Concept understanding — Absolute Value: Equations and Inequalities
Definition.∣x∣=x if x≥0, and ∣x∣=−x if x<0 -- the distance of x from 0 on the number line. Consequently ∣x∣≥0 always, and ∣x∣=∣−x∣.
Solving equations.∣u∣=r (with r≥0) splits into u=r or u=−r; if r<0 there is no solution, since an absolute value can never be negative. ∣u∣=∣v∣ splits into u=v or u=−v.
Solving inequalities -- the two master rules:
∣x∣<r⟺−r<x<r,∣x∣>r⟺x<−r or x>r,
proved by splitting into the cases x≥0 and x<0. Shifted forms: ∣x−a∣≤r⟺x∈[a−r,a+r]; ∣x−a∣≥r⟺x∈(−∞,a−r]∪[a+r,∞).
Algebraic identities.∣xy∣=∣x∣∣y∣; yx=∣y∣∣x∣ (y=0); the triangle inequality∣x+y∣≤∣x∣+∣y∣; and if ∣y+x∣=∣x−y∣ then xy=0.
Watch out
∣u∣≥(negative number) is true for EVERY real u; ∣u∣<(negative number) has NO solution -- always check the sign of the bound before mechanically unfolding a compound inequality. Also, dividing/multiplying by a negative constant while isolating ∣u∣ flips the inequality's direction, just as with any ordinary inequality.
Unfold each absolute-value inequality using ∣u∣<r⟺−r<u<r (and remember ∣u∣≥ a negative number is always true).
✓Final answer
(−4,10);
all x∈R;
[311,313];
(−7,7).
Step 1 (i).∣3−x∣<7⟺−7<3−x<7. Subtract 3: −10<−x<4. Multiply by −1 (flip): −4<x<10, i.e. x∈(−4,10).
Step 2 (ii). The left side ∣4x−5∣ is always ≥0, and 0≥−2, so ∣4x−5∣≥−2 holds automatically for every real x. Solution: all x∈R.
Step 3 (iii).3−43x≤41⟺−41≤3−43x≤41. Subtract 3: −413≤−43x≤−411. Multiply by −34 (flip both ends): 313≥x≥311, i.e. x∈[311,313].
Step 4 (iv).∣x∣−10<−3⟺∣x∣<7⟺−7<x<7, i.e. x∈(−7,7).
✓Final answer
x∈(−4,10);
x∈R;
x∈[311,313];
x∈(−7,7).
Unfold ∣u∣<r as a compound inequality −r<u<r, flipping direction on negative multiplication
Forgetting to flip the inequality signs when multiplying/dividing by a negative number.
Missing that ∣⋅∣≥(negative number) is true for all x -- treating (ii) as a normal bounded interval.