How do you solve #6=\frac{1}{3}(3)+b#?
1 Answer
Sep 16, 2017
Explanation:
#"note "1/3(3)=1/cancel(3)^1xxcancel(3)^1=1#
#rArr6=1+blarr" is the equation to be solved"#
#"subtract 1 from both sides"#
#6-1=cancel(1)cancel(-1)+b#
#rArrb=5" is the solution"#
