Rational Equations (Description & Examples)

Procedure of solving the Rational Equations:

  1. First of all, find out the LCD of all the Rational Expressions in the given equation.
  2. Then multiply both sides by the LCD.
  3. Solve the equation.
  4. Finally, check your solutions and throw out any that make the denominator zero.

You must be emphasized on step 4 as you can never have a denominator of zero in a fraction, you have to make sure that none of your solutions make your denominator zero. Any solution that cannot be included in the solution set called “Extraneous Solutions”. If all of the solutions of a Rational Equation are extraneous then the equation would have no solution which means that there is no value for the variable that make the equation a True Statement.

rational equations
Rational equations

Few examples of rational equations are given below: –

Example No.1:

Solve  \displaystyle \frac{{x-4}}{4}+\frac{x}{3}=6

Solution:

\displaystyle \frac{{x-4}}{4}+\frac{x}{3}=6

In order to solve this equation, we must follow the above mentioned steps which means that we find out the LCD. Here, LCD is 12, so, we multiply both sides by the LCD. We get,

\displaystyle 12\left( {\frac{{x-4}}{4}+\frac{x}{3}} \right)=6(12)

Now we solve the equation

\displaystyle \begin{array}{l}12\left( {\frac{{3(x-4)+4(x)}}{{12}}} \right)=72\\3(x-4)+4(x)=72\\3x-12+4x=72\\7x-12=72\\7x=72+12\\7x=84\\x=\frac{{84}}{7}\\x=12\\Check\,\,it\,\,now\\\frac{{x-4}}{4}+\frac{x}{3}=6\\\frac{{12-4}}{4}+\frac{{12}}{3}=6\\\frac{8}{4}+4=6\\2+4=6\\6=6\end{array}

Example No.2:

Solve  \displaystyle \frac{{3x}}{2}-\frac{x}{3}=7

Solution:

\displaystyle \frac{{3x}}{2}-\frac{x}{3}=7

In order to solve this equation, we must follow the above mentioned steps which means that we find out the LCD. Here, LCD is 6, so, we multiply both sides by the LCD. We get,

\displaystyle 6\left( {\frac{{3x}}{2}-\frac{x}{3}} \right)=7\times 6

Now we solve the equation

\displaystyle \begin{array}{l}6\left( {\frac{{9x-2x}}{6}} \right)=42\\9x-2x=42\\7x=42\\x=\frac{{42}}{7}\\x=6\\Check\,\,it\,\,now\\\frac{{3x}}{2}-\frac{x}{3}=7\\\frac{{3(6)}}{2}-\frac{6}{3}=7\\\frac{{18}}{2}-\frac{6}{3}=7\\\frac{{54-12}}{6}=7\\\frac{{42}}{6}=7\\7=7\end{array}

Example No.3:

Solve  \displaystyle \frac{7}{{4x}}-\frac{3}{{{{x}^{2}}}}=\frac{1}{{2{{x}^{2}}}}

Solution:

\displaystyle \frac{7}{{4x}}-\frac{3}{{{{x}^{2}}}}=\frac{1}{{2{{x}^{2}}}}

In order to solve this equation, we must follow the above mentioned steps which means that we find out the LCD. Here, LCD is 4x2, so, we multiply both sides by the LCD. We get,

\displaystyle 4{{x}^{2}}\left( {\frac{7}{{4x}}-\frac{3}{{{{x}^{2}}}}} \right)=\frac{1}{{2{{x}^{2}}}}(4{{x}^{2}})

Now we solve the equation

\displaystyle \begin{array}{l}4{{x}^{2}}\left( {\frac{{7x-12}}{{4{{x}^{2}}}}} \right)=2\\7x-12=2\\7x=12+2\\7x=14\\x=\frac{{14}}{7}\\x=2\\Check\,\,it\,\,now\\\frac{7}{{4x}}-\frac{3}{{{{x}^{2}}}}=\frac{1}{{2{{x}^{2}}}}\\\frac{7}{{4(2)}}-\frac{3}{{{{2}^{2}}}}=\frac{1}{{2{{{(2)}}^{2}}}}\\\frac{7}{8}-\frac{3}{4}=\frac{1}{8}\\\frac{{7-6}}{8}=\frac{1}{8}\\\frac{1}{8}=\frac{1}{8}\end{array}