I'm new here, and I was wondering if any of you could help me out with this little problem that is already getting on my nerves since I've been trying to solve it for hours.
Studying for my next test on inequalities with absolute values, I found this one:
$$ |x-3|-|x-4|<x $$
(I precisely found the above inequality on this website, here to be precise, but, the problem is that when I try to solve it, my answer won't be $(-1,+\infty)$, but $(1,7)$. I took the same inequalities that the asker's teacher had given to him and then put them on a number line, and my result was definitely not $(-1,+\infty)$
Here are the inequalities:
$$ x−3 < x−4 +x $$
$$ x−3 < −(x−4) +x $$
$$ −(x−3)<−(x−4)+x $$
And here are my answers respectively:
$$ x>1, \quad x>-1, \quad x<7 $$
I will really appreciate if anyone could help me out, because I'm already stressed dealing with this problem, that by the way, it is not demanding that I solve it, but you know, why not?
Best Answer
How to solve $|x-3|-|x-4|<x$? Let $f(x)=|x-3|-|x-4|$
I would begin by noting $|x-3|=x-3$ for $x \geq 3$ whereas $|x-3|=3-x$ for $x \leq 3$. Likewise, $|x-4|=x-4$ for $x \geq 4$ whereas $|x-4|=4-x$ for $x \leq 4$
if $x \geq 4$ then $x > 3$ hence $f(x)=x-3-(x-4)=1$ and we find $1<x$ which is true in this case. This puts $[4,\infty)$ in the solution set of the inequality.
if $x \leq 3$ then $x < 4$ hence $f(x)=3-x-(4-x)=-1$ and we face $-1<x$ which is true for each $x$ with $-1<x\leq 3$. This shows $(-1,3]$ is also in the solution set of the inequality.
if $3<x<4$ then $f(x) = x-3-(4-x)=2x-7$. Hence in this context we must solve $2x-7<x$ which gives $x<7$ which is true for each $x$ in the interval considered. This places $(3,4)$ in the solution set of the inequality.
Put all of this together, we obtain $(-1,3] \cup (3,4) \cup [4,\infty) = (-1, \infty)$ which is your answer. Usually I solve this sort of thing with a sign-chart method.