If  $|x-2|+|x-3|=7,$ then x =

Here $x=2$ and 3 are the critical points. When $x<2,|x-2|=-(x-2),|x-3|=-(x-3)$

$\therefore$The given equation reduces to $2-x+3-x=7$

$\Rightarrow x=-1<2$

\  $x=-1$ is a solution.

When $2\le x<3,|x-2|=x-2,|x-3|=-(x-3)$

\ The equation reduces to $x-2+3-x=7\Rightarrow 1=7$

\ No solution in this case.

When  $x\ge 3,$ the equation reduces to $x-2+x-3=7\Rightarrow x=6>3$

Hence we get, $x=6$ or –1

Trick : By inspection, we have that both the values $x=6,-1$ satisfy the given equation.