4.7 The modulus function
Definition 4.32. The modulus (or absolute value) of \(x\) is its size without regard to sign: \[|x|=\begin {cases} x & \text {if } x\geq 0,\\ -x & \text {if } x<0.\end {cases}\]
So \(|5|=5\) and \(|-5|=5\). Read \(|x|\) as the distance from \(x\) to \(0\) on the number line; that reading makes the rules below obvious rather than memorised.
Note 4.33. For \(k>0\): \[|f(x)|=k\iff f(x)=k \text { or } f(x)=-k\] \[|f(x)|<k\iff -k<f(x)<k\] \[|f(x)|>k\iff f(x)>k \text { or } f(x)<-k\] In words: an equation gives two cases, “less than” gives one interval between two bounds, and “greater than” gives two separate pieces.
Solution. (a) Two cases: \[2x-3=7\quad \implies \quad 2x=10\quad \implies \quad x=5\] \[2x-3=-7\quad \implies \quad 2x=-4\quad \implies \quad x=-2.\] So \(x=5\) or \(x=-2\).
(b) Now a single double inequality: \[-7<2x-3<7.\] Add \(3\) throughout, then divide by \(2\) — both positive operations, so no sign flips: \[-4<2x<10\quad \implies \quad -2<x<5.\] So the solution is \((-2,5)\).
Reading the two answers together. Part (a) gave the two endpoints and part (b) gives everything between them. That is the distance picture: \(|2x-3|\) is the distance from \(2x\) to \(3\), and asking for it to be less than \(7\) is asking for the values that sit strictly inside the two points where it equals \(7\).
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