Fractions Within Fractions: Work Your Way OUt

[RobbieM518]

Hi,

I have a question about the final example provided for this section. They give you a "Like this:" example problem for the "If you.." "then you..." at the end but they actually go about solving it. I am unsure if I have solved it correctly. Has anyone solved this problem. The way that I've worked it out I'm left with -1.

The problem is 1 in the numerator all over the sum y + (1 over (2 - (3 over y)

Please help me figure out if I've done it right.


Thanks,
Robbie

[Sage Pearce-Higgins]

When you post in the future, please be clear about the source and quote the whole problem. This is taken from Chapter 4 of the 6th edition of Foundation of GMAT Math.

The example I think you're mentioning is: simplify 1 / ( y + 1 / (2 - 3/y))
Following the logic of 'work your way out', start by simplifying (2 - 3/y). We can find the common denominator of y, and simplify this to (2y/y - 3/y), so it becomes (2y - 3)/y. Now we're in a good position to sort out 1 / ((2y - 3)/y), which equals y / (2y - 3).

So, our original fraction is now 1 / ( y + y / (2y -3)). Again, find the common denominator of the fraction, which is (2y - 3), so that y + y / (2y - 3) equals (y (2y - 3)) + y / (2y - 3), which equals (y + y(2y - 3)) / (2y - 3). Let's simplify the top part to give ourselves (y + 2y^2 - 3y) / (2y - 3), which equals (2y^2 - 2y) / (2y - 3).

Putting that back in our original fraction gives us 1 / ((2y^2 - 2y) / (2y - 3)), which we can again simplify by flipping to give (2y - 3) / (2y^2 - 2y) as a final answer. Phew!

If that doesn't satisfy you, then please write out the steps that you took so that I can see where we differ.