Solve the congruence given 2x=1(mod 3), x=3(mod 5), x=2(mod 8)

Here is what I did.

2x=4(mod 3)

x=2(mod 3)

x=2+3n

2+3n=3(mod 5)

3n=1(mod 5)

3n=6(mod 5)

n=2(mod 5)

then n=2+5k

x=2+3(2+5k)

x=2+6+15k

x=8(mod 15)

Let x=8+15k

8+15k=2(mod 8)

15k=-6(mod 8)

15k=4(mod 8)

15k=60(mod 8)

k=4(mod 8)

k=4+8m

x=8(mod 15)

x=8+15(4+8m)

x=8+15(4+8m)

x=8+60+120m

x=68+120m

x=68(mod 120)

Did I get that right?