Answer:
with
.
Explanation:
So we have:
and we are solving for
.
We want to get all terms that have x on one side and all terms without x on the opposing side. This is actually already done.
Now we are going to factor the side that has the x's. We are going to factor x out.

Now you have a number times x and you want x by itself. The inverse operation of multiplication is division. So we are going to divide both sides by
:

You could probably stop here but just in case....
can't be 0 because then you would have 10/0 which is not a number.
So if we solve
we can find out what we don't want and b to be:

Subtract b on both sides:

So we we don't want
and
to be the same value.