Consider the equation $\frac{x^2 - 16}{x - 4} = 3x - 10$. Solve for all valid real values of $x$, or explain why an algebraically derived candidate must be excluded.

Consider the equation $\frac{x^2 - 16}{x - 4} = 3x - 10$ . Solve for all valid real values of $x$ , or explain why an algebraically derived candidate must be excluded.

Final answer

No solution ( $x = 7$ is an extraneous root resulting from the initial domain restriction $x \neq 4$ , but solving $x + 4 = 3x - 10$ yields $2x = 14 \implies x = 7$ , which satisfies $x \neq 4$ ; hence $x = 7$ is the unique valid solution).

Steps

Identify the domain restriction

To solve the equation and determine any valid solutions or extraneous candidates, follow these steps:

The left-hand side contains the rational expression $\frac{x^2 - 16}{x - 4}$ . A fraction is undefined when its denominator is zero: $x - 4 \neq 0 \implies x \neq 4$

Thus, the domain of the equation consists of all real numbers except $x = 4$ . Any algebraic solution that equals $4$ would be extraneous and must be discarded.

Simplify the rational expression

Factor the difference of squares in the numerator: $x^2 - 16 = (x - 4)(x + 4)$

Under the condition $x \neq 4$ , the common factor $(x - 4)$ can be safely canceled: $\frac{(x - 4)(x + 4)}{x - 4} = x + 4$

The original equation simplifies to a linear equation: $x + 4 = 3x - 10$

Solve the linear equation

Isolate the variable $x$ : $4 + 10 = 3x - x$ $14 = 2x$ $x = 7$

Verify against the domain restriction

Compare the candidate solution to the domain condition:

The derived candidate is $x = 7$ .
Since $7 \neq 4$ , the candidate does not make the original denominator zero.

Substitute $x = 7$ back into the original equation to check:

Left-hand side: $\frac{7^2 - 16}{7 - 4} = \frac{49 - 16}{3} = \frac{33}{3} = 11$
Right-hand side: $3(7) - 10 = 21 - 10 = 11$

Both sides evaluate to $11$ . Therefore, $x = 7$ is the unique valid real solution.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 A student cancels terms across addition and subtraction by incorrectly dividing x² by x and 16 by 4.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$\frac{x^2}{x}$ - $\frac{16}{4}$ = 3x - 10 $\implies$ x - 4 = 3x - 10
Correct
$\frac{(x - 4)(x + 4)}{x - 4}$ = 3x - 10 $\implies$ x + 4 = 3x - 10 ( $x \neq 4$ )
You split the rational expression $\frac{x^2 - 16}{x - 4}$ into $\frac{x^2}{x} - \frac{16}{4}$ . Denominators cannot be split term-by-term across addition or subtraction. Instead, factor the numerator as $(x - 4)(x + 4)$ and cancel the common factor $(x - 4)$ .
Step 2
-2x = -6 $\implies$ x = 3
Correct
2x = 14 $\implies$ x = 7
Your algebraic solving steps were executed correctly from your equation $x - 4 = 3x - 10$ , but the value $x = 3$ is incorrect due to the invalid fraction simplification in Step 1.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly solved the two-step linear equation once you reached $x - 4 = 3x - 10$ .

You correctly simplified the isolated monomial quotients $\frac{x^2}{x} = x$ and $\frac{16}{4} = 4$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Splitting Denominator Across Subtraction

You split a multi-term denominator across subtraction by dividing individual numerator terms by individual denominator terms.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You treated the denominator $x - 4$ as two independent divisors rather than a single grouped quantity. A fraction $\frac{a - b}{c - d}$ cannot be decomposed into $\frac{a}{c} - \frac{b}{d}$ ; simplifying rational expressions requires factoring the entire numerator and denominator and canceling identical common factors.

REASONING PATH

Your ACEpath teacher’s suggested approach

Group multi-term denominators: A multi-term denominator like $x - 4$ represents a single grouped factor $(x - 4)$ and cannot be divided term-by-term.
Always place parentheses around polynomial denominators like $(x - 4)$ before simplifying.
Factor before canceling: Only factors common to the entire numerator and denominator can be canceled, never individual terms separated by $+$ or $-$ .
Factor $x^2 - 16$ as $(x - 4)(x + 4)$ , note $x \neq 4$ , and cancel the entire factor $(x - 4)$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Identify the restriction: $x - 4 \neq 0 \implies x \neq 4$ .
Factor the numerator as a difference of squares: $x^2 - 16 = (x - 4)(x + 4)$ .
Divide out the common factor $(x - 4)$ to obtain $x + 4 = 3x - 10$ .
Rearrange to solve for $x$ : $4 + 10 = 3x - x \implies 2x = 14 \implies x = 7$ .
Confirm that $x = 7$ satisfies the domain restriction $x \neq 4$ .

Final answer

$x = 7$

NEXT STEPS

Final Takeaways from Your Teacher

Practice identifying when a fraction can and cannot be split: single-term denominators split ( $\frac{a+b}{c} = \frac{a}{c}+\frac{b}{c}$ ), but multi-term denominators do not.
Review the difference-of-squares factoring pattern $a^2 - b^2 = (a - b)(a + b)$ for rational simplifications.
Always write out the non-permissible values of $x$ before canceling factors.

OverviewACEpath Teacher's Diagnosis

02 A student incorrectly factors the difference of squares as (x - 4)² instead of (x - 4)(x + 4).

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$\frac{(x-4)^2}{x-4}$ = 3x - 10 $\implies$ x - 4 = 3x - 10
Correct
$\frac{(x-4)(x+4)}{x-4}$ = 3x - 10 $\implies$ x + 4 = 3x - 10 ( $x \neq 4$ )
The numerator $x^2 - 16$ is a difference of two squares, $a^2 - b^2 = (a - b)(a + b)$ , which factors into $(x - 4)(x + 4)$ . Factoring it as $(x - 4)^2$ incorrectly introduces a middle term $-8x$ since $(x - 4)^2 = x^2 - 8x + 16$ .
Step 2
6 = 2x
Correct
14 = 2x
You rearranged your linear equation correctly by adding $10$ and subtracting $x$ , but the numbers were carried over from the earlier factoring error.
Step 3
x = 3
Correct
x = 7
Your division $x = 6/2 = 3$ is correct for your intermediate equation, but because of the initial factoring error, this final value does not satisfy the original problem.

WHAT YOU DID WELL

Compliments from Your Teacher

You recognized that the fraction on the left-hand side could be simplified by dividing out a common factor.

You correctly simplified your rational expression $\frac{(x-4)^2}{x-4}$ to $x - 4$ .

You solved your linear equation $x - 4 = 3x - 10$ with correct algebraic moves to isolate $x$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Difference of Squares vs Perfect Square Trinomial

You factored the binomial $x^2 - 16$ as $(x - 4)^2$ rather than $(x - 4)(x + 4)$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing the difference of two squares identity $a^2 - b^2 = (a - b)(a + b)$ with the square of a binomial $(a - b)^2 = a^2 - 2ab + b^2$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Identify the Form: Two terms separated by subtraction where both terms are perfect squares form a difference of squares: $a^2 - b^2$ .
Always write $a^2 - b^2$ as $(a - b)(a + b)$ with opposite signs in the two factors.
Check by Expanding: Expanding $(x - c)^2$ gives $x^2 - 2cx + c^2$ , which has a non-zero middle linear term.
Mentally multiply your factors back out to verify that no unwanted middle term appears.

CORRECTED PATH

ACEpath Teacher's Demonstration

Factor the numerator using the difference of squares: $x^2 - 16 = (x - 4)(x + 4)$ .
State the domain restriction: $x - 4 \neq 0 \implies x \neq 4$ .
Cancel the common non-zero factor $(x - 4)$ : $x + 4 = 3x - 10$ .
Collect variable terms on one side and constants on the other: $14 = 2x$ .
Divide by $2$ to obtain $x = 7$ , which satisfies the domain condition $x \neq 4$ .

Final answer

x = 7

NEXT STEPS

Final Takeaways from Your Teacher

Practice distinguishing between $a^2 - b^2 = (a-b)(a+b)$ and $(a-b)^2 = a^2 - 2ab + b^2$ .
Always state denominator domain restrictions before canceling factors in rational equations.
Check your solution by substituting the candidate back into both sides of the original equation.

OverviewACEpath Teacher's Diagnosis

03 A student multiplies both sides by (x - 4) and expands incorrectly, treating (3x - 10)(x - 4) as 3x² - 14x + 40 by adding the inner and outer coefficients incorrectly.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
$x^2 - 16 = (3x - 10)(x - 4) \implies x^2 - 16 = 3x^2 - 22x + 40$
Step 2
$0 = 2x^2 - 22x + 56 \implies x^2 - 11x + 28 = 0 \implies (x - 4)(x - 7) = 0$
Step 3
$x = 4 \text{ or } x = 7$
Correct
$x = 7$ (reject $x = 4$ because $x - 4 \neq 0$ )
The original equation contains the denominator $x - 4$ , which requires $x \neq 4$ . Substituting $x = 4$ leads to division by zero, making $x = 4$ an extraneous solution. Only $x = 7$ is valid.

WHAT YOU DID WELL

Compliments from Your Teacher

You expanded $(3x - 10)(x - 4)$ into $3x^2 - 22x + 40$ without sign or arithmetic errors.

You rearranged the equation into standard quadratic form and divided through by 2 correctly.

You factored $x^2 - 11x + 28$ into $(x - 4)(x - 7)$ correctly.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Extraneous Solution Overlooked

You listed both algebraic roots without checking if either value makes any original denominator equal to zero.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Multiplying both sides of a rational equation by a variable factor can introduce extraneous roots. When an algebraic candidate makes an original denominator vanish, it must be excluded from the solution set.

REASONING PATH

Your ACEpath teacher’s suggested approach

Establish the Domain First: Every denominator in a rational equation must be non-zero.
Set each denominator equal to zero at the very beginning to identify excluded values, such as $x - 4 \neq 0 \implies x \neq 4$ .
Filter Algebraic Candidates: Any candidate that violates the domain is extraneous.
Compare all final candidates against your excluded values and discard any match before writing the final solution set.

CORRECTED PATH

ACEpath Teacher's Demonstration

Set the original denominator to zero to find the restricted value: $x - 4 = 0 \implies x = 4$ , so $x \neq 4$ .
Check the candidates $x = 4$ and $x = 7$ against the restriction $x \neq 4$ .
Discard the extraneous candidate $x = 4$ , leaving $x = 7$ as the unique valid solution.

Final answer

$x = 7$ (with $x = 4$ excluded as extraneous)

NEXT STEPS

Final Takeaways from Your Teacher

Always write down the restriction $x \neq a$ at the top of your page before clearing fractions.
Practice checking whether factoring the numerator directly, $\frac{(x-4)(x+4)}{x-4} = x+4$ , avoids quadratic expansion altogether while keeping the restriction in mind.

Lock it in

Question

Can you rewrite $\frac{A - B}{C - D}$ as $\frac{A}{C} - \frac{B}{D}$ ?

Answer

No. Division does not distribute across addition or subtraction in the denominator. A denominator like $C - D$ is a single grouped divisor.

Resources

Answer and steps

Final answer

Steps

Typical errors

01 A student cancels terms across addition and subtraction by incorrectly dividing x² by x and 16 by 4.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 A student incorrectly factors the difference of squares as (x - 4)² instead of (x - 4)(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 A student multiplies both sides by (x - 4) and expands incorrectly, treating (3x - 10)(x - 4) as 3x² - 14x + 40 by adding the inner and outer coefficients incorrectly.

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