<s>[INST] 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?
 [/INST] 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.
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