Question

How do you solve \[\left| {2x - 3} \right| = \left| {x + 2} \right|\] ?

Answer 1

Hint: The given equation is in the form of absolute values. We will remove the absolute sign, consider the positive sign first, and solve it to get on of the solution. Then we will take the negative sign and solve it further to get the second solution. Thus, the solution set is the required solution.

Complete Step by Step Solution:
We are given that \[\left| {2x - 3} \right| = \left| {x + 2} \right|\]
So, removing modulus, we get
\[\left( {2x - 3} \right) = \pm \left( {x + 2} \right)\]
Now, we will consider the equality with only the positive sign, thus we get
\[ \Rightarrow \left( {2x - 3} \right) = + \left( {x + 2} \right)\]
Thus, we get
\[ \Rightarrow 2x - 3 = + x + 2\]
By rewriting the equation, we get
\[ \Rightarrow 2x - x = 2 + 3\]
Subtracting and adding the like terms, we get
\[ \Rightarrow x = 5\]
Now, we will consider the equality with only the negative sign, thus we get
\[\left( {2x - 3} \right) = - \left( {x + 2} \right)\]
Thus by rewriting the equation, we get
\[ \Rightarrow 2x - 3 = - x - 2\]
By rewriting the equation, we get
\[ \Rightarrow 2x + x = - 2 + 3\]
Subtracting and adding the like terms, we get
\[ \Rightarrow 3x = 1\]
Dividing both sides by 3, we get
\[ \Rightarrow x = \dfrac{1}{3}\]

Therefore, the solution set of \[\left| {2x - 3} \right| = \left| {x + 2} \right|\] is \[x = 5\] and \[x = \dfrac{1}{3}\].

Note: We know that the absolute value is defined as the non-negative integer without regard to its sign, it can be either a positive or negative integer. We will follow these steps to solve absolute value equality which contains two absolute values. We will write both the equations without absolute values. The first set of equations will set the quantity inside the bars on the left side of the equation equal to the quantity inside the bars on the right side of the equation and then solve the linear equation formed. The second set of equations will set the quantity inside the bars on the left side of the equation opposite to the quantity inside the bars on the right side of the equation and then solve the linear equation formed. Thus, the solution set for the linear equation.