Solving Rational Equations: Introduction - Purplemath

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Search Solving Rational Equations

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While adding and subtracting rational expressions can be a royal pain, solving rational equations is generally simpler, even if rational expressions are added within those equations. (Note that I'm not saying that solving rational equations is "simple"; I'm only saying that it's simpler.) This is because, as soon as you go from a rational expression (that is, something with no "equals" sign in it) to a rational equation (that is, something with an "equals" sign in the middle), you get a whole different set of tools to work with. In particular, once you have that "equals" sign in the middle, you then have two sides, which means that you can multiply through both of those sides of the equation, and this allows you to get rid of the denominators.

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Solving Rational Equations

• Solve the following equation:

This equation is so simple that I can solve it just by looking at it! How?

I have two fractions. These fractions have the same denominator. These fractions will be equal when their numerators are also the same, and only then. So I can equate the numerators and obtain my answer. Since the numerators are so simple, I immediately arrive at my answer:

x = 2

• Solve the following equation:

This equation has fractions on either side of the "equals" sign. The two fractions have the same denominator. The two fractions will be equal when their numerators are equal, so I can "equate" the numerators (that is, I can set them equal) and solve the resulting equation:

x − 3 = 4x + 12

−3 − 12 = 4x − x

−15 = 3x

−5 = x

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• Solve the following equation:

This equation has two fractions which are set equal to each other (which can be viewed as a proportion). There are three ways that I can solve this. I'll show each, and you can pick whichever you prefer.

Method 1: Converting to a common denominator:

I can convert to a common denominator of 15:

Now that I have "(one fraction) equal to (another fraction)", I can equate the numerators:

x − 1 = 6

x = 7

Method 2: Multiplying through by the common denominator:

The lowest common denominator is 15. Rather than converting the fractions to this denominator (something that would be required if I were adding or subtracting these rational fractions), I can instead multiply through (that is, multiply both sides of the equation) by 15. This gives me:

x − 1 = 2(3)

x − 1 = 6

x = 7

Method 3: Cross-multiplying:

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The term "cross-multiplying" isn't technical, and some instructors absolutely hate it. But it's a term you'll hear, and it stands for a technique which can be handy.

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Since this is an equation, I can multiply through by whatever I like. In particular, to get rid of the denominators, I can multiply through by those denominators. In this case, I'd multiply the 15 from the left-hand side's denominator, up against the 2 in the right-hand side's numerator; and I'd multiply the 5 from the right-hand side's denominator, up against the x − 1 in the left-hand side's numerator. In other words, I'd do this:

This process of "crossing" the "equals" sign with each denominator and multiplying each against its opposing numerator is what is meant by "cross-multiplying". It's a shorthand for "multiplying through by the common denominators, when there are only the two fractions set equal to themselves, and then simplifying what's left", and can be a nice shortcut.

The cross-multiplication gives me the following new (and linear) equation:

5(x − 1) = 15(2)

5x − 5 = 30

5x = 35

x = 7

So, by each of the methods, my answer is:

x = 7

Note: Cross-multiplying (that is, Method 3 above) works only if the equation has exactly one fraction on one side of the "equals" sign, set equal to exactly one fraction on the other side of the "equals". If either side of the equation has added (or subtracted) fractions, we must use Method 1 or Method 2.

• Solve the following equation:

This equation has subtracted fractions on the left-hand side, so I can't cross-multiply. Also, there's the new wrinkle of variables in the denominator. This means that I'll need to keep track of the values of x that would cause division by zero. These values cannot be part of my final answer. In this case, the denominators tell me that my answer will have the following restriction:

x ≠ −2, 0

Method 1: To solve this equation, I can convert everything to the common denominator of 5x(x + 2) and then compare the numerators:

At this point, the denominators are the same. So do they really matter? Not really — other than for saying what values x can't be, due to division-by-zero issues. At this point, the two sides of the equation will be equal as long as the numerators are equal. That is, all I really need to do now is solve the numerators:

15x − (5x + 10) = x + 2

10x − 10 = x + 2

9x = 12

Since x = 4/3 won't cause any division-by-zero problems in the fractions in the original equation, then this solution is valid.

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Method 2: The other method is to find the common denominator but, rather than converting everything to that denominator, I'll take advantage of the fact that I have an equation here. That is, I'll multiply through on both sides by that common denominator. This will get rid of the denominators. I've used colors below to highlight the parts that cancel off:

3(5x) − 1(5(x + 2)) = 1(x + 2)

15x − 5x − 10 = x + 2

10x − 10 = x + 2

9x = 12

Either way, my answer is the same:

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I view Method 2 as being quicker and easier, but this is only my personal preference. In my classroom experience, students have typically been fairly evenly divided in their preferences for Methods 1 and 2. You should use the method that works best for you.

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