How do I solve 49x+99y=9999? I'm trying to solve this equation.
I want to figure out how many product I need to sell at price $x$ and price $y$ to make $\$9,999$.
I went through the $2$-step equations and inequality videos via khan academy but still no luck. I want to use what I've learned so far to solve a hypothetical question on how I can make $\$9,999$ in a year drop shipping if I sell the same product $\$49$ in spring and $\$99$ in fall.
What section of algebra should I learn to solve this equation 49x+99y=9999?
The feedback helped me understand that this equation might be what I'm after:
1) 49x+99y=10000
2) x = 5000 / 49 = ~102 sales
3) y = 5000 / 99 = ~50 sales
4) 102 sales at \$49 and 50 sales at \$99 to get \$10,000.
^ What is the right equation for the above, so I can plug in numbers to solve for it next time?
 A: Divide by $99$ to get 
$$\frac{49x}{99} + y = 101$$
We want $\frac{49x}{99} $ to be an integer, so let $x = 99$. This gives us $49 + y = 101$, or $y = 52$.
So $(x,y) = (99, 52) $ is a solution.
In addition, we could let $x = 2×99 = 198$ to get
$98 + y = 101$ or $y = 3$, so
$(x, y)  = (198, 3) $ is another solution.
Trivially, we also have $(x,y) = (0, 101) $ as a solution
These are the only solutions, as $\text {gcd} (49, 99) = 1$, so $x $ must be a multiple of $99$ for $\frac{49x}{99} $ to be an integer. But $x = 3×99$ does not work, as this will exceed $9999$
A: If you want to sell those products all in fall then you need $101$ of them.
If you want to sell those products all in spring then you need approximately $204$ of them.
If you want to make, as is natural, all combinations of selling both in fall and spring then calculate your spring for every fall equal to $0,1,2,3,...101$ 
I do not see an easy way to handle all cases all at once, because we are talking here about $102$ possible equations.
