Find the set of solutions to this inequality?
$|x − 3| + |x − 6| < 5$
I have been taught to do it by treating $x$ in $3$ separate cases however I am not getting the correct answer. The answer is 'The set of real numbers $x$ such that $2 < x < 7.$ I am getting The set of real numbers $x$ such that $2 < x < 3$ or $6 < x < 7$.
Method
Case 1 $x<3$
$|x-3|= -(x-3)$
$|x-6|= -(x-6)$
$-(x-3)-(x-6)<5$
$-2x+9<5$
$-2x<-4$
$2x>4$
$x>2$
Because you have assumed $x<3$ the solutions is the intersection so $2<x<3$.
Case 2: $3<x<6$
$|x-3|= (x-3)$
$|x-6|= -(x-6)$
$(x-3)-(x-6)<5$
$3<5$
No solutions
Case 3: $x>6$
$|x-3|= (x-3)$
$|x-6|= (x-6)$
$(x-3)+(x-6)<5$
$2x-9<14$
$x<7$
Assumed $x>6$ therefore $6<x<7$.
So joint solutions the set of real numbers $x$ such that $2 < x < 3$ or $6 < x < 7$.
Where am I going wrong??
Case 2:
3 < x < 6
|x-3|= (x-3)
|x-6|= -(x-6)
(x-3)-(x-6)<5
3<5
3<5 is true, so 3 < x < 6 is also solution for |x − 3| + |x − 6| < 5
Finally we have |x − 3| + |x − 6| < 5 $\Leftrightarrow$ 2 < x < 7