)$. All the other solutions are then given by
$$(x,\ y)=\left(5 + 5k,\ 2-2k\right)$$
for $k\in \mathbb{Z}$.
A: Note that $2a$ must have as its unit digit $0, 2, 4, 6,$ or $8$ (because it's even!).
Similarly, note that $5b$ must have as its unit digit $0$ or $5$.
With a bit of thinking, you can see that for $2a + 5b$ to be $20$, we need to ensure that $2a$ has a unit digit of $0$ (hence $a$ is a multiple of $5$).
So let $a$ be a multiple of $5$. That is, let $a = 5k$, for some integer $k$.
Then $2a + 5b = 20$ becomes $10k + 5b = 20$, so that $5b = 20 - 10k$.
Dividing both sides of our last equation by $5$, we have $b = 4 - 2k$.
This gives you all the possible answers for $(a, b)$, namely, $(5k, 4-2k)$.
For example, when $k = 1$ you get $(5 \cdot 1, 4 - 2 \cdot 1) = (5, 2)$, which is the answer you came to.
A: 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.
