2.1 Absolute value equations

Absolute value equation (no variable on the other side)

$3|2x+1|-7=5$

• Step 1: Algebraically isolate the absolute value
  • $3|2x+1|=12$
  • $|2x+1|=4$
• Step 2: Think!
  • For the moment, forget about the quantity $2x+1$ ; just think of it as something . The absolute value of “something” is 4. So, in analogy to what we did before, the “something” can either be 4, or –4. So that gives us two possibilities...
  • $\mathrm{2x}+1=4$
  • $\mathrm{2x}+1=-4$
• Step 3: Algebraically solve (both equations) for x
  • $2x=3$ or $2x=-5$
  • $x=\frac{3}{2}$ or $x=-\frac{5}{2}$

So this problem has two answers: $x=\frac{3}{2}$ and $x=-\frac{5}{2}$

Absolute value equation (no variable on the other side)

$\frac{6\mid x-2\mid }{5}+7=4$

• Step 1: Algebraically isolate the absolute value
  $\frac{6|x-2|}{5}=-3$
  $6|x-2|=-15$
  $|x-2|=\frac{-15}{6}=\frac{-5}{2}$
• Step 2: Think! The absolute value of “something” is $-2\frac{1}{2}$ . But wait—absolute values are never negative! It can’t happen! So we don’t even need a third step in this case: the equation is impossible.
• No solution.

Absolute value equation with variables on both sides

$\mid \mathrm{2x}+3\mid =-\text{11}x+\text{42}$

We begin by approaching this in analogy to the first problem above, $\mid x\mid =\text{10}$ . We saw that $x$ could be either 10, or –10. So we will assume in this case that $\mathrm{2x}+3$ can be either $-\text{11}x+\text{42}$ , or the negative of that, and solve both equations.

Problem one

• $2x+3=-11x+42$
• $13x+3=42$
• $13x=39$
• $x=3$

Problem two

• $2x+3=-(-11x+42)$
• $2x+3=11x-42$
• $-9x+3=-42$
• $-9x=-45$
• $x=5$

So we have two solutions: $x=3$ and $x=5$ . Do they both work? Let’s try them both.

Problem one

1. $\mid 2(3)+3\mid =-\text{11}(3)+\text{42}$ .
   • $\mid 9\mid =9$ . $✓$

Problem two

1. $\mid 2(5)+3\mid =-\text{11}(5)+\text{42}$ .
   • $\mid \text{13}\mid =-\text{13}$ . $✗$