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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!

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  • $\begingroup$ 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. $\endgroup$
    – cardinal
    Mar 12, 2011 at 16:46
  • $\begingroup$ 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? $\endgroup$
    – cardinal
    Mar 12, 2011 at 16:48
  • $\begingroup$ @cardinal I thought there was a general direction to the solution besides the identities of absolute value. Thanks :) $\endgroup$
    – Cu7l4ss
    Mar 12, 2011 at 17:10

2 Answers 2

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In my answer here, I describe a method for solving multiple-absolute-value equations that can similarly be applied to inequalities. It is the same idea as what your teacher showed and what's discussed in the comments, but the particular organization of the work by dividing up a number line and working under each section may be helpful.

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

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    $\begingroup$ Your final answer is correct, but the steps between are strange. I guess they refer to OPs approach, but without any words / explanation this is just confusing $\endgroup$
    – Winther
    Oct 1, 2015 at 22:33

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