Solve the equation $|5 - 2x| - |x + 4| = 0$ for all real values of $x$. Ensure correct treatment of negative coefficients inside the absolute value bars.

Solve the equation $|5 - 2x| - |x + 4| = 0$ for all real values of $x$ . Ensure correct treatment of negative coefficients inside the absolute value bars.

Final answer

$x = \frac{1}{3}$ or $x = 9$

Steps

Isolate the absolute value expressions

To solve the equation $|5 - 2x| - |x + 4| = 0$ , follow these steps:

Add $|x + 4|$ to both sides of the equation: $|5 - 2x| = |x + 4|$

(Note on negative coefficients: Since $|-a| = |a|$ , you can also rewrite $|5 - 2x|$ as $|-(2x - 5)| = |2x - 5|$ . Both forms yield the exact same results.)

Apply the absolute value equality rule

For any real expressions $A$ and $B$ , the equation $|A| = |B|$ holds if and only if: $A = B \quad \text{or} \quad A = -B$

Setting $A = 5 - 2x$ and $B = x + 4$ , we split the problem into two separate linear equations.

Solve Case 1 ( $A = B$ )

$5 - 2x = x + 4$

Add $2x$ to both sides and subtract $4$ from both sides: $5 - 4 = x + 2x$ $1 = 3x$ $x = \frac{1}{3}$

Solve Case 2 ( $A = -B$ )

$5 - 2x = -(x + 4)$

Distribute the negative sign carefully across all terms inside the parentheses: $5 - 2x = -x - 4$

Add $2x$ to both sides and add $4$ to both sides: $5 + 4 = 2x - x$ $x = 9$

Verify the solutions

For $x = \frac{1}{3}$ : $|5 - 2\left(\frac{1}{3}\right)| - \left|\frac{1}{3} + 4\right| = \left|\frac{13}{3}\right| - \left|\frac{13}{3}\right| = 0 \quad \checkmark$
For $x = 9$ : $|5 - 2(9)| - |9 + 4| = |-13| - |13| = 13 - 13 = 0 \quad \checkmark$

Thus, the solutions are $x = \frac{1}{3}$ or $x = 9$ .

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student improperly negates the negative coefficient term inside the absolute value as positive without changing the constant term, rewriting |5 - 2x| as 5 + 2x.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
|5 - 2x| = |x + 4|
Step 2
5 + 2x = x + 4 $\text{ or }$ 5 + 2x = -(x + 4)
Correct
5 - 2x = x + 4 $\text{ or }$ 5 - 2x = -(x + 4)
When solving $|A| = |B|$ , you must split into $A = B$ or $A = -B$ while keeping the expressions inside the bars unchanged. You changed $-2x$ to $+2x$ , but $|5 - 2x|$ is equal to $|2x - 5|$ , not $5 + 2x$ .
Step 3
x = 4 - 5 = -1
Correct
5 - 2x = x + 4 $\implies$ 3x = 1 $\implies$ x = $\frac{1}{3}$
Your algebraic solving of $5 + 2x = x + 4$ was mathematically consistent, but the equation itself came from the error in Step 2.
Step 4
5 + 2x = -x - 4 $\implies$ 3x = -9 $\implies$ x = -3
Correct
5 - 2x = -x - 4 $\implies$ x = 9
Your algebraic expansion and solving for the second branch were mechanically sound, but propagated from the altered expression in Step 2.
Step 5
x = -1 $\text{ or }$ x = -3
Correct
x = $\frac{1}{3}$ $\text{ or }$ x = 9
The final values are incorrect because they were obtained from equations with an incorrectly rewritten absolute value expression.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly isolated the absolute value terms on opposite sides of the equation.

You recognized that an equation of the form $|A| = |B|$ splits into two cases: $A = B$ and $A = -B$ .

You correctly distributed the negative sign when expanding $-(x + 4)$ to $-x - 4$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Improper Sign Modification Inside Absolute Value

You turned the subtraction sign in $5 - 2x$ into addition when removing the absolute value bars.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing the absolute value sign-reversal property $|-E| = |E|$ with changing the sign of an individual term. Negating the inside of an absolute value negates every term: $|5 - 2x| = |-(5 - 2x)| = |2x - 5|$ , never $5 + 2x$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Keep the Inner Expression Intact: For $|A| = |B|$ , write $A = B$ or $A = -B$ without modifying any terms inside $A$ or $B$ .
Copy the expressions exactly as they appear inside the bars before solving each linear branch.
Full Negation Rule: If you choose to flip signs inside absolute value bars using $|E| = |-E|$ , you must multiply the entire expression by $-1$ .
Check: $|5 - 2x| = |-(5 - 2x)| = |-5 + 2x| = |2x - 5| \neq 5 + 2x$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Set up Case 1: $5 - 2x = x + 4$
Solve Case 1: $5 - 4 = x + 2x \implies 1 = 3x \implies x = \frac{1}{3}$
Set up Case 2: $5 - 2x = -(x + 4) = -x - 4$
Solve Case 2: $5 + 4 = 2x - x \implies x = 9$

Final answer

x = $\frac{1}{3}$ $\text{ or }$ x = 9

NEXT STEPS

Final Takeaways from Your Teacher

Practice splitting $|A| = |B|$ directly by copying $A$ and $B$ verbatim into $A = B$ and $A = -B$ .
Remember that absolute value bars do not act on individual minus signs inside an expression; you cannot change $5 - 2x$ to $5 + 2x$ .
Always plug your final values back into the original equation to verify that both sides evaluate to zero.

OverviewACEpath Teacher's Diagnosis

02 The student fails to distribute the negative sign to all terms when taking the opposite case, writing -(x + 4) as -x + 4.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
$|5 - 2x| = |x + 4|$
Step 2
$5 - 2x = x + 4 \implies 5 - 4 = 3x \implies 3x = 1 \implies x = \frac{1}{3}$
Step 3
$5 - 2x = -x + 4 \implies 5 - 4 = -x + 2x \implies x = 1$
Correct
$5 - 2x = -(x + 4) \implies 5 - 2x = -x - 4 \implies x = 9$
When negating the right-hand side, the negative sign must apply to the entire quantity using parentheses: $-(x + 4) = -x - 4$ . Omitting the parentheses left the constant $+4$ unchanged.
Step 4
$x = \frac{1}{3} \text{ or } x = 1$
Correct
$x = \frac{1}{3} \text{ or } x = 9$
Because of the distribution error in the negative case, your second solution was $x = 1$ instead of $x = 9$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly isolated the absolute value terms on opposite sides of the equation.

You recognized that an equation of the form $|A| = |B|$ splits into two linear cases.

You solved the first linear equation $5 - 2x = x + 4$ with precise algebraic steps.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Sign distribution error in absolute value split

You wrote $-x + 4$ instead of $-(x + 4)$ , forgetting to negate the constant term $+4$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Treating the negative case as simply prepending a minus sign to the leading term rather than taking the opposite of the entire polynomial expression using parentheses.

REASONING PATH

Your ACEpath teacher’s suggested approach

Use Parentheses for the Opposite Case: When $|A| = |B|$ , the second case is $A = -(B)$ , where the negative sign distributes to every term of $B$ .
Always write explicit parentheses around the expression when setting up $A = -(B)$ before simplifying.
Quick Verification via Substitution: Substituting your potential solutions back into the original equation checks both your arithmetic and signs.
Substitute $x = 1$ into $|5 - 2x| - |x + 4|$ : $|3| - |5| = -2 \neq 0$ , which immediately signals a sign error.

CORRECTED PATH

ACEpath Teacher's Demonstration

Apply parentheses to negate the entire expression: $5 - 2x = -(x + 4)$ .
Distribute the negative sign across both terms: $5 - 2x = -x - 4$ .
Add $2x$ and add $4$ to both sides: $5 + 4 = 2x - x$ .
Simplify to obtain the second solution: $x = 9$ .

Final answer

$x = \frac{1}{3}$ or $x = 9$

NEXT STEPS

Final Takeaways from Your Teacher

Review how to distribute a negative sign across a binomial before removing grouping symbols.
Practice always writing grouping parentheses when setting up the opposite case $A = -(B)$ in absolute value equations.
Check your candidate solutions by plugging them back into the original equation.

OverviewACEpath Teacher's Diagnosis

03 The student simply drops the absolute value bars without considering the negative case, finding only one solution.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
|5 - 2x| = |x + 4|
Step 2
-3x = 4 - 5 $\implies$ -3x = -1
Step 3
x = $\frac{1}{3}$
Correct
x = $\frac{1}{3}$ $\text{or}$ x = 9
You stated $x = \frac{1}{3}$ as the complete solution, omitting the negative case $5 - 2x = -(x + 4)$ which produces the second valid solution $x = 9$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly isolated the absolute value terms on opposite sides of the equation.

You accurately solved the first linear equation using valid algebraic manipulation.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Omitted Absolute Value Branch

You solved only the case where the expressions inside the absolute values are identical, forgetting that opposite quantities also have equal absolute values.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Treating $|A| = |B|$ as simply $A = B$ rather than the two-branch definition $A = B$ or $A = -B$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Two-branch rule for absolute value equality: For any algebraic expressions $A$ and $B$ , $|A| = |B| \iff A = B \text{ or } A = -B$ .
Whenever you see two absolute values equated, write down two separate branch equations before doing any algebra.
Parentheses when negating: The opposite case is $A = -(B)$ , where the negative sign distributes to every term.
Always place parentheses around the entire side when applying the negative sign: $5 - 2x = -(x + 4)$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Set up the second branch: $5 - 2x = -(x + 4)$ .
Distribute the negative sign: $5 - 2x = -x - 4$ .
Add $2x$ and add $4$ to both sides: $5 + 4 = 2x - x$ , which gives $x = 9$ .

Final answer

x = $\frac{1}{3}$ $\text{or}$ x = 9

NEXT STEPS

Final Takeaways from Your Teacher

Write out both branches, $A = B$ and $A = -B$ , side-by-side as soon as you isolate $|A| = |B|$ .
Substitute both solutions back into the original absolute value equation to verify they make the equation true.

Lock it in

Question

What does the equation $|A| = |B|$ split into when solving for real numbers?

Answer

$A = B$ or $A = -B$ . Keep the expressions $A$ and $B$ exactly as they are without altering any signs inside them.

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student improperly negates the negative coefficient term inside the absolute value as positive without changing the constant term, rewriting |5 - 2x| as 5 + 2x.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 The student fails to distribute the negative sign to all terms when taking the opposite case, writing -(x + 4) as -x + 4.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

03 The student simply drops the absolute value bars without considering the negative case, finding only one solution.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

Lock it in

Similar question