### Is it possible to solve for 2 unknown variables related by a single linear equation when it is known that x and y will be positive integers?

I heavily edited my post, so I am resubmitting it, and deleting the old one, to be sure you see the finalized version:

You can solve the equation for x and you can solve it for y, but you cannot solve each equation individually for both x and y as integers.

To solve the first equation for X:

4x + 25y = 1200

4x = -25y + 1200

x = -6.25y  + 300

To solve the first equation for y:

4x + 25y = 1200

25y = -4x + 1200

y = -.16x + 48

To solve the second equation for x:

80x + 28.96y = 50,930.24

80x = -28.96y + 50,930.24

x = -0.362y + 636.625

To solve the second equation for y:

80x + 28.96y = 50,930.24

28.96y = -80x + 50,930.24

y = -19.220x + 1758.640

If you want to find integer values, you would need to solve a "system of equations" which is simply a second equation that also uses the same variables, in the case of your example x and/or y. You simply solve one of the equations for one of the variables, then take that answer and plug it into the second equation. EDIT: I am assuming both of the equations you listed above are part of a system of equations, in other words that the x in the first equation equals the x in the second, and the same for the y variable. I did a little extra work by solving both equations for both variables, so it will be very easy to plug in the variables from the first equation into the second (or vice versa).

Since we have one solution for x

x = -6.25y  + 300

and another solution for x,

x = -0.362y + 636.625

if

x = x

then

-6.25y  + 300 = -0.362y + 636.625

which we can use to solve for y

-6.25y  + 300 = -0.362y + 636.625

-5.888y +300 = 636.625

-5.888y = 336.625

y = 57.171

note: I am rounding off at the thousandths place, this is not a precise answer

now we can plug that value of y into this equation

y = -.16x + 48

to get this:

57.171 = -.16x + 48

9.171 = -.16x

-x = 57.321

x = -57.321