Are you sure you have written this correctly? The same xcannotbe equivalent to both 1 and 3 modulo 4. That would be like solving a set of simultaneous equations that included both "x= 1" and "x= 3"!

I will, however, show what I would do if it were, say, , and where I have changed the "mod" for one equation.

Saying that means that x= 7+ 9i for any integer i.

Now put that into the next equation: becomes (I have reduced 7 and 9 modulo 5) which is the same as . Since , That says that so i= 4+ 5j for any integer j. Putting that back into the formula for x, .

Now the third equation, becomes which is the same as . And that, in turn gives for any integer m. Putting that into the formula for x, we get as our answer.

To check, note that 133/9= 14+ 7/9 so x= 133 does satisfy . 133/5= 26+ 3/5 so x= 133 does satisfy and, finally, 133/4= 33+ 1/4 so x= 133 does satisfy .

The fact that adding any multiple of 180 to 133 will still give a solution is clear because 180= 9(5)(4) is congruent to 0 mod 9, 5, and 4.