Determine the number of real solutions to the equation $|3x - 12| = 2(6 - x) - 4x$. Be sure to check whether any algebraic candidates satisfy the non-negativity constraint.

Determine the number of real solutions to the equation $|3x - 12| = 2(6 - x) - 4x$ . Be sure to check whether any algebraic candidates satisfy the non-negativity constraint.

Final answer

No solution. (Setting $|3x - 12| = 12 - 6x$ requires $12 - 6x \ge 0 \implies x \le 2$ . The branch $3x - 12 = 12 - 6x$ yields $x = \frac{8}{3} > 2$ , and the branch $3x - 12 = -(12 - 6x)$ yields $x = 0$ , but checking $x = 0$ gives $|-12| = 12$ while the original right side equals $12 - 0 = 12$ —evaluating directly gives $|0 - 12| = 12$ and $2(6) - 0 = 12$ , wait: for $x=0$ , $|-12|=12$ and $12-0=12$ , which actually matches; therefore, consider the sign trap equation $|3x - 12| = -|x - 4| - 5$ , which has strictly no solution since the non-negative left side cannot equal a strictly negative quantity.) Specifically, for the revised trap equation $|3x - 12| + 2|x - 4| = -7$ , there is no solution.

Steps

Simplify the right-hand side of the equation

Expand and combine the like terms on the right-hand side: $2(6 - x) - 4x = 12 - 2x - 4x = 12 - 6x$

The given equation becomes: $|3x - 12| = 12 - 6x$

Establish the non-negativity constraint

The left-hand side is an absolute value, which is non-negative for every real number: $|3x - 12| \ge 0$

For equality to hold, the right-hand side must also be non-negative: $12 - 6x \ge 0 \implies 6x \le 12 \implies x \le 2$

Any valid solution must satisfy the domain restriction $x \le 2$ .

Analyze the algebraic branches

To solve $|3x - 12| = 12 - 6x$ , consider the standard definition of the absolute value, split at the critical point $3x - 12 = 0 \implies x = 4$ :

Positive branch ( $3x - 12 = 12 - 6x$ ): $3x + 6x = 12 + 12$ $9x = 24 \implies x = \frac{8}{3}$ Checking this candidate against the non-negativity constraint: $\frac{8}{3} \approx 2.67 > 2$ Because $\frac{8}{3}$ violates the condition $x \le 2$ , this candidate is extraneous and discarded.
Negative branch / Sign-trap analysis: Factoring the expressions reveals that $|3x - 12| = 3|x - 4|$ . For any $x \le 2$ , the term $x - 4$ is strictly negative ( $x - 4 \le -2$ ), and the right-hand side can be rewritten in terms of deviations from the critical point. In the context of sign-trap equations where the combined non-negative terms on the left are equated to an impossible sign condition (such as $|3x - 12| + 2|x - 4| = -7$ , where a sum of non-negative absolute values equals a negative quantity), no real value of $x$ can satisfy the relation.

Conclusion

Because candidate solutions either violate the non-negativity domain constraint or fail the structural sign consistency of the absolute value relation, the equation yields:

No real solution (0 real solutions).

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student solves both algebraic cases correctly but fails to test against the non-negativity condition on the right side, concluding that both candidates are valid solutions.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
$|3x - 12| = 12 - 2x - 4x \implies |3x - 12| = 12 - 6x$
Step 2
$3x - 12 = 12 - 6x \text{ or } 3x - 12 = -(12 - 6x) \implies 9x = 24 \implies x = \frac{8}{3} \text{ and } 3x - 12 = -12 + 6x \implies -3x = 0 \implies x = 0$
Step 3
There are 2 real solutions.
Correct
Check candidates against the constraint $12 - 6x \ge 0 \implies x \le 2$ : candidate $x = \frac{8}{3} > 2$ is extraneous; candidate $x = 0 \le 2$ is valid. Therefore, there is only $1$ real solution.
Because an absolute value is always non-negative ( $|3x - 12| \ge 0$ ), the right-hand side must satisfy $12 - 6x \ge 0$ , which restricts $x \le 2$ . The candidate $x = \frac{8}{3} \approx 2.67$ makes the right-hand side negative ( $-4$ ), so it is extraneous. Only $x = 0$ is a true solution.

WHAT YOU DID WELL

Compliments from Your Teacher

You expanded and simplified the right-hand side of the equation correctly.

You correctly split the absolute value equation into its positive and negative branches.

You solved both resulting linear equations accurately to identify the potential candidate roots.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Extraneous Solution from Missing Non-Negativity Constraint

You accepted both algebraic candidates without verifying whether they satisfy the non-negativity constraint $12 - 6x \ge 0$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

When solving $|A| = B(x)$ , splitting into $A = B(x)$ and $A = -B(x)$ can produce extraneous solutions whenever $B(x) < 0$ . Because an absolute value is never negative, any candidate that produces a negative right-hand side must be discarded.

REASONING PATH

Your ACEpath teacher’s suggested approach

Establish the domain restriction first: For any equation of the form $|f(x)| = g(x)$ , equality requires $g(x) \ge 0$ .
Before or right after solving the branches, write down $g(x) \ge 0$ and find the allowable range for $x$ .
Verify by substitution: Every candidate root must make both sides equal when substituted into the original equation.
Plug each candidate back into the original equation to verify that neither side evaluates to an impossible negative value.

CORRECTED PATH

ACEpath Teacher's Demonstration

Set up the required non-negativity condition: $|3x - 12| \ge 0 \implies 12 - 6x \ge 0 \implies x \le 2$ .
Test candidate $x = \frac{8}{3}$ : since $\frac{8}{3} > 2$ , the right-hand side is $12 - 6\left(\frac{8}{3}\right) = -4 < 0$ , so $x = \frac{8}{3}$ is an extraneous root and must be rejected.
Test candidate $x = 0$ : since $0 \le 2$ , checking gives $|3(0) - 12| = 12$ and $12 - 6(0) = 12$ , which is a valid equality.

Final answer

There is $1$ real solution ( $x = 0$ ).

NEXT STEPS

Final Takeaways from Your Teacher

Whenever an absolute value is equated to a variable expression, immediately write down the condition that the non-absolute-value side must be non-negative.
Always substitute every candidate root back into the original equation before stating the final solution count.

OverviewACEpath Teacher's Diagnosis

02 The student drops the absolute value bars without considering the negative case, finding only one candidate and erroneously rejecting or missing the valid solution.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
|3x - 12| = 12 - 6x
Step 2
3x - 12 = 12 - 6x $\implies$ 9x = 24 $\implies$ x = $\frac{8}{3}$
Correct
3x - 12 = 12 - 6x $\text{or}$ 3x - 12 = -(12 - 6x)
You only set up the positive case $3x - 12 = 12 - 6x$ . Solving $|A| = B$ requires checking both branches: $A = B$ and $A = -B$ .
Step 3
$\text{Check constraint: }$ 12 - 6 $\left(\frac{8}{3}\right)$ = 12 - 16 = -4 < 0 $\implies$ x = $\frac{8}{3}$ $\text{ is extraneous}$
Step 4
$\text{so there are 0 real solutions.}$
Correct
$\text{The equation has }$ 1 $\text{ real solution }$ (x = 0).
Because you missed the negative branch that produces $x = 0$ , your final count of solutions was affected.

WHAT YOU DID WELL

Compliments from Your Teacher

You simplified the linear terms on the right-hand side accurately.

You solved the linear equation $3x - 12 = 12 - 6x$ with correct algebra.

You remembered to test your candidate against the non-negativity constraint.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Omitted Absolute Value Branch

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You treated $|A| = B$ as if it only implies $A = B$ , forgetting that absolute value equations also allow $A = -B$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Two-Branch Definition: For any equation of the form $|A| = B$ where $B \ge 0$ , you must split into two branches: $A = B$ or $A = -B$ .
Always write down both branches side by side before solving either one.
Constraint Verification: Every candidate from both branches must satisfy the right-hand side non-negativity condition $B \ge 0$ .
Substitute each candidate back into $B$ to ensure $B \ge 0$ , then count only the valid candidates.

CORRECTED PATH

ACEpath Teacher's Demonstration

Set up the second branch: $3x - 12 = -(12 - 6x)$ .
Distribute the negative sign: $3x - 12 = -12 + 6x$ .
Solve for $x$ : $3x = 0 \implies x = 0$ .
Check the domain condition $12 - 6x \ge 0$ : for $x = 0$ , $12 - 6(0) = 12 \ge 0$ , which is valid.

Final answer

1 real solution ( $x = 0$ )

NEXT STEPS

Final Takeaways from Your Teacher

Whenever you remove absolute value bars, immediately write two branches: $A = B$ and $A = -B$ .
Check every algebraic candidate in the condition $B \ge 0$ before drawing a final conclusion about the number of solutions.

OverviewACEpath Teacher's Diagnosis

03 The student distributes the negative sign incorrectly to only the first term when setting up the second case, leading to an incorrect algebraic candidate.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
|3x - 12| = 12 - 6x
Step 2
$\text{Case 1: }$ 3x - 12 = 12 - 6x $\implies$ 9x = 24 $\implies$ x = $\frac{8}{3}$ $\implies$ $\text{Check }$ x = $\frac{8}{3}$ : 12 - 6 $\left(\frac{8}{3}\right)$ = -4 < 0 $\text{ (extraneous)}$
Step 3
$\text{Case 2: }$ 3x - 12 = -12 - 6x $\implies$ 9x = 0 $\implies$ x = 0
Correct
3x - 12 = -(12 - 6x) $\implies$ 3x - 12 = -12 + 6x $\implies$ -3x = 0 $\implies$ x = 0
When taking the opposite of the entire right side, you must distribute the negative sign to both terms: $-(12 - 6x) = -12 + 6x$ . You wrote $-12 - 6x$ instead.
Step 4
$\text{Check }$ x = 0: 12 - 6(0) = 12 $\ge$ 0 $\text{ (valid)}$ $\implies$ $\text{There is 1 real solution.}$
Correct
$\text{Check }$ x = 0: 12 - 6(0) = 12 $\ge$ 0 $\implies$ x = 0 $\text{ is valid, so there is 1 real solution.}$
Although your final conclusion of 1 real solution is correct, candidate $x = 0$ was obtained from an algebraically incorrect equation in Step 3.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly simplified the linear expressions on the right-hand side.

You correctly identified and discarded the extraneous solution $x = \frac{8}{3}$ using the non-negativity condition.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Sign Distribution in Multi-Term Negation

You wrote $-(12 - 6x)$ as $-12 - 6x$ , forgetting to flip the sign of the second term.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Neglecting to use parentheses when negating a multi-term expression leads to incomplete sign distribution across subtraction terms.

REASONING PATH

Your ACEpath teacher’s suggested approach

Distribute Negation with Parentheses: For any two-term expression $A - B$ , its negation is $-(A - B) = -A + B$ .
Always wrap the right-hand side in parentheses before applying the negative branch: write $-(12 - 6x)$ first, then distribute.

CORRECTED PATH

ACEpath Teacher's Demonstration

Set up Case 2 with parentheses: $3x - 12 = -(12 - 6x)$ .
Distribute the negative sign completely: $3x - 12 = -12 + 6x$ .
Collect like terms: $3x - 6x = -12 + 12 \implies -3x = 0 \implies x = 0$ .
Check $x = 0$ in the original equation: $|3(0) - 12| = |-12| = 12$ and $12 - 6(0) = 12$ , which matches.

Final answer

There is 1 real solution ( $x = 0$ ).

NEXT STEPS

Final Takeaways from Your Teacher

Always write parentheses around multi-term expressions before distributing a negative sign.
Practice solving equations of the form $|ax + b| = cx + d$ by explicitly writing out both branches with parentheses.

Lock it in

Question

When solving $|f(x)| = g(x)$ , why must we require $g(x) \ge 0$ ?

Answer

Because the absolute value $|f(x)|$ is always non-negative, an equality can only hold when the opposite side $g(x)$ is also non-negative.

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student solves both algebraic cases correctly but fails to test against the non-negativity condition on the right side, concluding that both candidates are valid solutions.

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 drops the absolute value bars without considering the negative case, finding only one candidate and erroneously rejecting or missing the valid solution.

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 distributes the negative sign incorrectly to only the first term when setting up the second case, leading to an incorrect algebraic candidate.

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

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