nd all integers x such that 2x 1 mod 5 x 2 mod 6 and x 3 mod

nd all integers x such that 2x 1 (mod 5), x 2 (mod 6), and -x 3 (mod 7).

Solution

1.

2x=1 mod 5

We know, 2*3=6=1 mod 5

So we multiply the given equation by 3 and get

3*2x=3 mod 5

6x=x=3 mod 5

So, x=5m+3 ,where m is any integer

2.

x=2 mod 6

x=6m+2 where m is any integer

3.

-x=3 mod 7

x=-3 mod 7

x=7-3 mod 7

x=4 mod 7

x=7m+4 ,where m is any integer