Question

Find all real values of x that satisfy the following equation:
$|2x-4|+|1-x|=4-x$

Answer 1

Hint: First take 2 common out of absolute function operator of the first term of LHS. Then break the absolute function considering specific intervals of x one by one. Then it will form a linear equation in one variable which will be easy to solve.

Complete step-by-step answer:

We are given the equation as:
$|2x-4|+|1-x|=4-x$
We know that for any real number a and b we have
$|ab|=|a||b|$
Applying this concept in the first term of our equation with b = 2 and a = $x-2$.
$|2||x-2|+|1-x|=4-x\,\,\,\,\cdot \cdot \cdot \text{(i)}$
We know that for any nonnegative real number a,
$|a|=a$
So applying this in equation (i) we get
$2|x-2|+|1-x|=4-x\,\,\,\,\cdot \cdot \cdot \text{(ii)}$
Now for breaking the absolute function operator we find the values of x for each absolute function becoming 0.
$\begin{align}
  & x-2=0 \\
 & \Rightarrow x=2 \\
 & \text{and} \\
 & 1-x=0 \\
 & \Rightarrow x=1 \\
\end{align}$
So there are two critical values of x.
Now considering three cases of x corresponding to two critical values
Case1: $x\le 1$
Then
$x-2\le 1-2=-1$
We know that for any real number a ,
$|a|=\left\{ \begin{array}{*{35}{l}}
   a & \text{for }a\ge 0 \\
   -a & \text{for }a<0 \\
\end{array} \right.$
So $|x-2|=-(x-2)=2-x$
Also, $x\le 1$
   $\begin{align}
  & \Rightarrow 1-x\ge 0 \\
 & \Rightarrow |1-x|=1-x \\
\end{align}$
Putting simplified value in equation (ii) we get
$\begin{align}
  & 2(2-x)+1-x=4-x \\
 & \Rightarrow 4-2x-3=0 \\
 & \Rightarrow 2x=1 \\
 & \Rightarrow x=\dfrac{1}{2} \\
\end{align}$
Now, $x=\dfrac{1}{2}\le 1$. So our solution comes in the defined range. Hence $x=\dfrac{1}{2}$is the solution.
Case 2: $1\le x\le 2$
In this case,
$\begin{align}
  & x-2\le 0 \\
 & \Rightarrow |x-2|=-(x-2)=2-x \\
 & \text{and} \\
 & 1-x\le 0 \\
 & \Rightarrow |1-x|=-(1-x)=x-1 \\
\end{align}$
So we get our equation as
$\begin{align}
  & 2(2-x)+x-1=4-x \\
 & \Rightarrow 4-2x+2x-5=0 \\
 & \Rightarrow 4=5 \\
\end{align}$
It is not possible that 4 = 5. So, no value of x in this range.
Case 3: $x\ge 2$
In this case,
$\begin{align}
  & x-2\ge 0 \\
 & \Rightarrow |x-2|=x-2 \\
 & \text{and} \\
 & 1-x\le 0 \\
 & \Rightarrow |1-x|=-(1-x)=x-1 \\
\end{align}$
So we get our equation as
$\begin{align}
  & 2(x-2)+x-1=4-x \\
 & \Rightarrow 2x-4+2x-5=0 \\
 & \Rightarrow 4x=9 \\
 & \Rightarrow x=\dfrac{9}{4} \\
\end{align}$
We see that $x=\dfrac{9}{4}\ge 2$. So this value of x is accepted.
Hence overall $x=\dfrac{1}{2}$ and $x=\dfrac{9}{4}$ is the solution of the given equation.

Note: This question requires breaking the modulus function and then solving the problem. Students might get wrong in identifying the critical points and then the whole solution may get wrong.