Solving inequality with two absolute values

Hey, ! In my pre-calculus class the teacher showed the solution of the following example: \begin{align} \vert x-3 \vert \lt \vert x - 4 \vert + x \end{align} He started by stated the domains needed to be checked:

\begin{aligned} \lbrack 4, +\infty ) \newline \lbrack 3, 4 ) \newline ( -\infty, 3) \end{aligned}

Which I don't have a based lead on how he come to these domains and then deducted the following inequalities ( for each domain respectively): \begin{align} x-3 \lt x - 4 + x \newline x-3 \lt -(x-4) + x \newline -(x-3) \lt -(x-4) + x \end{align}

The final solution was \begin{aligned} (-1, +\infty ) \end{aligned}

Now I cannot understand how he deducted the domains and the right inequalities(there is one missing possibility): \begin{align} -(x-3) \lt x-4 + x \end{align}

It may be apparent but I still cannot wrap my mind around it. Thanks!

• the domains you need to check are based on the definition of the absolute value function. Let $a$ be a constant. Then if $x < a$, $|x - a| = -(x-a)$ and if $x \geq a$, then $|x - a| = x - a$. (Why?) This should help you understand how the domains were chosen. A little more thought will reveal while your last case does not need to be considered. – cardinal Mar 12 '11 at 16:46
• also it helps to visualize these things. Rewrite your first inequality as $|x - 4| - |x - 3| + x$ > 0. Graph the function on the left-hand side. When is it (strictly) above the $x$-axis? – cardinal Mar 12 '11 at 16:48
• @cardinal I thought there was a general direction to the solution besides the identities of absolute value. Thanks :) – Cu7l4ss Mar 12 '11 at 17:10

We have $x>1$: $$x>-1$$ $$x>7/3$$ $$x<7$$ so solution set is $x>-1$.