Question

How do you solve $ 5x - 4 = \left| {2x + 1} \right| $ and find any extraneous solution?

Answer 1

Hint: The given equation consists of a modulus function, and in order to solve this equation the first step should be to open the modulus function. Modulus function opens with a positive sign when its argument is greater than or equal to zero and with a negative sign when its argument is less than zero. So make two cases for the domain of the arguments of the modulus function and solve for $ x $ and finally take the union of solutions from both cases.

Complete step by step solution:
To solve the given equation $ 5x - 4 = \left| {2x + 1} \right| $ , we will first make two cases for the domain of argument of modulus function in order to open it as follows
Case I: When $ 2x + 1 < 0 \Rightarrow 2x < - 1 \Rightarrow x < - \dfrac{1}{2} $
In this case argument of modulus is negative, so it will open with negative sign
 $
   \Rightarrow 5x - 4 = - (2x + 1) \\
   \Rightarrow 5x - 4 = - 2x - 1 \\
  $
Adding $ 4\;{\text{and}}\;2x $ to both sides, we will get
 $
   \Rightarrow 5x - 4 + 4 + 2x = - 2x - 1 + 4 + 2x \\
   \Rightarrow 7x = 3 \\
  $
Now, dividing both sides with $ 7 $
 $
   \Rightarrow \dfrac{{7x}}{7} = \dfrac{3}{7} \\
   \Rightarrow x = \dfrac{3}{7} \\
  $
But wait a second, here the solution $ x = \dfrac{3}{7} $ does not satisfying the condition of its case $ \Rightarrow x < - \dfrac{1}{2} $
So $ x = \dfrac{3}{7} $ is an extraneous solution.
Case II: When $ 2x + 1 \geqslant 0 \Rightarrow 2x \geqslant - 1 \Rightarrow x \geqslant - \dfrac{1}{2} $
In this case argument of modulus is positive, so it will open with positive sign
 $
   \Rightarrow 5x - 4 = (2x + 1) \\
   \Rightarrow 5x - 4 = 2x + 1 \\
  $
Adding $ 4 $ and subtracting $ 2x $ from both sides, we will get
 $
   \Rightarrow 5x - 4 + 4 - 2x = 2x + 1 + 4 - 2x \\
   \Rightarrow 3x = 5 \\
  $
Now, dividing both sides with $ 3 $
 $
   \Rightarrow \dfrac{{3x}}{3} = \dfrac{5}{3} \\
   \Rightarrow x = \dfrac{5}{3} \\
  $
So we have solved the equations for both possibilities.
Taking union of both we will get
 $ x \in \left\{ {} \right\} \cup \left\{ {\dfrac{5}{3}} \right\} \Rightarrow x = \dfrac{5}{3} $
Therefore $ x = \dfrac{5}{3} $ is the required solution of the given equation.

Note: The symbol $ \left\{ {} \right\} $ represents a set with zero elements or null and when we take union with it or another set then the result equals the other set. And we have taken null for the solution of the first case because that was an extraneous solution.