# Solve for X issue

• Apr 1st 2013, 08:50 AM
cmf0106
Solve for X issue
Problem: If $\displaystyle y(x-1)=z$ then x = ?

According to the solution to the problem$\displaystyle x = \frac{z}{y} + 1$

I am having a hard time finding this solution the first thing I do is distribute the y.
Thus$\displaystyle y(x-1)=z$ then becomes $\displaystyle xy-y = z$

But that step doesn't seem correct, would someone please guide me through the problem?

Thanks
• Apr 1st 2013, 09:08 AM
Plato
Re: Solve for X issue
Quote:

Originally Posted by cmf0106
Problem: If $\displaystyle y(x-1)=z$ then x = ?

According to the solution to the problem$\displaystyle x = \frac{z}{y} + 1$

$\displaystyle y(x-1)=z\\(x-1)=\frac{z}{y}\\x=\frac{z}{y}+1$
• Apr 1st 2013, 09:34 AM
cmf0106
Re: Solve for X issue
@Plato, thanks the solution. However when I compare it to another equation I am confused.

For instance in this equation

$\displaystyle L = a + (n-1)d$ solve for $\displaystyle n$
$\displaystyle L = a + nd - d$ here we distribute the $\displaystyle d$
... few steps further and we get
$\displaystyle \frac{L-a+d}{d}=n$

Compared to the original problem

$\displaystyle y(x-1)=z$ solve for $\displaystyle x$
here if we distribute the y $\displaystyle xy-y=z$
$\displaystyle xy=z+y$
divide each side by y

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.
• Apr 1st 2013, 08:02 PM
semouey161
Re: Solve for X issue
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.