Henry,
If you have one equation with two unknowns then there may be many solutions. In your example you give the solution x = 9 and y = 10 but there are other solutions. For example x = 3 and y = 1, or x = 5 and y = 4. In fact I'll try to convince you that there are infinitely many integer solutions.
In your equation
3x = 2y + 7
the left side is divisible by 3 so the right side must be divisible by 3. If 7 is divided by 3 it leaves a remainder of 1 so, since the right side is divisible by 3, 2y divided by 3 must have a remainder of 2. That means that y divided by 2 has a remainder of 1. In other words, y must be one more than a multiple of 3, that is y can be 1, 4, 7, 10, 13,...
For each such choice of y, both sides of the equation are divisible by 3 and hence if you divide both sides by 3 you get an integer value for x, and this value of x, together with the chosen value of y is a solution to the equation.
For example if y = 52 (1 more than a multiple of 3) then 2y + 7 = 111 and ^{111}/_{3} = 37 so x = 37 and y = 52 is a solution.
Penny
