Many Ways to Solve a Proportion

Last week we looked at a set of special rules for working with proportions, which have been largely replaced by the more general “tool” of algebra (the “Swiss army knife” of problem solving, which can do the job of many specialized tools), though the latter can still be useful. We still find that many students memorize a related special tool, cross-multiplication, and miss out on the great flexibility algebra provides. Here we’ll focus on the variety of ways you can solve these.


I’ll start with a question from 2002:

Why Does Cross Multiplication Work?

I am a sixth grade teacher in PA. We just started working on proportions in math and a few students want to know why we cross multiply to solve a proportion like: 

  3/15 = n/30 

where we cross multiply 3 x 30= 90 then this means that 15n = 90. To solve this, I divide 90 by 15 and get 6 = n.  

Why does the rule a/b = c/d also mean ad = bc?  Why can we solve proportions by cross multiplying?  And I apologize if you have answered this before.  Thank you so much!

Cross multiplication means multiplying diagonally opposite terms in the proportion, like this in Kimberly’s example:

$$\require{cancel}\frac{3}{15}\xcancel{=}\frac{n}{30}$$ $$3\times 30=n\times 15$$ $$90=15n$$ $$n=\frac{90}{15}=6$$

We’ll see other ways to solve this soon; but the immediate question is, why can we do that?

I answered:

Hi, Kimberly. 

There was a time when this rule was considered a wonderful piece of magic, called the "rule of three"; students just learned to do it without expecting to understand why it worked. That makes me sad!

The Rule of Three, an old version of cross-multiplication, will be our topic next week. Math isn’t just a set of magical incantations to recite because an authority tells you to do so; we need reasons for what we do. When we know how and why it works, we are more likely to remember it correctly and to apply it appropriately.

The students in this class are likely learning just the beginnings of algebra, such as the use of a variable in the problem, and using letters for all four terms of a proportion. I wanted to explain in a way that would both show a little of the power of algebra, but also not overwhelm them – though the main target of my answer was the teacher!

With algebra, it's almost obvious, and certainly not something special. I'll try to express this in a way that students who don't know algebra (or don't realize how much of it they have already seen) can follow.

Remember that if you have two equal quantities and multiply them by the same amount, the products will again be equal. So if we multiply the fractions a/b and c/d by b, the results are equal:

     a         c
    --- * b = --- * b
     b         d

which can be written as

    a = ----

Now we can multiply both fractions by d:

    ad = ---- * d

which, of course, means

    ad = bc

To make this more concrete, consider our proportion $$\frac{3}{15}=\frac{n}{30}.$$ We can first multiply both sides by the first denominator, 15, canceling to get $$\cancel{15}\times\frac{3}{\cancel{15}}=15\times\frac{n}{30}$$ $$3=\frac{15n}{30},$$ and then multiply both sides by the second denominator (without doing the arithmetic, so we can see what’s happening), to get $$30\times 3=\cancel{30}\times\frac{15n}{\cancel{30}}$$ $$30\times 3=15n$$ Cross-multiplication is just a shortcut for multiplying the equation by both denominators.

But if you look closely at what I just did, you may notice that we could actually save effort by multiplying by only one denominator, since the 15 just has to be divided back out. This gives us the first of several alternative methods we’ll see:

Alternative 1: Multiply by one denominator

I personally prefer not to cross multiply, but just to multiply by whichever denominator helps. In your example, 3/15 = n/30, I would just multiply both sides by 30 and get 

  n = 30*3/15 

    = 30/15*3 

    = 2*3 

    = 6.

Here I saw that I could divide 30 by 15 before doing the multiplication (the cancel-first strategy that was discussed here), and never have to work with large numbers.

Kimberly replied,

Dear Dr. Peterson,

I would really like to thank you for responding to my question.  It is really going to be a big help on Monday when I go into school. 
I have used this web site for other math activities and problems. It is nice to know that there is such a reliable resource for teachers and students of mathematics.

Teachers are always among our favorite users!

Means and extremes

Here is a question, from 2001, about an older way to talk about cross-multiplication:

Means and Extremes

My class is currently learning about proportions, and has asked the question, "Why are certain parts of the proportion called the means and the other parts called the extremes?"

Can you assist us in clarifying why these particular terms are used and how they apply to the ratios in a proportion?

This relates to an old way to describe cross-multiplication: “the product of the means equals the product of the extremes“. This particularly related to the old notation for proportions, using “horizontal” ratio notation, in which the last example might have been written as “3:15::n:30″, with “::” playing the role of the equal sign, and read as “3 is to 15 as n is to 30″.

I explained:

Hi, Jill.

As you know, the "means" are the inside terms, b and c, and the "extremes" are the outside terms, a and d, in the proportion we might write in any of these ways:

    a : b :: c : d

    a : b = c : d

    a/b = c/d

     a     c
    --- = ---
     b     d

Whichever way we write it, except the last, b and c are on the inside, and a and d are on the outside.

When written in the vertical form, it is harder to see an inside and an outside (except when you read it aloud), and easier to see them as forming a cross.

The word "mean" comes through French from Latin "medius," meaning "middle." (It is used in several ways in math, all related to the middle.)

The word "extreme" comes from Latin "extremus," the superlative form of "exterus," meaning "outside"; so it means "outermost."

Now it should make sense!

You don’t need to know the Latin, of course; but if you use these terms, you need to know the ideas of “middle” and “outside”. We also use the word “extreme” to refer to the largest or smallest value of something – the farthest you can get from the “middle”.

Using this formulation, our proportion looks like $$\underbrace{3:\underbrace{15::n}:30},$$ where the means are 15 and n, and the extremes are 3 and 30. Setting the means equal to the extremes, we have, as before, $$15n=3\cdot30.$$

More alternatives

For yet more flexibility, consider this question from 2014, suggesting an alternative method:

Proportion Solution Option Profusion

The standard algorithm to solve a proportion is to cross multiply and then isolate the variable.

Wouldn't it be easier to scale up or down accordingly?

For example, with the standard algorithm for solving a proportion,

       9/5 = n/20
      9*20 = 5*n
       180 = 5*n
     180/5 = 5*n/5
        36 = n
My easier alternative exploits a little proportional reasoning at the outset. I know the denominator, when multiplied by 4, gives me 20. So why don't I just multiply the other numerator by 4 as well to get 36 and make it an equivalent ratio? 

       9/5 = n/20
   9*4/5*4 = 36/20

Saves me a lot of steps, doesn't it?

Why would I want to use the standard algorithm when it takes longer?

Vladimir recognizes that solving this proportion is equivalent to asking, what fraction with 20 in the denominator is equivalent to \(\frac{9}{20}\)? To do that, you just multiply the numerator by the same number you have to multiply the denominator by. It’s a good method … for this particular problem!

I congratulated him, and then took it further:

Hi, Vladimir.

There are MANY different ways to solve a proportion, each appropriate to a different kind of problem. What troubles me is when students have only been taught one algorithm, and don't even think there is any other way to think about it. In fact, they cross-multiply by reflex when they see a proportion. Clearly, you are thinking for yourself, which is the only way to really learn math. Wonderful!

A key to good problem solving is to see special features of the problem at hand that offer shorter and better ways than a routine approach.

You're right that this "standard algorithm" is often far from the easiest. The reason it is taught, I imagine, is that it may be the only method that works for any proportion without having to do any special thinking. So if you are going to learn only one way (or, rather, *remember* only one way), it might as well be that one.

(Actually, that's wrong. You'll see below a much easier method that is just as general. Over-reliance on the standard algorithm probably has more to do with tradition, or ease of explanation to children.)

We’ll see next week that the tradition is very old indeed. I started by examining when Vladimir’s method (which I’ll call alternative 2) works best:

Alternative 2: Make an equivalent fraction

Your method works very well when one denominator (or numerator) is a multiple of the other. It also happens to be the same way you find equivalent fractions before you learn any algebra:

       --- = ---
        5     20

It wouldn't work so well if one were a decimal multiple of the other, though you could still do the equivalent:

       9.1     x
       --- = -----
       5.3   23.85

   9.1*4.5     x
   ------- = -----
   5.3*4.5   23.85

         x = 9.1*4.5
           = 40.95

In other words, his method is essentially what students are taught to do when they need to make an equivalent fraction using a common denominator (for addition of fractions, say), which is a multiple of the existing denominator. We just fill in the blank by multiplying the numerator by the same quantity as the denominator. If those numbers aren’t so compatible, the method still works, but is considerably more awkward. Here, I had to divide 23.85 by 5.3 to get 4.5 – and it could have been a lot worse! I made this problem to be relatively easy to do this way.

Note that this method still requires a multiplication and a division; but we are doing the division first because (in his example) it is easy to see.

Alternative 3: Multiply by the divisor

Another method is to multiply both sides by the divisor of x, and then simplify the resulting fraction (or, if you want a decimal, carry out the multiplication and division on the other side to get the answer):

        9     x
       --- = ---
        5     20

      20*9   20*x
      ---- = ----
        5     20

       4*9 = x

         x = 36

This is really the same as alternative 1 above, but viewed algebraically. As before, I chose to cancel the 20 and the 5 before multiplying; otherwise, this would have been the same work as cross-multiplying, namely multiplying \(20\times9\) and dividing by 5. To put it another way, like Vladimir, I recognized an easy division and did it first, rather than the multiplication.

Aside: What if the variable is in the denominator?

None of my examples here will involve problems where the unknown is on the bottom, like this: $$\frac{5}{9}=\frac{20}{x}$$ This is a case where students will often fall back on cross-multiplying: $$5x=9\cdot20=180$$ $$x=\frac{180}{5}=36$$

But there is a simple way, again: Take the reciprocal of each side. (After all, if \(x=y\), then \(\frac{1}{x}=\frac{1}{y}\).) So in my example, we would just flip everything, to get $$\frac{9}{5}=\frac{x}{20}$$

This is, of course, the same problem we just solved the easy way. So a quick flip turns the awkward problem into a simpler one.

This is particularly useful when we use the Law of Sines in trigonometry, which can be written in two forms: $$\frac{a}{\sin A}=\frac{b}{\sin B}$$ and $$\frac{\sin A}{a}=\frac{\sin B}{b}$$ Rather than always use one form, I use the first (“side over sine”) when the unknown is a side length, and the second (“sine over side”) when the unknown is an angle, so that the unknown is always on top. I recently showed this in tutoring a student face to face, and it saved her from a major struggle.

Alternative 4: Clear fractions using the LCD

Or, you can think of the proportion as an equation involving fractions, and eliminate fractions by multiplying by the least common denominator (LCD). This method is equivalent to the last method for your example, so I'll use a different one:

        9     x
       --- = ---
        15    25

        9        x
    75*--- = 75*---
        15       25

       5*9 = 3x

        45 = 3x

      45/3 = x

         x = 15

That is equivalent to cross-multiplying when the LCD is the product of the denominators, but involves smaller numbers in this case.

Here, the Least Common Multiple of 15 and 25 is 75. Using this keeps all numbers as small as possible; but that comes at the cost of having to find the LCD, which isn’t always easy. On the other hand, as usual in simplifying fractions, you don’t need to worry about finding the best number to use, but can approach it step by step, perhaps first multiplying by one of the denominators, and then seeing what is left. (If you do that, though, you’ll often find yourself really using another of our methods!)

Aside: Simplify first

Note in this example that I could have made the work easier by first simplifying the left side:

        9     x
       --- = ---
        15    25

        3     x
       --- = ---
        5     25

        3        x
    25*--- = 25*---
        5        25

       5*3 = x

         x = 15

School exercises using nice numbers often benefit from doing this; you may have noticed that it applies to our first example as well: $$\frac{3}{15}=\frac{n}{30}$$ becomes $$\frac{1}{5}=\frac{n}{30}$$ so that multiplying by 30 leads directly to the answer, $$n=\frac{30}{5}=6$$

This is less common in real life.

Check by simplifying

If you want to check whether a proportion is true, rather than solve one, then there are even more methods. Rather than cross-multiply, for example, if the numbers are all integers, then you can simplify each fraction and see if they come out the same.

Cross-multiplication is commonly first introduced as a way to check whether two fractions are equal, or equivalently whether an equation is a [true] proportion. For example, you might be given $$\frac{45}{67}=\frac{86}{128}$$ and asked whether it is a proportion (that is, whether it is true). The cross-products are \(45\times128=5760\) and \(67\times86=5762\); since these are close but not equal, you would say that the numbers are not [exactly] in proportion, but nearly so. With some effort, you could instead determine that both are in lowest terms, so the fractions are not equal. On the other hand, given $$\frac{45}{78}=\frac{75}{130}$$ you would find that you can simplify both to \(\frac{15}{26}\), so they are equivalent fractions and the equation is true. Is this easy? That depends on the numbers, and whether you have a calculator handy!

Check using a common denominator

Another similar approach would go in the opposite direction: Rather than try to reduce both fractions (hopefully to the same thing), we might write both fractions using the same denominator (hopefully with the same numerator, too). If we use the LCD to do this, it can take some work:  Given $$\frac{45}{78}=\frac{75}{130}$$ we would have to see that the least common multiple of \(78=2\times3\times13\) and \(130=2\times5\times13\) is \(2\times3\times5\times13=390\), so $$\frac{45}{78}=\frac{45}{78}\times\frac{5}{5}=\frac{225}{390}$$ while $$\frac{75}{130}=\frac{75}{130}\times\frac{3}{3}=\frac{225}{390}$$ and they are equal.

But we don’t need to use the Least Common Denominator; we can use what I call the “Obvious Common Denominator”, which is just the product of the two denominators. That gives bigger numbers, but saves thought. Here is the work: $$\frac{45}{78}=\frac{45}{78}\times\frac{130}{130}=\frac{5,850}{10,140}$$ while $$\frac{75}{130}=\frac{75}{130}\times\frac{78}{78}=\frac{5,850}{10,140}$$ There are, of course, the same. And do you notice that the new numerators are the cross-products? This is another way to see why cross-multiplication works!


So the answer to your question is: use any method you see that will be reasonably efficient for the particular problem you are working on. Never stick to one algorithm for ANYTHING! People who don't want to think will stick to what they know (and maybe they'll be perfectly happy that way, if they don't have to solve too many proportions in their lives). On occasion, you will want to cross-multiply; but you'll do far better if you think first about whether it's the best way for the circumstances.

So keep up the good thinking!

By thinking first, you can often save a lot of work; but also, you are sharpening you skills and building “number sense”, which can be useful in other areas of math. This is not just about proportions!

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