Suppose we have to work with approximate numbers up to the second order, i.e., with two decimal places. Now consider $a=5/3=1.\bar6\approx1.67$.
Now we have to solve $$b=1+a\tag1$$ so we have to insert $a\approx1.67$ in that equation.
However, I have doubts about how an approximately could be written:
$$b\boldsymbol{\color{red}=}1+1.67\implies b\approx2.67\tag2$$
$$b\boldsymbol{\color{red}\approx}1+1.67\implies b\approx2.67.\tag3$$
Note that I have doubts when the value is already entered, not in $(1)$.
What would you do?
Thanks!