Question

If $|x-2|+|x-6|=3$, then find the value(s) of $x$.

Answer 1

Given: $|x-2|+|x-6|=3$

Case 1: For $x<2$

$-(x-2)-(x-6)=3$

$-x+2-x+6=3$

$-2x+8=3$

$-2x=3-8$

$-2x=-5$

$\Rightarrow x=\frac{5}{2}$

This is not possible since $x<2$

Case 2: For $2\le x<6$

$(x-2)-(x-6)=3$

$x-2-x+6=3$

$4=3$ No solution

Case 3: For $x\ge 6$

$x-2+x-6=3$

$2x-8=3$

$2x=3+8$

$\Rightarrow x=\frac{11}{2}$

This is not possible since $x\ge 6$

Therefore, no value of x satisfies the given equation.