Question 1186833
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Let's take the equation shown by the other tutor and go through the process of solving it to answer the question.<br>
{{{(2/9)((3/5)b-30)-38=54}}}<br>
{{{(2/9)((3/5)b-30)=54+38=92}}}<br>
{{{((3/5)b-30)=(9/2)(92)=9(46)=414}}}<br>
{{{(3/5)b=414+30=444}}}<br>
{{{b=(5/3)(444)=5(148)=740}}}<br>
Now let's work the problem backwards; we will see that we do the same calculations that we made in the solution above.<br>
He finished with 54 books.
Before that, he gave away 38 books; so before that he had 54+38=92 books.
Before that, he sold 7/9 of the books he had, so what he had left was 2/9 of his books.  So the number he had before he sold 7/9 of his books was 92*(9/2)=414.
Before that, he gave away 30 books; so before that he had 414+30=444 books.
Before that, he sold 2/5 of his books, so what he had left was 3/5 of his books.  So the number he had before he sold 2/5 of his books was 444*(5/3)=740.<br>
We do the same calculations using either method -- so why show two different methods?<br>
Imagine a similar problem where, instead of selling some books and giving away some books twice, he does that four or five times.  Solving the problem by writing and solving an equation similar to the one the other tutor used would get VERY messy; but working the problem backwards one step at a time would be much easier.<br>
So practice solving problems like this by both methods, knowing that for more complicated problems the solution will be easier if you work the problem backwards.<br>