## Problem

The letters \(w\), \(x\), \(y\), and \(z\) each represent a different positive integer greater than \(3\). If we know that \[\dfrac{1}{w-3}=\dfrac{1}{x+1}=\dfrac{1}{y+2}=\dfrac{1}{z-2}\] then write \(w\), \(x\), \(y\), and \(z\) in order from the letter that represents the smallest integer to the letter that represents the largest integer.

## Solution

**Solution 1:**

Since the fractions are all equal and they all have a numerator of \(1\), that means that their denominators must all be equal. So \(w-3=x+1=y+2=z-2\).

Now let’s suppose that \(w=10\). Then \(w-3=10-3=7\).

So \(7=x+1=y+2=z-2\). We can make the following conclusions.

Since \(7=x+1\), that means \(x=7-1=6\).

Since \(7=y+2\), that means \(y=7-2=5\).

Since \(7=z-2\), that means \(z=7+2=9\).

So when \(w=10\), we have \(x=6\), \(y=5\), and \(z=9\). We can see that \(x\) is four less than \(w\), \(y\) is five less than \(w\), and \(z\) is one less than \(w\). So when we write these in order from smallest to largest, we get \(y\), \(x\), \(z\), \(w\).

**Solution 2:**

As with Solution 1, we notice that since the fractions are all equal and they all have a numerator of \(1\), that means that their denominators must all be equal.

So \(w-3=x+1=y+2=z-2\). Let’s add \(3\) to each expression.

\[\begin{array}{c c c c c c c} w-3 & = & x+1 & = & y+2 & = & z-2\\ \textcolor{red}{\downarrow {\scriptstyle +3}} & & \textcolor{red}{\downarrow {\scriptstyle +3}} & & \textcolor{red}{\downarrow {\scriptstyle +3}} & & \textcolor{red}{\downarrow {\scriptstyle +3}} \\ w & = & x+4 & = & y+5 & = & z+1 \end{array}\]

From this we can make the following conclusions.

Since \(w=z+1\), that means \(w\) is \(1\) more than \(z\), so \(w>z\).

Since \(z+1=x+4\), that means \(z\) is \(3\) more than \(x\), so \(z>x\).

Since \(x+4=y+5\), that means \(x\) is \(1\) more than \(y\), so \(x>y\).

So when we write these in order from smallest to largest, we get \(y\), \(x\), \(z\), \(w\).