Question

How do you solve\[\dfrac{x}{{2x + 1}} + \left( {\dfrac{1}{4}} \right) = \dfrac{2}{{2x + 1}}\] and find any extraneous solutions?

Answer 1

Hint: We need to solve the equation to get an extraneous solution, which is a root of a transformed equation that is not a root of the original equation because it was excluded from the domain of the original equation. To check whether it is an extraneous solution or not by substituting the value into the original equation.

Complete step by step solution:
The given algebraic expression is performed to find an extraneous solution whether it occurs or not.
\[\dfrac{x}{{2x + 1}} + \left( {\dfrac{1}{4}} \right) = \dfrac{2}{{2x + 1}}\]
Take LCM on both sides of the equation, we have
\[\dfrac{{4x}}{{4(2x + 1)}} + \left( {\dfrac{{1(2x + 1)}}{{4(2x + 1)}}} \right) = \dfrac{{2(4)}}{{4(2x + 1)}}\]
By performing multiplication on both sides by\[4(2x + 1)\], we get
\[4x + 2x + 1 = 8\]
To simplify the algebraic expression, we have
\[
  6x + 1 - 8 = 0 \\
  6x - 7 = 0 \;
 \]
 To find the value of \[x\],
\[
  6x = 7 \\
  x = \dfrac{7}{6} \;
 \]
Therefore, the value of \[x = \dfrac{7}{6}\] is not an extraneous solution.
So, the correct answer is “\[x = \dfrac{7}{6}\]”.

Note: To check whether it is an extraneous solution or not by substituting the value into the original equation. When the values are applied into the equation of both sides of is equal its extraneous otherwise not.
Hence, there is no extraneous solution

To determine if a solution is extraneous, we simply plug the solution into the original equation. If it makes a true statement, then it is not an extraneous solution.