How To Solve Rational Equations?


By Robert O

How to define rational equations?

Equations with one or more expressions are rational equations. Such equations may involve additions, subtractions, multiplications, and divisions. Rational expressions are ratios of two numbers, numerator and denominator, where denominator is never zero.

How do I solve rational equations?

Learning math is by doing practice through examples. So, we will be looking at how to solve rational equations using examples. Please take note that every example includes something new that you should learn about this topic. Your knowledge of simplifying algebraic fractions is very here.

Example 1:

What is the value of x in the equation



You can approach this problem in two ways. One is cross-multiplication, and two, multiplying both sides by a common denominator.


    \[\frac{2}{3}=\frac{x}{3} \equiv 6=3 x\]

    \[3 x=6\]


Using a common denominator:

    \[3 \times \frac{2}{3}=\frac{x}{3} \times 3\]


Example 2:

Solve for the value of x in

    \[\frac{x-3}{9}=\frac{4 x+12}{9}\]


The equation has 9 as a common denominator. We can cross-multiply or multiply both sides by this common denominator. But if you have such a problem with a common denominator, then we can simply drop those denominators and equate only the numerator.

    \[x-3=4 x+12\]

By moving like terms to one side:

    \[x-4 x=12+3\]

    \[-3 x=15\]


Example 3:

Find the value of x in

    \[\frac{2}{x}+\frac{1}{6}=\frac{3}{4 x}\]


Cross-multiplication does not work here since we have an addition operation on the left side of the equation. In this case, we first collect like terms on one side and reduce it to a single fraction.

    \[\frac{2}{x}-\frac{3}{4 x}=-\frac{1}{6}\]

    \[\frac{8-3}{4 x}=-\frac{1}{6}\]

    \[\frac{5}{4 x}=-\frac{1}{6}\]

We can now apply cross multiplication to solve the problem.

    \[\frac{5}{4 x}=-\frac{1}{6}\]

    \[4 x=-30\]


Example 4:

Find the value of x in



In this example, we should not rush to say that denominators are the same because they are not. We have 3 as a whole number on the left side. To proceed, multiply every term by the denominator, which is x+2.

    \[(x+2) \times \frac{1}{x+2}+3(x+2)=\frac{1}{x+2} \times(x+2)\]

By simplification:


    \[3 x+6=0\]


Example 5:

Find the value of x in

    \[\frac{x+4}{x^{2}+8 x+16}=\frac{1}{2 x+3}\]


We don’t have any like terms in this case. So, we proceed to factorize the quadratic expression at the bottom of the left-hand side fraction.

    \[\frac{x+4}{x^{2}+8 x+16}=\frac{1}{2 x+3}\]

    \[\frac{x+4}{(x+4)(x+4)}=\frac{1}{2 x+3}\]

    \[\frac{1}{x+4}=\frac{1}{2 x+3}\]


    \[2 x+3=x+4\]


Example 6:

What is x in



We can do cross-multiplication or multiply both sides of the equation by the LCM, which is


    \[(x-4)(x+4) \times \frac{x+1}{x+4}=\frac{x+1}{x-4} \times(x-4)(x+4)\]


    \[x^{2}+x-4 x-4=x^{2}+x+4 x+4\]

The term


on both sides cancels out. By grouping like terms:

    \[-3 x=8\]


Example 7:

Find all the values of x in the equation:

    \[\frac{2}{x^{2}-2 x}=\frac{1}{x-2}-\frac{2}{x^{2}}\]


The first step is to factor our x for the denominator of the fraction on the left side of the equation.


Next, we multiply every term with this x.


The LCM of the denominators is x(x-2). We use it to remove the numerator by multiplying every term by it.

    \[x(x-2) \frac{3}{x-2}=x(x-2) \frac{x}{x-2}-\frac{2}{x} \times x(x-2)\]

    \[3 x=x^{2}-2 x+4\]

Realize that this results in a quadratic equation. Rearrange it in the form of

    \[a x^{2}+b x+c=0\]

    \[x^{2}-5 x+4=0\]

Using any method of your choice, solve for x in the above equation. Using quadratic formula to solve for x:

    \[x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\]

    \[x=\frac{5 \pm \sqrt{25-16}}{2}=\frac{5 \pm 3}{2}\]

    \[x=1 { or } 4\]


You can solve any rational equation if you use the right steps and apply your knowledge of other topics. Be keen to manipulate the letters. But with regular practice, you can always get it right.

About the Author

This lesson was prepared by Robert O. He holds a Bachelor of Engineering (B.Eng.) degree in Electrical and electronics engineering. He is a career teacher and headed the department of languages and assumed various leadership roles. He writes for Full Potential Learning Academy.