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Exercise 2.2 · Q3

Q.Solve −3∣x∣+5≤−2-3|x|+5\le-2 and graph the solution set in a number line.

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Concept understanding — Absolute Value: Equations and Inequalities

Definition. ∣x∣=x|x|=x if x≥0x\ge0, and ∣x∣=−x|x|=-x if x<0x<0 -- the distance of xx from 00 on the number line. Consequently ∣x∣≥0|x|\ge0 always, and ∣x∣=∣−x∣|x|=|-x|.

Solving equations. ∣u∣=r|u|=r (with r≥0r\ge0) splits into u=ru=r or u=−ru=-r; if r<0r<0 there is no solution, since an absolute value can never be negative. ∣u∣=∣v∣|u|=|v| splits into u=vu=v or u=−vu=-v.

Solving inequalities -- the two master rules:

∣x∣<r  ⟺  −r<x<r,∣x∣>r  ⟺  x<−r or x>r,|x|<r \iff -r<x<r, \qquad |x|>r \iff x<-r \text{ or } x>r,

proved by splitting into the cases x≥0x\ge0 and x<0x<0. Shifted forms: ∣x−a∣≤r  ⟺  x∈[a−r,a+r]|x-a|\le r\iff x\in[a-r,a+r]; ∣x−a∣≥r  ⟺  x∈(−∞,a−r]∪[a+r,∞)|x-a|\ge r\iff x\in(-\infty,a-r]\cup[a+r,\infty).

Algebraic identities. ∣xy∣=∣x∣∣y∣|xy|=|x||y|; ∣xy∣=∣x∣∣y∣\left|\dfrac xy\right|=\dfrac{|x|}{|y|} (y≠0y\ne0); the triangle inequality ∣x+y∣≤∣x∣+∣y∣|x+y|\le|x|+|y|; and if ∣y+x∣=∣x−y∣|y+x|=|x-y| then xy=0xy=0. …

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