Consider the parameter-dependent linear equation $(k^2 - 9)x = k - 3$, where $k$ is a real parameter. Analyze the asymptotic limit of the unique solution $x(k)$ as $k \to 3$, and determine the number of solutions to the equation at the exact boundary values $k = 3$ and $k = -3$.

Consider the parameter-dependent linear equation $(k^2 - 9)x = k - 3$ , where $k$ is a real parameter. Analyze the asymptotic limit of the unique solution $x(k)$ as $k \to 3$ , and determine the number of solutions to the equation at the exact boundary values $k = 3$ and $k = -3$ .

Final answer

For $k \notin \{-3, 3\}$ , the unique solution is $x = \frac{1}{k + 3}$ , with $\lim_{k \to 3} x(k) = \frac{1}{6}$ . At the exact boundary $k = 3$ , the equation reduces to $0x = 0$ (infinitely many solutions). At the exact boundary $k = -3$ , the equation reduces to $0x = -6$ (no solution).

Steps

Write the Equation in Standard Form $Ax = B$

A linear equation in one variable can be represented as: $Ax = B$

For the given equation: $(k^2 - 9)x = k - 3$

Factoring the coefficient of $x$ using the difference of squares gives: $(k - 3)(k + 3)x = k - 3$

Here, the coefficient is $A = (k - 3)(k + 3)$ and the constant term is $B = k - 3$ .

Find the Unique Solution $x(k)$ for $k \notin \{-3, 3\}$

A linear equation $Ax = B$ has a unique solution if and only if the coefficient of $x$ is non-zero ( $A \neq 0$ ): $(k - 3)(k + 3) \neq 0 \implies k \neq 3 \text{ and } k \neq -3$

When $k \notin \{-3, 3\}$ , we can divide both sides by $(k^2 - 9)$ : $x(k) = \frac{k - 3}{(k - 3)(k + 3)}$

Since $k \neq 3$ , the factor $(k - 3)$ can be canceled from the numerator and denominator: $x(k) = \frac{1}{k + 3}$

Evaluate the Asymptotic Limit as $k \to 3$

Using the simplified form of $x(k)$ valid for all $k \neq 3$ : $\lim_{k \to 3} x(k) = \lim_{k \to 3} \frac{1}{k + 3} = \frac{1}{3 + 3} = \frac{1}{6}$

Even though the original coefficient vanishes at $k = 3$ , the unique solution approaches $\frac{1}{6}$ as $k$ gets arbitrarily close to $3$ .

Analyze the Exact Boundary Value $k = 3$

Substitute $k = 3$ directly into the original equation: $(3^2 - 9)x = 3 - 3$ $0x = 0$

For any real value of $x$ , $0 \cdot x = 0$ is always true.
Therefore, at $k = 3$ , the equation is an identity and has infinitely many solutions.

Analyze the Exact Boundary Value $k = -3$

Substitute $k = -3$ directly into the original equation: $((-3)^2 - 9)x = -3 - 3$ $0x = -6$

The left-hand side is $0$ for any real $x$ , but the right-hand side is $-6 \neq 0$ .
Because $0 = -6$ is a contradiction, there is no solution at $k = -3$ .

Pitfall to Avoid

Do not confuse the limit of the solution function, $\lim_{k \to 3} x(k) = \frac{1}{6}$ , with the solution set at the exact point $k = 3$ . Evaluating at $k = 3$ yields the form $0x = 0$ , which is satisfied by all real numbers, not merely $x = \frac{1}{6}$ . Division by zero is undefined, so the expression $x = \frac{1}{k+3}$ does not represent a single solution at $k = 3$ .

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student cancels the terms in the numerator and denominator incorrectly by forgetting that factoring k² - 9 yields 1/(k + 3) rather than k + 3, resulting in an incorrect limit of 6.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
x(k) = $\frac{k^2 - 9}{k - 3}$ $\implies$ x(k) = $\frac{(k - 3)(k + 3)}{k - 3}$ = k + 3
Correct
x(k) = $\frac{k - 3}{k^2 - 9}$ = $\frac{k - 3}{(k - 3)(k + 3)}$ = $\frac{1}{k + 3}$ (k $\notin$ $\{$ -3, 3 $\}$ )
When solving $(k^2 - 9)x = k - 3$ , you must divide both sides by the coefficient of $x$ , which is $k^2 - 9$ . Dividing by the right-hand side inverted the fraction.
Step 2
$\lim_{k $\to$ 3}$ x(k) = 3 + 3 = 6_
Correct
$\lim_{k $\to$ 3} $x(k) = \$ \lim_{k \to 3}\frac{1}{k + 3} $=$ \frac{1}{3 + 3} $=$ \frac{1}{6}$_
Your limit calculation followed your expression $x(k) = k + 3$ . Evaluating the limit with the correct function gives $\frac{1}{6}$ .
Step 3
k = 3 $\text{ gives }$ (9 - 9)x = 3 - 3 $\implies$ 0x = 0 $\implies$ $\text{Infinitely many solutions}$
Step 4
k = -3 $\text{ gives }$ (9 - 9)x = -3 - 3 $\implies$ 0x = -6 $\implies$ 0 $\text{ solutions}$

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly substituted the parameter values into the original equation to determine the number of solutions at both boundaries.

You correctly identified that $0x = 0$ corresponds to infinitely many solutions and $0x = -6$ corresponds to no solution.

You correctly factored the difference of squares $k^2 - 9 = (k - 3)(k + 3)$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Reciprocal / Fraction Inversion Error

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

When isolating the variable in an equation of the form $Ax = B$ , you divided $A$ by $B$ instead of computing $x = \frac{B}{A}$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Isolate the Variable: To solve $Ax = B$ when $A \neq 0$ , always divide both sides by $A$ : $x = \frac{B}{A}$ .
Identify the multiplier attached to $x$ , and place that entire term in the denominator.
Keep Factors in Their Correct Position: When simplifying $\frac{k - 3}{(k - 3)(k + 3)}$ , canceling $k - 3$ leaves $1$ in the numerator, not in the denominator.
Write out the canceled factor explicitly as $\frac{1 \cdot (k - 3)}{(k - 3)(k + 3)} = \frac{1}{k + 3}$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Divide both sides of $(k^2 - 9)x = k - 3$ by $k^2 - 9$ : $x(k) = \frac{k - 3}{k^2 - 9}$
Factor the denominator using the difference of squares: $x(k) = \frac{k - 3}{(k - 3)(k + 3)} = \frac{1}{k + 3} \quad (k \neq 3, -3)$
Compute the limit as $k \to 3$ : $\lim_{k \to 3} x(k) = \frac{1}{3 + 3} = \frac{1}{6}$

Final answer

For $k \notin \{-3, 3\}$ , the unique solution is $x = \frac{1}{k + 3}$ , with $\lim_{k \to 3} x(k) = \frac{1}{6}$ . At $k = 3$ , there are infinitely many solutions. At $k = -3$ , there is no solution.

NEXT STEPS

Final Takeaways from Your Teacher

Practice double-checking the denominator when isolating a variable in parametric linear equations.
Work through exercises where rational expressions simplify with remaining factors in the denominator.

OverviewACEpath Teacher's Diagnosis

02 The student confuses the indeterminate form 0x = 0 with having no solution, mistakenly believing that dividing by zero always implies no solution exists.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
x(k) = $\frac{k - 3}{k^2 - 9}$ = $\frac{1}{k + 3}$ $\implies$ $\lim_{k $\to$ 3}$ \frac{1}{k + 3} = \frac{1}{6}_
Step 2
(3^2 - 9)x = 3 - 3 $\implies$ 0x = 0 $\implies$ x = $\frac{0}{0}$ $\text{ which is undefined}$ $\implies$ $\text{No solution}$
Correct
(3^2 - 9)x = 3 - 3 $\implies$ 0x = 0 $\implies$ $\text{Infinitely many solutions (all real numbers }$ x $\text{)}$
You substituted $k = 3$ correctly to reach $0x = 0$ , but you then attempted to solve for $x$ by dividing by $0$ . Instead of dividing, interpret the statement $0 \cdot x = 0$ : it is true for any real value of $x$ , so the equation has infinitely many solutions.
Step 3
((-3)^2 - 9)x = -3 - 3 $\implies$ 0x = -6 $\implies$ $\text{No solution}$

WHAT YOU DID WELL

Compliments from Your Teacher

You simplified the parameter-dependent solution $x(k)$ and computed its limit as $k \to 3$ accurately.

You substituted both boundary values into the original linear equation correctly.

You correctly interpreted $0x = -6$ as a contradiction having no solution.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Misinterpreting the indeterminate equation 0x = 0

You divided by zero to write $x = \frac{0}{0}$ instead of directly reading the truth set of the linear equation $0x = 0$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You treated division by the coefficient as a mandatory step to isolate $x$ , which led to an undefined fraction $\frac{0}{0}$ and the false conclusion that no solution exists. In a linear equation of the form $Ax = B$ , when $A = 0$ and $B = 0$ , the equation $0x = 0$ is an identity satisfied by every real number, yielding infinitely many solutions.

REASONING PATH

Your ACEpath teacher’s suggested approach

Evaluate 0x = B Directly Without Dividing: For any equation in the form $0 \cdot x = B$ , never divide by zero.
Check the value of $B$ : if $B = 0$ , any real number $x$ satisfies the equation (infinitely many solutions); if $B \neq 0$ , no real number satisfies the equation (no solution).

CORRECTED PATH

ACEpath Teacher's Demonstration

Recognize that $0 \cdot x = 0$ holds true for every real value of $x$ .
Conclude that the solution set is all real numbers $\mathbb{R}$ , which means there are infinitely many solutions at $k = 3$ .

Final answer

Infinitely many solutions at $k = 3$ .

NEXT STEPS

Final Takeaways from Your Teacher

Review the three possible solution types for standard linear equations $Ax = B$ when $A = 0$ .
Practice identifying identities ( $0x = 0$ ) versus contradictions ( $0x = c$ where $c \neq 0$ ) without dividing by zero.

OverviewACEpath Teacher's Diagnosis

03 The student treats 0x = -6 as having a unique solution x = 0 by incorrectly dividing 0 by -6 instead of dividing -6 by the coefficient 0.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
x(k) = $\frac{k - 3}{(k-3)(k+3)}$ = $\frac{1}{k+3}$ $\implies$ $\lim_{k $\to$ 3}$ x(k) = \frac{1}{3+3} = \frac{1}{6}_
Step 2
0x = 0 $\implies$ $\text{Infinitely many solutions}$
Step 3
((-3)^2 - 9)x = -3 - 3 $\implies$ 0x = -6
Step 4
x = $\frac{0}{-6}$ = 0 $\implies$ $\text{One solution, }$ x = 0
Correct
0x = -6 $\implies$ 0 = -6 $\text{ (contradiction)}$ $\implies$ $\text{No solution}$
To isolate $x$ from $Ax = B$ , one divides $B$ by the coefficient $A$ , which would be $\frac{-6}{0}$ (undefined), not $\frac{0}{-6}$ . Since $0 \cdot x = 0$ for all real $x$ , the left-hand side is always $0$ , making $0 = -6$ impossible.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly factored the difference of squares and evaluated the limit as $k \to 3$ .

You correctly identified that $0x = 0$ results in infinitely many solutions.

You correctly substituted $k = -3$ to simplify the equation to $0x = -6$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

division by zero / inverted ratio in Ax = B

Inverting the ratio $\frac{B}{A}$ to $\frac{A}{B}$ when solving $0x = -6$ , writing $x = \frac{0}{-6} = 0$ instead of recognizing division by zero.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing the coefficient $A$ and constant $B$ when isolating $x$ , and failing to interpret $0x = B$ as the contradiction $0 = B$ when $B \neq 0$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Standard Form Inspection: For any linear equation in the form $Ax = B$ , if $A = 0$ and $B \neq 0$ , the equation simplifies to $0 = B$ , which is false for all $x$ .
Always evaluate the product $0 \cdot x = 0$ directly instead of attempting to divide by zero.

CORRECTED PATH

ACEpath Teacher's Demonstration

Evaluate the left-hand side for any real $x$ : $0 \cdot x = 0$ .
Compare both sides: $0 = -6$ , which is a contradiction.
Conclude that no real value of $x$ satisfies the equation.

Final answer

At $k = -3$ , there is no solution.

NEXT STEPS

Final Takeaways from Your Teacher

Review the three cases for the linear equation $Ax = B$ : unique solution ( $A \neq 0$ ), infinitely many solutions ( $A = 0, B = 0$ ), and no solution ( $A = 0, B \neq 0$ ).
Practice solving boundary-value parameter equations by directly evaluating the left-hand side rather than dividing by zero.

Lock it in

Question

When solving $A \cdot x = B$ for $x$ (with $A \neq 0$ ), which term belongs in the denominator?

Answer

$A$ belongs in the denominator: $x = \frac{B}{A}$ . Always divide by the coefficient attached to the variable.

Resources

Answer and steps

Final answer

Steps

Pitfall to Avoid

Typical errors

01 The student cancels the terms in the numerator and denominator incorrectly by forgetting that factoring k² - 9 yields 1/(k + 3) rather than k + 3, resulting in an incorrect limit of 6.

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 confuses the indeterminate form 0x = 0 with having no solution, mistakenly believing that dividing by zero always implies no solution exists.

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 treats 0x = -6 as having a unique solution x = 0 by incorrectly dividing 0 by -6 instead of dividing -6 by the coefficient 0.

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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