The solution 1 transforms the equation into

1.x=2- 15t

2. y =-1+11t t t is an integer.

Since x is an integer, 7- 15y should be a multiple of 1 1. It is observed that x0=2 and y0=- 1 are a set of integer solutions of this equation, so the solution of the equation is

Solution 2 is first investigated at11x+15y =1,which can be easily obtained by observation.

1 1×(-4)+ 15×(3)= 1,

therefore

1 1×(-4×7)+ 15×(3×7)=7,

X0=-28，Y0 = 2 1。 Therefore,

1.x=-28- 15t

2. y = 21+11t t is an integer.

It can be seen that unconstrained binary linear indefinite equations usually have countless sets of integer solutions. Because of the difference of special solutions, the form of the solution of the same indefinite equation can be different, but the solutions contained are all the same. If the parameter t in the solution is properly substituted, it can be transformed into the same form.