Lesson 1 of 4 · about 20 minutes · Win: solve any of the six equation types from your notes, including the “no solution” traps.
1. What |x| means
The absolute value (or modulus) |x| is the distance of x from zero, so it is never negative. (abs.pdf p.1)
Definition (your notes): |x| = x if x ≥ 0, and |x| = −x if x < 0.
Another definition: |x| = √(x²).
Useful: |a − b| = |b − a|, e.g. |2 − 5| = |5 − 2| = 3.
Note: −x is positive when x is negative: |−6| = −(−6) = 6.
2. Solving equations: the recipe
Get the absolute value alone on one side (notes: “Get absolute value alone!”).
Restriction check: if |…| = a negative number → no solution. If the right side contains x, write the restriction “RHS ≥ 0”.
Split into two cases: inside = RHS or inside = −(RHS).
Check every answer against the restriction / original equation.
abs.pdf p.2: “If |x| = 4 then x = 4 or x = −4. If |x| = −4 then there is no solution (restriction).”
3. The six examples from your notes, worked
Example (abs.pdf p.2)
Working
Answer
1. |x + 3| = 5
x+3 = 5 or x+3 = −5
x = 2 or x = −8
2. 2|x − 4| = 3
Alone first: |x−4| = 3/2 → x−4 = ±3/2
x = 11/2 or x = 5/2
3. |x − 3| = 2x − 1
Restriction 2x−1 ≥ 0 ⇒ x ≥ ½. x−3 = 2x−1 ⇒ x = −2 ✘ (breaks restriction) x−3 = −(2x−1) ⇒ 3x = 4 ⇒ x = 4/3 ✔
x = 4/3 only
4. |x| = |x + 2|
x = x+2 (impossible: 0 = 2) or x = −(x+2) ⇒ 2x = −2
x = −1
5. |x + 4| + 6 = 4
Alone: |x+4| = −2 < 0
No solution
6. |3x − 4| + 4 = x
Alone: |3x−4| = x−4, restriction x ≥ 4. 3x−4 = x−4 ⇒ x = 0 ✘; 3x−4 = −(x−4) ⇒ x = 2 ✘
No solution
Check example 3: |4/3 − 3| = 5/3 and 2(4/3) − 1 = 5/3 ✔.
4. Inequalities (short; not in your abs.pdf pages)
Your absolute value notes (abs.pdf) do not contain inequalities. If your teacher covered them, use this standard rule (Paul’s Online Notes) and write answers in the interval notation from your Term 2 inequality notes (term2.pdf pp.2–7): |u| < k (k>0) ⇒ −k < u < k (“between”) |u| > k ⇒ u < −k or u > k (“outside”)
e.g. |x − 2| < 3 ⇒ −1 < x < 5, i.e. x ∈ (−1; 5). Source: Paul’s Online Notes – Absolute Value Inequalities