Porisma

Worked solution

Keep the parameter when translating a modulus interval

Self-Practice 4: solve ∣2x+a∣<5|2x+a|<5, leaving the answer in terms of a.

CJC · Lecture · Self-Practice 4

Translate the interval, then divide by a positive constant

−5<2x+a<5-5<2x+a<5. Subtract a and divide by 2: (−5−a)/2<x<(5−a)/2(-5-a)/2<x<(5-a)/2, retaining the printed answer. The inequality stays strict and its direction is preserved because 2>0.

How did this part go?

Common mistakes

The parameter shifts both endpoints; it does not change the positive divisor 2.

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