# Solving Rational Equations · Examples

Listen up, fraction fans! In today’s lesson, you will learn and practice solving rational equations. As you will see, these are any equation involving a fraction, also known as a rational number in math talk!

By the end, you will know the difference between rational and irrational numbers and have two tricks for solving rational equations.

You could even tackle one of the tricky challenges to form a rational equation using the Pythagorean theorem, or to simplify an expression involving some radicals!

Contents

## What is a Rational Equation?

A rational equation is simply an equation involving a rational number.

A ratio-nal number can be written as a ratio of two integers – an irratio-nal number cannot.

Most of the numbers you know and love such as $$\Large\frac{2}{7}$$, $$\Large\frac{1}{2}$$ and $$-\Large\frac{20817}{43}$$ are rational. Some common irrational numbers are π, $$\sqrt{2}$$ and Euler’s number, e. These cannot be written as a fraction of integers.

Numberphile has an interesting video about All the Numbers, which categorizes number types, including rational and irrational numbers.

Technically speaking, basic equations like x+2=5 are rational because each term is a rational number. However, the rational equations you will solve today won’t be so easy!

An example of what you will more likely see in an exam is something like this:

$$\Large\frac{x}{6}\normalsize + \Large\frac{2}{x}\normalsize = \Large\frac{5}{4}$$

Each term is shown as a fraction.

Rational equations can also include radicals:

$$\Large\frac{\sqrt{x}}{2}\normalsize + 6= 7$$

Or other operations such as division:

$$\Large\frac{x^2+3}{3x}\normalsize \div \Large\frac{5}{4}\normalsize = 13$$

Luckily, the technique you learn now will work for every type of rational equation!

## How to Solve Rational Equations

The method to solve these equations is pretty much the same for every type of rational equation. You’ll see questions of varying difficulty in this lesson; don’t be afraid to tackle the challenges later on!

### Step 1: The Denominator Elimination Round!

First, you need to deal with the elephant in the room: what should you do with the denominators!?

Solving rational equations is just like solving any other equation once you complete this step.

If it’s a simple case, where you have one fraction being equal to one other fraction, you can cross multiply.

Multiply both sides by the values of both denominators. In this example, both sides are multiplied by 3, then 5.

$$\Large\frac{x+4}{3}\normalsize = \Large\frac{2}{5}$$

The 3 cancels with the left denominator and the 5 cancels with the right denominator, leaving you with 5(x+4)=3×2.

See why it’s called cross multiplying?

The product of the left denominator and right numerator equals the product of the right denominator and left numerator!

The more general way to deal with the denominators is to find their lowest common multiple (LCM). This is the smallest number which all denominators divide neatly, leaving no remainder.

If you cannot find the LCM by inspection – if you cannot “just see it” – you need to factor every denominator like you would with a polynomial.

If you have more than one constant term, you may need to find their prime factors.

The LCM is the smallest combination of each denominator’s factors.

You’ll now see a worked example to illustrate!

Solve:

$$\Large\frac{x^2+1}{x}\normalsize-\Large\frac{2x^2 +1}{3x}\normalsize = \Large\frac{x+1}{4}$$

Remember, you can only cross multiply when each side has only one fraction, so in this case, your first step is to find the LCM.

The only factors of 3x you know for certain are 3 and x. The only factor you know of x is just x, and 4 is a constant so you can use it as it is.

Write down each denominator’s polynomial factors into rows, with similar terms lined up in the same column.

You need to include both 3 and 4 because neither is a factor of the other. You don’t need both copies of x because x is a factor of itself! So the LCM is 12x.

You might find another example of finding the LCM with the same technique helpful.

You’re now ready to eliminate the denominators by multiplying both sides by the LCM.

### Step 2: Simplify the Equation

Multiply each term by the LCM. Continuing from the last example, you have:

\eqalign{ 12x\left(\frac{x^2\mathord{+}1}{x}\normalsize\mathord{-}\frac{2x^2\mathord{+}1}{3x}\right)&=12x\left(\frac{x\mathord{-}2}{4}\right)\\ \\ 12(x^2\mathord{+}1)\mathord{-}4(2x^2\mathord{+}1) &= 3x(x\mathord{-}2)\\ \\ 12x^2\mathord{+}12\mathord{-}8x^2\mathord{-}4 &= 3x^2\mathord{-}6x\\ \\ x^2\mathord{+}6x\mathord{+}8 &= 0 }

You now have a regular equation with no fractions, which should be familiar ground!

### Step 3: Solve the Equation

Solving rational equations usually produces a simple polynomial equation. Hopefully, you’ve solved lots of these before!

You could complete the square, factor the terms by inspection, or use the quadratic formula.

This example can be solved by factoring the polynomial, having found that x+2 and x+4 are factors.

x2 + 6x + 8 = 0
(x + 2)(x + 4) = 0
x = {-2, -4}

You could also solve the equation by completing the square:

(x + 3)2 – 1 = 0
(x + 3)2 = 1
x + 3 = ±1
x = {-2, -4}

Or by using the quadratic formula with a=1, b=6 and c=8:

\eqalign{ x&=\frac{-b \pm \sqrt{b^2 -4ac}}{2a}\\ x&=\frac{-6 \pm \sqrt{36-32}}{2}\\ x&=\frac{-6 \pm \sqrt{4}}{2}\\ x&=-3\pm1\\ x&=\{-2,-4\} }

Each way of solving the simplified rational equation is valid, but you will find that some are quicker than others!

### Step 4: Check Every Solution

It is important to check that your solutions are complete, meaning you’ve found all of them and that they don’t give any weird numbers when substituted into the original equation.

In the worked example, you were left with a quadratic equation and found two distinct roots.

Quadratic equations either have two distinct solutions, one repeated solution, or no real solution so the solution x=-2 or x=-4 is complete.

You must be careful that none of the rational terms in the original equation have a zero in the denominator.

Do this by going back to the beginning and substituting your answers into the denominators!

The denominators in the worked example are 3x, x, and 4. Replacing x with -2 or -4 doesn’t give you zero in any of them, so you’re safe here!

A solution that gives a zero-denominator is not allowed. That’s because dividing by zero is “illegal” in math!

Any number divided by zero gives an error on a calculator. Ever wondered why that is?

## Practice

This is your time to shine – try solving rational equations for yourself and, if you’re feeling confident, tackle the challenges too.

As they say, practice makes perfect! Use the worked example for guidance if you get stuck.

### Question 1

Find x in the following rational equation:

$$\Large\frac{3x^2 -2}{5}\normalsize = \Large\frac{15}{3}$$

The equation is two equal fractions so you can cross-multiply. You could also simplify $$\Large\frac{15}{3}\normalsize$$ to 5, but this does not change the final answer.

3 · (3x2 – 2) = 5 · 15
9x2 – 6 = 75
9x2 – 81 = 0

Each term is divisible by 9. Simplify the equation by dividing both sides by 9:

x2 – 9 = 0

This form is called the difference of two squares because it can be factored like this:

(x-3)(x+3) = 0

So the solution is x=±3.

These must be all the solutions because quadratic equations have a maximum of two distinct real roots.

Neither denominator in the original rational equation has an x term, so substituting any value for x makes no difference to their values – there is no chance of them being zero!

This means the solutions x=3 and x=-3 are valid.

### Question 2

Solve the following rational equation:

$$\Large\frac{3}{x-5}\normalsize + \Large\frac{20}{x^2-25}\normalsize = – \Large\frac{2}{x+5}$$

There are three fractions so you cannot cross-multiply.

See that the second denominator is the difference of two squares?

(x – 5)(x + 5)

Multiply each term by the LCM and simplify.

So its solution is -5, right?… STOP RIGHT THERE! Don’t forget, we can’t divide by zero!

If you put x=±5 into the original equation, at least one of the denominators is always zero, so the original equation has no solutions.

### Challenge 1

Can you spot the mistake in the following example? Hint: there has been some cheating with radicals!

\eqalign{ \frac{\sqrt{2x-1}}{2}+ \frac{\sqrt{1-x}}{2}&= \frac{\sqrt{x }}{5}\\ 10\left(\frac{\sqrt{2x-1}}{2}+\frac{\sqrt{1-x}}{2}\right) &=10\left(\frac{\sqrt{x }}{5}\right)\\ 5\sqrt{2x-1} + 5\sqrt{1-x} &= 2\sqrt{x} \\ 5\sqrt{2x-1} + 5\sqrt{1-x} &= 2\sqrt{x} \\ 5\sqrt{2x-1+1-x} &= 2\sqrt{x} \\ 5\sqrt{x} &= 2\sqrt{x} \\ x&=0 }

If you need a refresher on radicals, check out our lesson on multiplying them. That will get you on the right track!

The mistake is that radicals cannot be subtracted like normal terms.

$$5\sqrt{2x-1} + 5\sqrt{1-x}$$
$$\neq$$
$$5\sqrt{2x-1+1-x}$$

Instead, you must square both sides of the equation to remove the radical. Similar terms can then be combined as usual.

Still confused? You can find lots of interactive questions on Lumen Learning. Radicals often pop up in rational equations, so getting comfortable with radicals is super helpful for exam success!

### Challenge 2

Find the value of x, by using the Pythagorean theorem on the following right-angled triangle:

If you need a refresher on the Pythagorean theorem or are interested in the man himself, check out our lesson. Do the worksheets and you’ll be acing triangle questions in no time!

The Pythagorean theorem states that:

a2 + b2 = c2

Where c is the length of the hypotenuse, and a and b are the other side lengths.

This gives the rational equation:

$$\left(\Large\frac{x-1}{2}\right)^2\normalsize + \left(\Large\frac{2\sqrt{x}}{3}\right)^2\normalsize = \left(\Large\frac{x}{2}\right)^2$$

Simplifying, you find:

$$\Large\frac{x^2-2x+1}{4}\normalsize + \Large\frac{4x}{9}\normalsize = \Large\frac{x^2}{4}$$

The LCM is 36 so the denominators are removed by dividing each term by this:

\eqalign{ 9x^2 – 18x + 9 + 16x &= 9x^2\\ -2x&=-9\\ x&=\frac{9}{2} }

It’s always fun when different areas of math link together!

## To Sum Up (Pun Intended!)

In today’s lesson on solving rational equations, you first saw the difference between rational and irrational numbers.

Rational numbers are “nice” because they can be written as a fraction of integers. Remember that all integers are rational because they can be written with a denominator of 1!

Irrational numbers are a little more abstract. They include weird but incredibly beautiful numbers like π and e, which cannot be written as a fraction of integers.

Rational equations are solved by eliminating the denominator in every term, then simplifying and solving as normal.

Denominators can be removed by cross-multiplication if there is only one fraction on either side or by finding the LCM if the equation is more complicated.