🎯⭐ INTERACTIVE LESSON

Absolute Value Equations

Learn step-by-step with interactive practice!

← Back to Standard Lesson

Absolute Value Equations - Complete Interactive Lesson

Part 1: What Absolute Value Really Means

📏 Absolute Value Equations

Part 1 of 5 — What Absolute Value Really Means

Topics in This Part

Section
Absolute Value as Distance
Why Two Answers?
The Core Rule: $\lvert x\rvert = a$

🔑 Key Concept: $\lvert x\rvert$ is the distance from $x$ to $0$ on the number line. Distance is never negative — and two different numbers can sit the same distance from zero. That single idea drives every equation in this lesson.

Absolute Value as Distance

The absolute value of a number is how far it is from $0$ , ignoring direction:

$\lvert 5\rvert = 5 \qquad \lvert -5\rvert = 5 \qquad \lvert 0\rvert = 0$

Both and sit from zero, so they share the same absolute value.

Concept Check 🎯

The Core Rule

Because two numbers share each positive distance, an absolute value equation splits into two cases:

$\boxed{\;\lvert x\rvert = a \;\Longrightarrow\; x = a \;\text{ or }\; x = -a\;}\quad (a > 0)$

Pick the Two Cases 🔽

For each equation, choose the correct pair of cases.

From Cases to Numbers

Once you've written the two cases, each one is just an ordinary equation to solve. For $\lvert x\rvert = 4$ the cases are already solved ( $x = 4$ or $x = -4$ ). For $∣ x - 1 ∣ =$ you solve each branch:

Set Up the Two Cases 🧮

For each equation, enter the larger solution in the first box and the smaller in the second.

1) $\lvert x\rvert = 4$ : larger $= \,?$ , smaller $= \,?$ 2) $∣ x - 1 ∣$ : larger , smaller

Part 2: Solving the Two Cases

📏 Absolute Value Equations

Part 2 of 5 — Solving the Two Cases

🔑 The Plan: Split $\lvert \text{expression}\rvert = a$ into two ordinary linear equations — one positive, one negative — and solve each. Two branches, two answers.

The Method

To solve $\lvert mx + b\rvert = a$ (with ):

Part 3: Isolate First, Then Watch for Special Cases

📏 Absolute Value Equations

Part 3 of 5 — Isolate First, Then Watch for Special Cases

🔑 Golden Rule: You may only split into two cases once the absolute value stands completely alone on one side. Anything multiplied, added, or subtracted outside the bars must be undone first.

Isolate the Absolute Value

If something is attached outside the bars, peel it away first — exactly like solving any linear equation.

Worked Example: $3\lvert x + 1\rvert - 2 = 7$

Add $2$ :

Part 4: Variables on Both Sides & Extraneous Solutions

📏 Absolute Value Equations

Part 4 of 5 — Variables on Both Sides & Extraneous Solutions

🔑 New Wrinkle: When a variable sits on the right (like $\lvert x - 2\rvert = 3x$ ), splitting still works — but one branch can produce a value that doesn't actually check. Those are extraneous solutions, and you must throw them out.

Why Extraneous Solutions Appear

The split $\lvert A\rvert = B \Rightarrow A = B$ or secretly (an absolute value can't equal a negative). When contains a variable, a branch might hand you an that makes negative — impossible. So:

Part 5: Mixed Practice & Mastery Check

📏 Absolute Value Equations

Part 5 of 5 — Mixed Practice & Mastery Check

You can now (1) read $\lvert x\rvert$ as distance, (2) split into two cases, (3) isolate first and handle the no-solution / one-solution specials, and (4) reject extraneous roots. Let's put it together.

Quick Reference

Situation	What to do
$\lvert x\rvert = a,\; a>0$

        |-- 5 units --|-- 5 units --|
  ◄─────┼─────────────┼─────────────┼─────►
       -5             0             +5

💡 Read it as a question: " $\lvert x\rvert = 5$ " asks "what numbers are 5 units from zero?" The answer is both $5$ and $-5$ .

Expression	Value
$\lvert 7\rvert$	$7$
$\lvert -7\rvert$	$7$
$\lvert -\tfrac{3}{4}\rvert$	$\tfrac{3}{4}$
$-\lvert 6\rvert$	$-6$ (the minus is outside* the bars)*

Example: $\lvert x\rvert = 9 \Rightarrow x = 9$ or $x = -9$ .

The inside doesn't have to be a single letter — it can be a whole expression:

$\lvert x - 3\rvert = 5 \;\Longrightarrow\; x - 3 = 5 \;\text{ or }\; x - 3 = -5$

Solving each branch gives $x = 8$ or $x = -2$ . (Both are exactly $5$ units from $3$ .)

⚠️ Don't average the bars away. $\lvert x - 3\rvert$ is not $x - 3$ and not $|x| - 3$ . Keep the whole expression inside the bars until you split into the two cases.

$x - 1 = 6 \Rightarrow x = 7 \qquad x - 1 = -6 \Rightarrow x = -5$

Try the next drill to lock in the full setup-and-solve flow.

Write both cases: $mx + b = a \qquad \text{and} \qquad mx + b = -a$
Solve each linear equation separately.
Check by substituting back — both should make the inside come out to $\pm a$ .

Worked Example: $\lvert 2x - 3\rvert = 7$

$2x - 3 = 7 \;\Rightarrow\; 2x = 10 \;\Rightarrow\; x = 5$ $2x - 3 = -7 \;\Rightarrow\; 2x = -4 \;\Rightarrow\; x = -2$

✅ Check: $\lvert 2(5)-3\rvert = \lvert 7\rvert = 7$ ✓ and $\lvert 2(-2)-3\rvert = \lvert -7\rvert = 7$ ✓

Worked Example: $\lvert 3x + 6\rvert = 12$

$3x + 6 = 12 \;\Rightarrow\; 3x = 6 \;\Rightarrow\; x = 2$ $3x + 6 = -12 \;\Rightarrow\; 3x = -18 \;\Rightarrow\; x = -6$

Solutions: $x = 2$ or $x = -6$ .

💡 Notice the symmetry. Both solutions sit the same distance from the value that makes the inside zero. Here $3x+6=0$ at $x=-2$ , and $2$ and $-6$ are each away from .

Concept Check 🎯

A Guided Walk-Through

Let's slow down on $\lvert 3x + 6\rvert = 12$ to see every blank get filled. The inside is $3x+6$ , so the two cases set that whole expression equal to $+12$ and $-12$ :

Positive case: $3x + 6 = 12$
Negative case: $3x + 6 = -12$

Solve each like a normal one-step-then-divide problem. The next check has you assemble exactly these pieces.

Build the Solution 🔽

You're solving $\lvert 3x + 6\rvert = 12$ . Choose what goes in each blank.

Reporting One Root

Some problems ask for just one of the two roots — often the larger. Solve both cases as usual, then compare:

$\lvert x + 4\rvert = 10 \;\Rightarrow\; x = 6 \;\text{ or }\; x = -14 \;\Rightarrow\; \text{larger} = 6$

The drill below asks for the larger root each time, so finish both branches before you answer.

Solve It 🧮

Solve each equation. Enter the larger solution.

1) $\lvert x + 4\rvert = 10 \Rightarrow x = \,?$ (larger root) 2) $\lvert 2x - 8\rvert = 4 \Rightarrow x = \,?$ (larger root)

\;3\lvert x + 1\rvert = 9

Divide by

3

\;\lvert x + 1\rvert = 3

Now split into two cases:

x + 1 = 3 \;\Rightarrow\; x = 2 \qquad x + 1 = -3 \;\Rightarrow\; x = -4

Solutions: $x = 2$ or $x = -4$ .

⚠️ Never split too early. Splitting $3\lvert x+1\rvert-2=7$ before isolating gives wrong cases. Get to $\lvert x+1\rvert = 3$ first.

Concept Check 🎯

Two Special Cases

Once isolated, look at the number on the right:

Equation form	Number of solutions	Why
$\lvert x\rvert = a,\; a > 0$	Two	two points at distance $a$
$\lvert x\rvert = 0$	One	only $0$ is $0$ units from $0$
$\lvert x\rvert = a,\; a < 0$	None	distance can't be negative

Example: $\lvert x - 4\rvert = 0$

A distance of $0$ means the inside is $0$ : $\;x - 4 = 0 \Rightarrow x = 4$ . Just one solution.

Example: $2\lvert x - 5\rvert + 8 = 2$

Isolate: $2\lvert x-5\rvert = -6 \Rightarrow \lvert x - 5\rvert = -3$ . An absolute value can never equal $-3$ , so there is .

💡 Spot it fast: after isolating, if the right side is negative, stop — no solution. If it's exactly zero, there's exactly one solution.

How Many Solutions? 🔽

Isolate first (in your head), then pick the number of solutions.

Putting It Together

The full routine now has three moves:

Isolate the absolute value (undo anything outside the bars).
Check the right side — negative means no solution, zero means one solution.
Split into two cases and solve.

The drill below mixes a normal two-solution problem, a negative-coefficient isolate, and a zero special case.

Isolate, Then Solve 🧮

1) $3\lvert x + 1\rvert - 2 = 7$ . Enter the larger solution. 2) $-4\lvert x - 1\rvert = -20$ . Enter the larger solution. 3) $\lvert x - 4\rvert = 0$ . Enter the only solution.

⚠️ Always substitute every candidate back into the original equation. Keep only the ones that make a true statement. Discard the rest as extraneous.

Worked Example: $\lvert x + 5\rvert = 2x - 1$

Case 1 (positive): $x + 5 = 2x - 1 \Rightarrow 6 = x$ , so $x = 6$ . Check: $\lvert 6+5\rvert = \lvert 11\rvert = 11$ and $2(6)-1 = 11$ . ✓ Keep.

Case 2 (negative): $x + 5 = -(2x - 1) = -2x + 1 \Rightarrow 3x = -4 \Rightarrow x = -\tfrac{4}{3}$ . Check: right side $= 2(-\tfrac{4}{3}) - 1 = -\tfrac{11}{3}$ , which is negative. An absolute value can't be negative. ✗ Reject.

Only solution: $x = 6$ .

Concept Check 🎯

A Case Where BOTH Branches Survive

Not every problem has an extraneous root — you still must check to be sure.

Worked Example: $\lvert 2x + 1\rvert = x + 4$

Case 1: $2x + 1 = x + 4 \Rightarrow x = 3$ . Check: $\lvert 2(3)+1\rvert = \lvert 7\rvert = 7$ and $3 + 4 = 7$ . ✓

Case 2: $2x + 1 = -(x + 4) = -x - 4 \Rightarrow 3x = -5 \Rightarrow x = -\tfrac{5}{3}$ . Check: and . ✓

Both work: $x = 3$ or $x = -\tfrac{5}{3}$ .

💡 Takeaway: Solve both branches, then let the check decide. Sometimes you keep both, sometimes one, sometimes none.

Keep or Reject? 🔽

For $\lvert x - 2\rvert = 3x$ , the two branches give $x = -1$ and $x = \tfrac{1}{2}$ . Decide each, then count.

Your Extraneous-Root Checklist

When the right side holds a variable, finish with these two questions for every candidate:

Does substituting it into the original make a true statement?
Is the resulting right side non-negative? (An absolute value can't equal a negative number.)

If either answer is no, the root is extraneous — cross it out. Now try a problem with one rejected root and one with both roots valid.

Solve and Check 🧮

Solve, discard any extraneous roots, and enter the valid solution(s).

1) $\lvert x + 5\rvert = 2x - 1$ . Enter the only valid solution. 2) $\lvert 2x + 1\rvert = x + 4$ . Enter the larger valid solution.

⚠️ Two habits that prevent most mistakes: isolate the bars before splitting, and substitute every candidate back into the original equation.

Mixed Practice 🎯

One Reliable Order of Operations

Whenever you meet an absolute value equation, run the same checklist:

Isolate the bars on one side.
Inspect the right side: negative → no solution; zero → one solution; positive → continue.
Split into the positive and negative cases and solve each.
Check every candidate in the original (especially when a variable is on the right).

The next drill rehearses an isolate-first problem and an extraneous-root problem back to back.

One More Round 🧮

1) $2\lvert x\rvert + 1 = 11$ . Enter the positive solution. 2) $\lvert x + 5\rvert = 2x - 1$ (check for extraneous!). Enter the valid solution.

Ready for the Exit Quiz

You've practiced every move: reading $\lvert x\rvert$ as distance, splitting into two cases, isolating before splitting, the no-solution / one-solution specials, and rejecting extraneous roots. The three questions below pull from all of it — take your time and check each answer.

Exit Quiz ✅

Answer all three to finish the lesson.

\lvert 2(-\tfrac{5}{3})+1\rvert = \lvert -\tfrac{7}{3}\rvert = \tfrac{7}{3}

-\tfrac{5}{3} + 4 = \tfrac{7}{3}

Absolute Value Equations

Absolute Value Equations - Complete Interactive Lesson

Part 1: What Absolute Value Really Means

📏 Absolute Value Equations

Topics in This Part

Absolute Value as Distance

The Core Rule

From Cases to Numbers

Part 2: Solving the Two Cases

📏 Absolute Value Equations

The Method

Part 3: Isolate First, Then Watch for Special Cases

📏 Absolute Value Equations

Isolate the Absolute Value

Worked Example: 3∣x+1∣−2=73\lvert x + 1\rvert - 2 = 73∣x+1∣−2=7

Part 4: Variables on Both Sides & Extraneous Solutions

📏 Absolute Value Equations

Why Extraneous Solutions Appear

Part 5: Mixed Practice & Mastery Check

📏 Absolute Value Equations

Quick Reference

Worked Example: ∣2x−3∣=7\lvert 2x - 3\rvert = 7∣2x−3∣=7

Worked Example: ∣3x+6∣=12\lvert 3x + 6\rvert = 12∣3x+6∣=12

A Guided Walk-Through

Reporting One Root

Two Special Cases

Example: ∣x−4∣=0\lvert x - 4\rvert = 0∣x−4∣=0

Example: 2∣x−5∣+8=22\lvert x - 5\rvert + 8 = 22∣x−5∣+8=2

Putting It Together

Worked Example: ∣x+5∣=2x−1\lvert x + 5\rvert = 2x - 1∣x+5∣=2x−1

A Case Where BOTH Branches Survive

Worked Example: ∣2x+1∣=x+4\lvert 2x + 1\rvert = x + 4∣2x+1∣=x+4

Your Extraneous-Root Checklist

One Reliable Order of Operations

Ready for the Exit Quiz

Worked Example: $3\lvert x + 1\rvert - 2 = 7$

Worked Example: $\lvert 2x - 3\rvert = 7$

Worked Example: $\lvert 3x + 6\rvert = 12$

Example: $\lvert x - 4\rvert = 0$

Example: $2\lvert x - 5\rvert + 8 = 2$

Worked Example: $\lvert x + 5\rvert = 2x - 1$

Worked Example: $\lvert 2x + 1\rvert = x + 4$