four times the larger of two integers is 94 more than three times the smaller

Question

we have to write and solve the equation

Answer 1

Let the integers be a and b with a>b. 4a=3b+94. To solve this for integer solutions we know that 3b+94 must be divisible by 4, so divide 3b+94 by 4 and note the remainders only: 3b+2. This number must be divisible by 4. So b=2, 6, 10, ..., 4n+2, where n is an integer. If b=2 a=(6+94)/4=25; if b=6 a=28, if b=10 a=31, etc.

The solution is b=4n+2 and a=(12n+100)/4=3n+25, where n is an integer (positive or negative). 3n+25>4n+2 so n<23, because a is bigger than b by definition.The maximum value of b is 90 (when n=22) and a=91. There's no minimum value.