Since we just looked at a complicated rational inequality, let’s look at some simpler rational *equations*, first a linear equation with fractions, and then truly rational equations, in which the variable(s) appear in the denominator. This discussion dealt with a common confusion I’ve seen in students.

## The problem

The question came from Fairooz in 2017:

Multiplication of Top and Bottom vs. Multiplication Across Left and Right Hi there. This is quite an easy question, so I'm probably missing a very basic concept here. Given x/2 + y/5 = 1, we would usually make a common denominator of 10, right? What I don't understand is that, when we multiply the first denominator by 5 and the second denominator by 2 and do the same with the numerators, how come we don't multiply the 1 by 10, as well? 5x/10 + 2y/10 = 10Aren't we supposed to multiply everythingin an equation if we've done that to one of the terms?

Fairooz has been taught to solve $$\frac{x}{2} + \frac{y}{5} = 1$$ by rewriting it as $$\frac{5x}{10} + \frac{2y}{10} = 1,$$ multiplying numerator and denominator of each fraction to produce a common denominator of 10. He wonders why the right-hand side is left unchanged. After all, whatever you do to one side, you have to do to the other, right?

### Combining fractions first

I answered:

You are confusing two procedures (as many students do) by not focusing on WHAT you are actually doing, and WHY you do what you do. There are two main ways to handle this kind of equation.One approachis what you suggest, tochange each term to use a common denominator. You rewrote x/2 as 5x/10, and y/5 as 2y/10, SO THAT YOU CAN ADD THEM. Note that you didn't multiply the left side by 10; you just changed the denominators by multiplying both the top and bottom of each fraction. Each term still has the SAME VALUE. Therefore, you can't change the value on the other side, either; it must also retain the same value, 1.

He has not multiplied the left-hand side by 10, but merely rewritten it as an **equivalent expression**, leaving the value unchanged. The next step will be to add the fractions: $$\frac{5x + 2y}{10} = 1.$$

If you DID want to change the right side,doing the same thingwould mean just writing it, too, with a DENOMINATOR of 10: 5x/10 + 2y/10 = 10/10 That would be a valid thing to do, and perhaps even useful. Both sides still have the same value, but now they can be combined: (5x + 2y)/10 = 10/10 There are several ways you could go from here; one is just to multiply both sides by 10, canceling the denominators: 5x + 2y = 10 (We didn't really have to write 10/10, and I rarely would; multiplying by 10 changes the 1 to 10 anyway. I did this just to show what youmightdo on the right side if you felt compelled to.)

When we talk about “doing the same thing to both sides”, we mean **applying some operation to the value on each side** (for example, multiplying both sides by 10). It is not necessary to rewrite both sides of an equation in the same way, as that doesn’t change their values. The important thing is that the values must remain equal (for any value of the variable).

Note, by the way, that we weren’t told what the goal is. If this were an equation with only one variable, we would probably be trying to solve for that variable. Here, probably the goal is to rewrite the equation in some particular form (such slope-intercept form, which means solving for *y*) in order to graph it.

### Clearing fractions first

The other primary methodis to MULTIPLY both sides by 10, which willCLEAR FRACTIONS. Here, we are no longer trying to keep the values the same (which required multiplying both numerator and denominator by the same number), but are working with the entire equation, keeping it equivalent bymultiplying both sides by the same number. And we have a different goal: not just making it possible to add the terms, but eliminating fractions entirely. So we don't put a 10 in the denominator. Our goal is to be able to cancel the denominators with the 10. It looks like this (writing out more steps than we usually do once we are familiar with it): 10(x/2 + y/5) = 10(1) 10*x/2 + 10*y/5 = 10 5x + 2y = 10 This is the same result we got before, without having to do so much with fractions. From here, of course, you can just solve for y as usual, with no fractions to complicate things.

In normal printed form, the work looks like this:

$$10\cdot\frac{x}{2} + 10\cdot\frac{y}{5} = 10\cdot1$$

$$5x + 2y = 10$$

This is the same result as before, obtained more quickly and without having to add fractions. Now we can do whatever we set out to do, with a simpler equation. If we’re solving for *y*, we can subtract 5*x* and then divide by 2:

$$2y = 10 – 5x$$

$$y = \frac{10 – 5x}{2} = \frac{10}{2} – \frac{5x}{2} = 5 – \frac{5}{2}x$$

I find that many students, having been taught one of these ways, confuse that way with the other, ending up with a hybrid result that is invalid. The first way has them face the fractions immediately, while the second tries to eliminate fractions as soon as possible. Both focus on the fractions as the main feature of the equation.

### Just do it!

But there’s a **third way**, which books often don’t point out: Rather than simplify things to avoid fractions, we could just grit our teeth and do the same work we’d do if there were no fractions:

Now, you don't really have to do any of that. You could just do all the usual work, keeping the fractions. It's just a little more error-prone for some students: x/2 + y/5 = 1 y/5 = 1 - x/2 5*y/5 = 5(1 - x/2) y = 5 - 5x/2 And we're done! In this case, clearing fractions is not really all that helpful. We usually teach it just becauseso many students fear fractions, and aremore likely to make mistakesmanipulating them. I hope that helps. The key is not to just mechanically multiply by something because you've done that in other problems, but to pay attention to what you are doing.If you multiply one sideby something, then by all means -- as you said --multiply the other side, too. Butif you only change the formof one side, then do only things on the other side thatwill not change its value. And don't think that you are multiplying when you are not.

Again, here is the work this way:

$$\frac{x}{2} + \frac{y}{5} = 1$$

$$\frac{y}{5} = 1 – \frac{x}{2}$$

$$5\cdot\frac{y}{5} = 5\cdot1 – 5\cdot\frac{x}{2}$$

$$y = 5 – \frac{5}{2}x$$

## More examples

Fairooz evidently had just given us a simple example to illustrate the issue, and now gave us the “real” problem:

Ahh, I understand. Thank you. Just to clarify this method with another question -- the main one, which confused me in the first place ... Given y = (c - bc)/(b + 1) + c We know that the c is equivalent to c/1. Before, when I thought about multiplying the denominator and numerator by (b + 1), I believed I had to do that to the left side (y), too. Now I know toleave the y as it is, correct? One last question. Say I were given Y/1 = 2/X + 3. Would I multiply the 3 (=3/1) with X (both numerator and denominator) but then also multiply the Y/1 with X as well (numerator and denominator) so that they all have acommon denominator?

Both of these examples are rational equations with variables in the denominator; but it is still unclear what the goal is, as both have *y* alone on the left.

### Simplifying a solution

The first problem is an equation that is already solved for *y*. I replied, about leaving *y* as it is:

Right. You don't NEED to do anything on the left. You already have y alone, so you don't want to change it at all. What you are doing on the right is NOT multiplying; there, your goal is just tosimplify: y = (c - bc)/(b + 1) + c y = (c - bc)/(b + 1) + c(b + 1)/(b + 1) y = (c - bc)/(b + 1) + (bc + c)/(b + 1) y = (c - bc + bc + c)/(b + 1) y = (2c)/(b + 1) In this case, you wouldn't want to actually multiply by the LCD (b + 1), because that would force you to change the left side. But you could do that if you didn't like fractions: y = (c - bc)/(b + 1) + c y(b + 1) = (c - bc) + c(b + 1) y(b + 1) = c - bc + bc + c y(b + 1) = 2c y = (2c)/(b + 1)

In fact, there is really no need to rewrite the right-hand side at all, in order to make this a valid solution to a problem that said to solve for *y*. But it did turn out to be a significant simplification, so it would be worth doing as part of a larger problem. Here, again, is the work the first way (combining fractions):

$$y = \frac{c – bc}{b + 1} + c$$

$$y = \frac{c – bc}{b + 1} + \frac{c(b + 1)}{b + 1} = \frac{c – bc}{b + 1} + \frac{bc + c}{b + 1} = \frac{c – bc + bc + c}{b + 1} = \frac{2c}{b + 1}$$

And here is the second way (clearing fractions):

$$y = \frac{c – bc}{b + 1} + c$$

$$y(b + 1) = \frac{c – bc}{b + 1}(b + 1) + c(b + 1)$$

$$y(b + 1) = c – bc + bc + c$$

$$y(b + 1) = 2c$$

$$y = \frac{2c}{b + 1}$$

Note that I kept my eye on the goal, to solve for *y*, so I didn’t distribute the left-hand side. In effect, I “unsolved” and then solved, because the unsolved form was easier to work with.

### Simplifying again

How about the second question, about making a common denominator throughout the equation \(\displaystyle\frac{y}{1} = \frac{2}{x} + 3\)? This is really just the same issue as the other, as the left-hand side is really just *y*.

No, I wouldn't quite do that; there is no reason here to rewrite the left side. Since y is alone there, you only need tosimplify the right side. I would do this: y = 2/x + 3 y = 2/x + 3x/x y = (2 + 3x)/x

Here I just simplified:

$$y = \frac{2}{x} + 3 = \frac{2}{x} + \frac{3x}{x} = \frac{2 + 3x}{x}$$

There is no need to be working with the entire equation.

If I wanted toavoid fractions, however, I might be willing to change the left side. Since x is the only denominator you have (that is not 1), I would multiply the entire left side and the entire right side by x -- that is, multiply every term: y = 2/x + 3 xy = 2/x*x + 3x xy = 2 + 3x Then I could solve for y by dividing by x: y = (2 + 3x)/x Here, it was already solved; but I temporarily "unsolved" it in order to avoid what might be risky work. Now, if the goal in these problems was to solve for x rather than for y, everything would change! Remember, keep the goal in mind, rather than blindly following some routine.

Here we had:

$$y = \frac{2}{x} + 3$$

$$xy = \frac{2}{x}\cdot x + 3x$$

$$xy = 2 + 3x$$

$$y = \frac{2 + 3x}{x}$$

Fairooz was satisfied:

I understand. Thank you.

## Solving for *x*

Let’s look at that last thing I said, which would change everything: We’ll try solving $$y = \frac{2}{x} + 3$$ for *x* in terms of *y* this time. Now we’re solving a genuine, full-fledged rational equation. I’ll try several methods.

### Clearing fractions first

The second method I discussed above is what textbooks in my experience generally teach as “the” way to solve a rational equation.

To get *x* out of the denominator (which we have to do eventually, no matter what method we follow), I’d multiply every term (both sides) by *x*:

$$y = \frac{2}{x} + 3$$

$$xy = \frac{2}{x}\cdot x + 3x$$

$$xy = 2 + 3x$$

Now I want to solve for *x*, so I get all terms containing *x* on the same side, and then factor it out:

$$xy – 3x= 2$$

$$x(y – 3)= 2$$

$$x = \frac{2}{y – 3}$$

The factoring step (equivalent to combining like terms) is the one students most often miss when they first see this kind of problem!

### Combining fractions first

This time, I’ll combine terms on the right-hand side first; many students reflexively do this when faced with fractions (having been conditioned to simplify any expression they see), so rather than forbid this, I tell them how to do it *right* if they find themselves doing it. (Sometimes, in fact, this method turns out to require less writing, so it is not a bad method at all.)

We want to only simplify:

$$y = \frac{2}{x} + 3 = \frac{2}{x} + \frac{3x}{x} = \frac{2 + 3x}{x}$$

Now we still have to get the *x* out of the denominator, by multiplying both sides:

$$xy = x\cdot\frac{2 + 3x}{x}$$

$$xy = 2 + 3x$$

The last steps will be the same as before.

### Just do it

As in the examples above, we don’t *have* to focus on the fractions, even though a rational equation is all about fractions. This will be a surprise to some! I can just ignore the fact that *x* is on the bottom, and solve as if it were not, just isolating the term by subtracting 3:

$$y = \frac{2}{x} + 3$$

$$y – 3= \frac{2}{x}$$

Now I have a single fraction on the right-hand side, and I don’t want *x* to stay in the denominator. What can I do? I can just **take the reciprocal of each side**! This isn’t often taught explicitly as one of the “things you can do to both sides”, but if two values are equal, clearly their reciprocals are, too:

$$\frac{1}{y – 3} = \frac{x}{2}$$

And now we just have to multiply by 2:

$$\frac{2}{y – 3} = x$$

Pick the way that fits your own thinking best!

Math is a toolbox, and which tool you use at a given moment depends on what you see that needs doing. If you see a screw, use your screwdriver. (And if it turns out that you can’t easily get that screw out without doing something else first, just back up and start again. You can’t break anything.)