Question

How do you solve \[\left| 2x-1 \right|=3\]?

Answer 1

Hint: This type of questions is based on the concept of modulus functions. To solve \[\left| 2x-1 \right|=3\], we first have to consider the modulus function. According to the definition of modulus function, we get two cases, that are 2x-1=3 and -(2x-1)=3. Consider 2x-1=3 and add 1 on both the sides of the equation. do necessary calculations and divide the equation by 2 to get the value of x. Now, consider -(2x-1)=3. Use distributive property to expand the equation and subtract the equation by 1. Divide the whole equation by 2 and find the value of x.

Complete step by step solution:
According to the question, we are asked to solve \[\left| 2x-1 \right|=3\].
We have been given the equation is \[\left| 2x-1 \right|=3\]. --------(1)
First, we have to consider the modulus function.
According to the definition of modulus function, we know that
\[\left| x \right|=\left\{ \begin{align}
  & \text{x , x > 0} \\
 & \text{-x , x < 0} \\
\end{align} \right.\]
Here, we have to take two cases.
CASE (I)
When 2x-1>0,
Then, we get \[\left| 2x-1 \right|=2x-1\].
Therefore, we get
2x-1=3
Now, we have to add 1 on both the sides of the equation.
\[\Rightarrow 2x-1+1=3+1\]
We know that terms with the same magnitude and opposite signs cancel out. On cancelling 1, we get
2x=3+1
\[\Rightarrow 2x=4\]
We can express the equation as
\[2x=2\times 2\]
Now, let us divide the whole equation by 2. We get
\[\dfrac{2x}{2}=\dfrac{2\times 2}{2}\]
We find that 2 are common in both the numerator and denominator of LHS and RHS. On cancelling 2, we get
x=2
CASE (II)
When 2x-1<0.
Then, we get \[\left| 2x-1 \right|=-\left( 2x-1 \right)\].
Therefore, we get
-(2x-1)=3
We have to now use distributive property, that is \[a\left( b+c \right)=ab+ac\].
Here, a=-1, b=2x and c=-1.
\[\Rightarrow -2x-\left( -1 \right)=3\]
On further simplifications, we get
\[-2x+1=3\]
Subtract 1 from both the sides of the equation. We get
\[-2x+1-1=3-1\]
We know that terms with the same magnitude and opposite signs cancel out. On cancelling 1, we get
\[-2x=3-1\]
\[\Rightarrow -2x=2\]
We find that 2 are common in both the numerator and denominator of LHS and RHS. On cancelling 2, we get
-x=1
On cancelling the negative sign from the LHS, we get
x=-1

Therefore, the values of x in the equation \[\left| 2x-1 \right|=3\] are -1 and 2.

Note: We should always consider two cases for the modulus functions. Avoid calculation mistakes based on sign conventions. We should know the conditions of the modulus function to solve this type of question.