Let’s solve \[\frac{x}{2}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}\hspace{0.33em}{=}\hspace{0.33em}{42}\] two different ways.

The first way is to remove the “2” from the denominator and get the “*x*” by itself. We can do this by multiplying by “2” so the “2”s will cancel. But when you multiply a side of an equation by a number, you must multiply the whole side. So, multiplying both sides of our equation by 2 gives:

{\frac{x}{2}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}\hspace{0.33em}{=}\hspace{0.33em}{42}\hspace{0.33em}\Longrightarrow\hspace{0.33em}{2}{(}\frac{x}{2}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}{)}\hspace{0.33em}{=}\hspace{0.33em}{2}{(}{42}{)}}\\

{\Longrightarrow\hspace{0.33em}\frac{\rlap{/}{2}x}{\rlap{/}{2}}\hspace{0.33em}{+}\hspace{0.33em}{2}{(}{3}{x}{)}\hspace{0.33em}{=}\hspace{0.33em}{84}\hspace{0.33em}\Longrightarrow{x}\hspace{0.33em}{+}\hspace{0.33em}{6}{x}\hspace{0.33em}{=}\hspace{0.33em}{84}}

\end{array}\]

Now notice I used the distributive property and properties of fractions to do the above work. We are now left with two like terms on the left, which you remember, can be added together (remember there is an implied “1” in front of *x*. So now we have

\]

So what do we do here? To get *x* by itself, let’s divide both sides by 7:

Let’s try a different way that’s a bit quicker. Remember that \[\frac{1}{2}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{5}\hspace{0.33em}{\mathrm{and}}\hspace{0.33em}\frac{x}{2}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{2}{x}\]. So,

\[\frac{x}{2}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}\hspace{0.33em}{=}\hspace{0.33em}{42}\hspace{0.33em}\Longrightarrow\hspace{0.33em}{0}{.}{5}{x}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}\hspace{0.33em}{=}\hspace{0.33em}{42}\]

Now the two terms on the left are like terms and we can add them together to get 3.5*x* = 42. Just as before, we can remove the “3.5” in front of the *x* be dividing both side by 3.5. It doesn’t matter if that is a decimal number – the same rules of cancelling apply:

The same answer but a bit quicker since we didn’t have to work with the fraction first.