4 Important Types of Absolute Value Equations

4 Important Types of Absolute Value Equations

There are 4 main types of absolute value equations regarding whether there are;

  • absolute value and a static value
  • absolute value and an expression involving unknown pronumerals
  • two absolute values on both sides
  • two absolute values and a value

Type 1: One Absolute Value and a Constant

Solve \( | x-2 | = 5 \).

\( \begin{aligned} \displaystyle
x-2 = 5 &\text{ or } x-2 = -5 \\
\therefore x = 7 &\text{ or } x = -3
\end{aligned} \)

Type 2: One Absolute Value and a Linear Expression

Solve \( | x-1 | = 2x+4 \).

\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
(|x-1|)^2 &= (2x+4)^2 &\color{red} \text{squaring both sides} \\
(x-1)^2 &= (2x+4)^2 \\
x^2-2x + 1 &= 4x^2 + 16x + 16 \\
3x^2 + 18x + 15 &= 0 \\
x^2 + 6x + 5 &= 0 \\
(x+5)(x+1) &= 0 \\
x = -5 &\text{ or } x = -1 \\
\color{red} \text{Test } x = -5 \\
\text{LHS} &= |-5-1| \\
&= |-6| \\
&= 6 \\
\text{RHS} &= 2 \times -5 +4 \\
&= -6 \\
\text{LHS} &\ne \text{RHS} \\
\therefore x &\ne -5 \\
\color{red} \text{Test } x = -1 \\
\text{LHS} &= |-1-1| \\
&= |-2| \\
&= 2 \\
\text{RHS} &= 2 \times -1 +4 \\
&= 2 \\
\text{LHS} &= \text{RHS} \\
\therefore x &= -1
\end{aligned} \)

Type 3: Two Absolute Value Expressions

Solve \( | x-1 | = |3-x| \).

\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
x-1 &= \pm(3-x) \\
x-1 &= 3-x \color{red} \cdots (1) \\
2x &= 4 \\
x &= 2 \\
x-1 &= -3+x \color{red} \cdots (2) \\
x-x &= -2 \\
0 &= -2 &\color{red} \text{no solution} \\
\therefore x &= 2 &\color{red} \text{from (a) and (2)}
\end{aligned} \)

Type 4: Two Absolute Value Expressions and a Constant

Solve \( | x+2 | + |x-3| = 7 \).

\( \begin{aligned} \displaystyle \require{AMSsymbols}\require{color}
-(x + 2)-(x-3) &= 7 &\color{red} \text{for } x \lt -2 \\
-x-2-x+3 &= 7 \\
-2x &= 6 \\
x &= -3 &\color{red} \text{this is OK for } x \lt -2 \\
(x + 2)-(x-3) &= 7 &\color{red} \text{for } -2 \le x \lt 3 \\
x+2-x+3 &= 7 \\
5 &\ne 7 &\color{red} \text{ no solution}\\
(x + 2) + (x-3) &= 7 &\color{red} \text{for } x \ge 3 \\
x+2+x-3 &= 7 \\
2x &= 8 \\
x &= 4 &\color{red} \text{this is OK for } x \ge 3 \\
\therefore x &= -3 \text{ or } 4
\end{aligned} \)

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