Mathematics - Additional - 0606 CIE

Modulus Equations

Akopọ

The modulus, or absolute value, of a number is its distance from zero, so it is never negative. That single idea, distance, is the key to solving every modulus equation. Strip away the bars and you are left with two ordinary equations to consider, one for each sign.

In this lesson you will solve equations of the form \(|ax+b|=c\), \(|ax+b|=cx+d\) and \(|ax+b|=|cx+d|\). The guiding rule is simple: if \(|\text{something}|=c\) then that something is \(+c\) or \(-c\). You will also learn the vital habit of checking your solutions, because the modulus can quietly create answers that do not work.

Awọn Afojusun

  1. Solve equations involving the modulus, such as |ax + b| = c, |ax + b| = cx + d and |ax + b| = |cx + d|, using algebraic or graphical methods.

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Akọ̀wé Ẹ̀kọ́

Modulus expresses size without direction: how far, how much error, how big a difference, regardless of sign. Solving modulus equations trains you to split a problem into clean cases and to check answers, two habits that pay off across the whole syllabus and in any work involving tolerances.

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  1. Solve |2x - 1| = 5. A. x = 3 or x = -2 B. x = 2 or x = -3 C. x = 3 only D. x = -2 only Answer: A
  2. The equation |A| = c (with c >= 0) is equivalent to: A. A = c only B. A = c or A = -c C. A = -c only D. A = 0 Answer: B
  3. Which value is never possible for |x|? A. 0 B. 5 C. -3 D. 2.5 Answer: C
  4. Solve |x - 1| = |2x + 1|. A. x = -2 or x = 0 B. x = 2 or x = 0 C. x = 1 or x = -1 D. x = -2 only Answer: A
  5. When solving |x - 2| = 2x - 7, x = 3 is rejected because: A. it does not solve case 2 B. it makes the right-hand side negative C. it makes |x - 2| negative D. it is not an integer Answer: B

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