Quote:

Originally Posted by

**cmf0106**

We get an incorrect answer. This problem seems to stem from distributing the y in $\displaystyle y(x-1)$

Why in one example does distributing the variable over the parenthesis result in the correct answer, and in another problem distributing the variable over the parenthesis results in a wrong answer.

This comes down to using a bit of reasoning, and not simply mechanically applying techniques. So let's experiment a little. The claim is that distributing results in the correct answer in the example. Let's try it without and see if we can still get the correct answer:

$\displaystyle L=a+(n-1)d$

$\displaystyle L-a=(n-1)d$

$\displaystyle \frac{L-a}{\ d} =n-1$

$\displaystyle \frac{L-a}{\ d} +1 =n$

At this point, it appears different. Recall, however, that anything over itself is equal to 1, and since we have a fraction with a denominator of $\displaystyle d$ , I can substitute $\displaystyle \frac{d}{d}$ instead of the 1. Therefore:

$\displaystyle \frac{L-a}{\ d} + \frac{d}{d} =n$

$\displaystyle \frac{L-a+d}{\ d}=n$

So the same answer was obtained using a method that didn't require distributing at first. So the strategy is to ask yourself, "What is it that I'm after?" and then figure out if distributing will simplify the process, or make it worse. In the original problem, there was no need to distribute the $\displaystyle y$.

By the way, $\displaystyle xy=z+y$ can be rewritten as follows:

$\displaystyle xy=z+y$

$\displaystyle x=\frac{z+y}{y}$

$\displaystyle x=\frac{z}{y} + \frac{y}{y}$

$\displaystyle x=\frac{z}{y} + 1$

So $\displaystyle xy=z+y$ was only a few steps away from showing that $\displaystyle x =$ something, as the question prompt required.