3(a+1)-5=3a-2
"A" could be any real number.
Start with 3(a+1)-5=3a-2.
3(a+1)−5=3a−2
3a+3+−5=3a+−2
3a+−2=3a−2
−2=−2
0=0
Getting 0=0 means that any real numbers will satisfy the value of a.