A student attempts to solve the equation $\frac{4x - 8}{x - 2} = 3x - 2$ by first dividing both sides by $(x - 2)$ after factoring the numerator as $4(x - 2)$, concluding that $x = 2$ is a valid solution. Rigorously analyze the equation over the real domain: isolate the variable on both sides, state whether $x = 2$ is an extraneous root or a valid solution, and determine the complete, true solution set.

A student attempts to solve the equation $\frac{4x - 8}{x - 2} = 3x - 2$ by first dividing both sides by $(x - 2)$ after factoring the numerator as $4(x - 2)$ , concluding that $x = 2$ is a valid solution. Rigorously analyze the equation over the real domain: isolate the variable on both sides, state whether $x = 2$ is an extraneous root or a valid solution, and determine the complete, true solution set.

Final answer

No real solution (the prospective solution $x = 2$ is extraneous due to division by zero in the original equation's domain restriction $x \neq 2$ , resulting in the empty set $\emptyset$ ).

Steps

Identify the Domain of the Equation

Before performing any algebraic operations on a rational equation, determine any values of the variable that make a denominator equal to zero.

The given equation is: $\frac{4x - 8}{x - 2} = 3x - 2$

The denominator on the left-hand side is $x - 2$ . Setting the denominator to zero gives: $x - 2 = 0 \implies x = 2$

Because division by zero is undefined, the domain of the equation is restricted to all real numbers except $2$ : $x \neq 2$

Simplify the Equation Under the Domain Restriction

Assuming $x \neq 2$ , the term $(x - 2)$ is non-zero, allowing us to factor the numerator and cancel the common factor: $\frac{4(x - 2)}{x - 2} = 3x - 2$

Canceling $(x - 2)$ yields: $4 = 3x - 2$

Solve the Linear Equation

Isolate the variable term $3x$ by adding $2$ to both sides: $4 + 2 = 3x$ $6 = 3x$

Divide both sides by $3$ : $x = 2$

Check for Extraneous Solutions

Compare the candidate solution $x = 2$ with the domain restriction established in Step 1:

The algebraic simplification yielded $x = 2$ .
However, the original equation requires $x \neq 2$ .

Substituting $x = 2$ into the original expression results in: $\frac{4(2) - 8}{2 - 2} = \frac{0}{0}$ which is undefined.

Therefore, $x = 2$ is an extraneous root created by removing the denominator during simplification.

Conclusion

Since the only algebraic candidate is extraneous, the original equation has no real solution, and the complete solution set is the empty set: $\emptyset$

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student ignores the restriction that the denominator cannot be zero, accepting the algebraic result x = 2 as a valid solution.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$\frac{4(x-2)}{x-2}$ = 3x - 2 $\implies$ 4 = 3x - 2
Step 2
3x = 6 $\implies$ x = 2
Step 3
x = 2 $\text{ is a valid solution because substituting gives }$ 4 = 4.
Correct
$\text{Substitute }$ x = 2 $\text{ into the original denominator: }$ 2 - 2 = 0 $\implies$ $\text{undefined. Thus, }$ x = 2 $\text{ is an extraneous solution.}$
You substituted $x = 2$ into the simplified expression $4 = 3x - 2$ instead of the original equation. In the original equation, substituting $x = 2$ produces $\frac{0}{0}$ , which is undefined.
Step 4
$\text{Solution set: } \lbrace 2 \rbrace$
Correct
$\text{Solution set: } \emptyset \text{ (or no real solution)}$
Because your verification concluded that $x = 2$ was valid, your final solution set included an extraneous root instead of being the empty set.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly factored $4(x - 2)$ from the numerator.

You canceled the common factor to produce the correct linear equation $4 = 3x - 2$ .

You solved the linear equation $3x = 6$ accurately to obtain $x = 2$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Extraneous solution from division by zero

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You validated the candidate root against the simplified linear equation instead of testing it against the original rational equation and its non-zero denominator condition ( $x \neq 2$ ).

REASONING PATH

Your ACEpath teacher’s suggested approach

State Denominator Restrictions First: Any variable expression in a denominator must never equal zero: $\text{denominator} \neq 0$ .
Before simplifying or canceling any factors, write down all excluded values by setting each denominator equal to $0$ .
Always Check Against the Original Equation: Candidate roots must satisfy the original, unsimplified equation, not an algebraically modified version.
Substitute the candidate value directly into the initial expression to confirm it does not cause division by zero.

CORRECTED PATH

ACEpath Teacher's Demonstration

Check the original denominator with the candidate root: when $x = 2$ , the denominator is $x - 2 = 2 - 2 = 0$ .
Recognize that division by zero is undefined, which makes $x = 2$ an extraneous root.
Since $x = 2$ is the only algebraic candidate and it is extraneous, discard it.

Final answer

No real solution (Solution set: $\emptyset$ )

NEXT STEPS

Final Takeaways from Your Teacher

Always write down domain exclusions (such as $x - 2 \neq 0 \implies x \neq 2$ ) at the very first step before canceling common algebraic factors.
Whenever you test candidate solutions, plug them back into the original equation rather than any simplified intermediate step.

OverviewACEpath Teacher's Diagnosis

02 The student multiplies both sides by (x - 2) and factors the resulting quadratic into two roots, failing to reject x = 2 as extraneous.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
4x - 8 = (3x - 2)(x - 2) $\implies$ 4x - 8 = 3x^2 - 8x + 4
Step 2
3x^2 - 12x + 12 = 0 $\implies$ 3(x - 2)^2 = 0 $\implies$ x = 2
Step 3
x = 2 $\text{ is a valid solution since it satisfies the expanded quadratic equation.}$
Correct
x = 2 $\text{ makes the original denominator }$ x - 2 = 0, $\text{ so it is an extraneous root.}$
Candidate solutions must satisfy the original equation, not merely an equation derived after clearing denominators. At $x = 2$ , the denominator $x - 2$ equals $0$ , making the original expression undefined.
Step 4
$\text{Solution set: } \{2\}$
Correct
$\text{Solution set: } \emptyset$
Because the only candidate root $x = 2$ is extraneous, the original equation has no real solution.

WHAT YOU DID WELL

Compliments from Your Teacher

You expanded the product $(3x - 2)(x - 2)$ into $3x^2 - 8x + 4$ without algebraic errors.

You collected like terms correctly to form $3x^2 - 12x + 12 = 0$ .

You recognized and factored the perfect square trinomial to isolate $x = 2$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Extraneous Solution Ignored

Testing the candidate root $x = 2$ in the derived quadratic equation instead of verifying it against the original denominator.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Treating an equation obtained after multiplying by a variable factor as strictly equivalent to the original rational equation without enforcing the domain restriction $x \neq 2$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Establish the domain first: Before multiplying through by any denominator containing a variable, set each denominator $\neq 0$ .
Write $x - 2 \neq 0 \implies x \neq 2$ at the very top of your solution.
Always check prospective roots in the original equation: Multiplying by an expression that can equal zero may introduce extraneous solutions.
Substitute any algebraic candidate back into the original denominators before declaring the final solution set.

CORRECTED PATH

ACEpath Teacher's Demonstration

State the domain restriction from the original equation: $x - 2 \neq 0 \implies x \neq 2$ .
Evaluate the candidate $x = 2$ : substituting $x = 2$ gives $\frac{4(2) - 8}{2 - 2} = \frac{0}{0}$ , which is undefined.
Conclude that $x = 2$ is extraneous, leaving no valid real solutions.

Final answer

No real solution (the prospective solution $x = 2$ is extraneous, so the solution set is $\emptyset$ ).

NEXT STEPS

Final Takeaways from Your Teacher

Record the non-permissible values of the variable as the very first step in rational equations.
Never verify solutions using a downstream derived equation; always substitute back into the original formulation.

Lock it in

Question

What is the very first step you should always take before simplifying or solving a rational equation?

Answer

Identify all restricted values that make any denominator equal to zero ( $x \neq \text{value}$ ).

Resources

Answer and steps

Final answer

Steps

Conclusion

Typical errors

01 The student ignores the restriction that the denominator cannot be zero, accepting the algebraic result x = 2 as a valid solution.

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 multiplies both sides by (x - 2) and factors the resulting quadratic into two roots, failing to reject x = 2 as extraneous.

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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