Determine all values of the parameter $m \in \mathbb{R}$ for which the system of two equations $\begin{cases} (m - 2)x + 3y = 6 \\ 4x + (m + 2)y = 12 \end{cases}$ has: (a) infinitely many solutions, (b) no solution, and (c) exactly one unique solution.

Final answer

(a) Infinitely many solutions when $m = 4$ ; (b) No solution when $m = -4$ ; (c) Exactly one unique solution when $m \in \mathbb{R} \setminus \{-4, 4\}$ .

Steps

Find the determinant of the coefficient matrix

To determine the number of solutions for the linear system, we examine the relationship between the coefficients of the two equations: $\text{Equation 1: } (m - 2)x + 3y = 6$ $\text{Equation 2: } 4x + (m + 2)y = 12$

A system of two linear equations has a single unique solution if and only if the determinant of its coefficient matrix, $D$ , is non-zero:

$D = (m - 2)(m + 2) - (3)(4)$

Expanding and simplifying:

$D = (m^2 - 4) - 12 = m^2 - 16$

Setting $D = 0$ :

$m^2 - 16 = 0 \implies (m - 4)(m + 4) = 0 \implies m = 4 \quad \text{or} \quad m = -4$

Analyze each case

Case 1: Exactly one unique solution ( $D \neq 0$ )

When $D \neq 0$ , the coefficient lines are not parallel and intersect at exactly one point. Thus, the system has a unique solution whenever:

$m \neq 4 \quad \text{and} \quad m \neq -4 \quad (\text{or } m \in \mathbb{R} \setminus \{-4, 4\})$

Case 2: Test $m = 4$

Substitute $m = 4$ into the original equations:

Equation 1: $(4 - 2)x + 3y = 6 \implies 2x + 3y = 6$
Equation 2: $4x + (4 + 2)y = 12 \implies 4x + 6y = 12$

Dividing Equation 2 by $2$ gives $2x + 3y = 6$ , which is identical to Equation 1. Since both equations represent the exact same line, every point on this line satisfies the system. Therefore, the system has infinitely many solutions when $m = 4$ .

Case 3: Test $m = -4$

Substitute $m = -4$ into the original equations:

Equation 1: $(-4 - 2)x + 3y = 6 \implies -6x + 3y = 6 \implies -2x + y = 2$
Equation 2: $4x + (-4 + 2)y = 12 \implies 4x - 2y = 12 \implies 2x - y = 6 \implies -2x + y = -6$

The left-hand sides are identical, but the right-hand sides are different ( $2 \neq -6$ ). Geometrically, these equations represent two distinct parallel lines that never intersect. Therefore, the system has no solution when $m = -4$ .

Summary

(a) Infinitely many solutions: $m = 4$
(b) No solution: $m = -4$
(c) Exactly one unique solution: $m \in \mathbb{R} \setminus \{-4, 4\}$

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets to test the specific values of m in the full system when D = 0, mistakenly concluding that both critical values lead to infinitely many solutions.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
D = (m-2)(m+2) - 3 $\cdot$ 4 $\implies$ D = m^2 - 4 - 12 = m^2 - 16 $\implies$ m^2 - 16 = 0 $\implies$ m = 4 $\text{ or }$ m = -4
Step 2
$D \neq 0$ $\implies$ $m \neq 4$ $\text{ and }$ m $\neq$ -4
Step 3
$\text{Infinitely many solutions when }$ m = 4 $\text{ or }$ m = -4
Correct
$\text{Substitute }$ m = 4: \begin{cases} 2x + 3y = 6 \ 4x + 6y = 12 \end{cases} ( $\text{dependent}$ ) $\implies$ $\text{infinitely many solutions when }$ m = 4
A determinant of zero only indicates that the lines are parallel or coincident; it does not automatically mean infinitely many solutions. You need to substitute each critical value into the original system to check the constant terms.
Step 4
$\text{No values of } m$
Correct
$\text{Substitute }$ m = -4: \begin{cases} -6x + 3y = 6 \ 4x - 2y = 12 \end{cases} $\implies$ \begin{cases} -2x + y = 2 \ -2x + y = -6 \end{cases} ( $\text{inconsistent}$ ) $\implies$ $\text{no solution when }$ m = -4
Because you assigned both critical values to the infinitely many solutions case, you concluded that no values of $m$ yield no solution. In fact, $m = -4$ makes the lines parallel and non-intersecting.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly calculated the determinant of the coefficient matrix as $D = m^2 - 16$ .

You correctly determined that a unique solution exists when $D \neq 0$ , giving $m \neq 4$ and $m \neq -4$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Conflating zero determinant with dependent system

Assuming that $D = 0$ always means infinitely many solutions without checking the constant terms.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

A zero determinant ( $D = 0$ ) only tells you that the lines share the same slope (parallel lines). It does not distinguish whether the lines are coincident (same $y$ -intercept, infinitely many solutions) or distinct parallel lines (different $y$ -intercepts, no solution). You must test each critical value back in the full system.

REASONING PATH

Your ACEpath teacher’s suggested approach

Branching at $D = 0$: $D = 0$ means the system is either inconsistent (no solution) or dependent (infinitely many solutions).
Always substitute each critical value of the parameter back into both original equations to compare the full equations.
Ratio Comparison: For $\begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases}$ , if $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ , there are infinitely many solutions; if $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ , there is no solution.
Check whether the ratio of the coefficients of $x$ and $y$ also matches the ratio of the constant terms.

CORRECTED PATH

ACEpath Teacher's Demonstration

For $m = 4$ : Equation 1 is $2x + 3y = 6$ ; Equation 2 is $4x + 6y = 12$ . Dividing Equation 2 by $2$ gives $2x + 3y = 6$ . The equations are identical, so there are infinitely many solutions when $m = 4$ .
For $m = -4$ : Equation 1 is $-6x + 3y = 6 \implies -2x + y = 2$ ; Equation 2 is $4x - 2y = 12 \implies 2x - y = 6 \implies -2x + y = -6$ . Since $2 \neq -6$ , the equations are contradictory, so there is no solution when $m = -4$ .

Final answer

(a) Infinitely many solutions when $m = 4$ ; (b) No solution when $m = -4$ ; (c) Exactly one unique solution when $m $\in$ $\mathbb{R}$ $\setminus$ $\{$ -4, 4 $\}$ .

NEXT STEPS

Final Takeaways from Your Teacher

Review the geometric meaning of $D = 0$ : it means the lines are parallel, but whether they coincide or never meet depends on the constants.
Practice checking both critical values by direct substitution into the augmented system.

OverviewACEpath Teacher's Diagnosis

02 The student incorrectly calculates the determinant by adding instead of subtracting the off-diagonal product, yielding m = ±√28 instead of ±4.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
D = (m - 2)(m + 2) + 12 $\implies$ D = m^2 - 4 + 12 = m^2 + 8
Correct
D = (m - 2)(m + 2) - (3)(4) = m^2 - 4 - 12 = m^2 - 16
When calculating the determinant of a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , the formula is $ad - bc$ . You added $3 \times 4 = 12$ instead of subtracting it, leading to $m^2 + 8$ instead of $m^2 - 16$ .
Step 2
m^2 + 8 = 0 $\text{ has no real solutions}$ $\implies$ $\text{No real values of }$ m
Correct
m^2 - 16 = 0 $\implies$ (m - 4)(m + 4) = 0 $\implies$ m = 4 $\text{ or }$ m = -4
Because your determinant expression was $m^2 + 8$ , solving $D = 0$ produced no real roots, preventing you from finding the critical values $m = 4$ and $m = -4$ .
Step 3
D = m^2 + 8 $\neq$ 0 $\text{ for all real }$ m $\implies$ $\text{All real numbers }$ m
Correct
$D \neq 0$ $\implies$ m^2 - 16 $\neq$ 0 $\implies$ m $\in$ $\mathbb{R}$ $\setminus$ $\{$ -4, 4 $\}$
Your reasoning that a unique solution occurs when $D \neq 0$ is correct, but the conclusion was affected by the incorrect determinant expression.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly recognized that setting the coefficient determinant $D = 0$ identifies when a system does not have a unique solution.

You correctly expanded the difference of squares $(m - 2)(m + 2) = m^2 - 4$ .

You accurately solved your resulting equation $m^2 + 8 = 0$ over the real numbers.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Determinant Formula Sign Error

You added the product of the off-diagonal entries instead of subtracting it ( $ad + bc$ instead of $ad - bc$ ).

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing the sign in the $2 \times 2$ determinant formula $\det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc$ , which led to a positive constant term rather than a negative one.

REASONING PATH

Your ACEpath teacher’s suggested approach

2x2 Determinant Definition: For a matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , the determinant is always main diagonal minus anti-diagonal: $ad - bc$ .
Always write down the explicit subtraction $(a)(d) - (b)(c)$ before simplifying.
Testing Critical Parameter Values: When $D = 0$ , substitute each critical value back into both original equations to distinguish between parallel lines (no solution) and coincident lines (infinitely many solutions).
Check the ratios $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ for infinite solutions versus $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ for no solution.

CORRECTED PATH

ACEpath Teacher's Demonstration

Set $D = 0$ : $m^2 - 16 = 0 \implies m = 4$ or $m = -4$ .
For (c), a unique solution requires $D \neq 0$ , so $m \in \mathbb{R} \setminus \{-4, 4\}$ .
Substitute $m = 4$ into the system: $\lbrace 2x + 3y = 6 ,\quad 4x + 6y = 12$ . Dividing the second equation by $2$ yields identical equations, so (a) has infinitely many solutions when $m = 4$ .
Substitute $m = -4$ into the system: $\lbrace -6x + 3y = 6 ,\quad 4x - 2y = 12$ . Simplifying gives $-2x + y = 2$ and $2x - y = 6 \implies -2x + y = -6$ . Since $2 \neq -6$ , the lines are parallel and distinct, so (b) has no solution when $m = -4$ .

Final answer

(a) Infinitely many solutions when $m = 4$ ; (b) No solution when $m = -4$ ; (c) Exactly one unique solution when $m $\in$ $\mathbb{R}$ $\setminus$ $\{$ -4, 4 $\}$ .

NEXT STEPS

Final Takeaways from Your Teacher

Double-check the sign in $ad - bc$ whenever you compute a $2 \times 2$ determinant.
After finding the roots where $D = 0$ , always plug them back into the original system to separate the 'parallel' case from the 'coincident' case.

OverviewACEpath Teacher's Diagnosis

03 The student inverts the conditions between infinitely many solutions and no solution when evaluating the critical values.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$D = m^2 - 16 = 0 \implies m = 4 \text{ or } m = -4$
Step 2
$\text{For } m = -4:\; -6x + 3y = 6 \text{ and } 4x - 2y = 12 \implies -2x + y = 2 \text{ and } 2x - y = 6 \implies \text{The lines are the same, so infinitely many solutions when } m = -4$
Correct
$\text{For } m = -4:\; -2x + y = 2 \text{ and } 2x - y = 6 \implies 2x - y = -2 \text{ and } 2x - y = 6$ , which are distinct parallel lines ( $2 \neq -6$ ), so there is no solution.
The simplified equations $-2x + y = 2$ and $2x - y = 6$ rewrite as $2x - y = -2$ and $2x - y = 6$ . Because the left-hand sides match but the constant terms differ ( $-2 \neq 6$ ), the lines are distinct parallel lines that never intersect, meaning there is no solution.
Step 3
$\text{For } m = 4:\; 2x + 3y = 6 \text{ and } 4x + 6y = 12 \implies \text{The lines are parallel, so no solution when } m = 4$
Correct
$\text{For } m = 4:\; 2x + 3y = 6 \text{ and } 4x + 6y = 12 \implies 2x + 3y = 6 \text{ and } 2x + 3y = 6$ , which are identical coincident lines, so there are infinitely many solutions.
Dividing the second equation $4x + 6y = 12$ by $2$ gives $2x + 3y = 6$ , which is identical to the first equation. Because both equations describe the exact same line, every point on the line is a solution, giving infinitely many solutions.
Step 4
$D \neq 0 \implies m \neq 4 \text{ and } m \neq -4$

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly computed the determinant of the coefficient matrix and solved for its roots.

You correctly identified that the system has a unique solution when $D \neq 0$ .

You showed clear intermediate steps by simplifying the equations for each critical value.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Inverted Coincident and Parallel Line Conditions

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You confused coincident lines with distinct parallel lines. Identical equations (proportional coefficients and constant terms) share all points and yield infinitely many solutions, whereas equations with identical variable combinations but different constants represent parallel lines and yield no solution.

REASONING PATH

Your ACEpath teacher’s suggested approach

Constant Term Comparison: If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ , the lines coincide (infinitely many solutions). If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ , the lines are strictly parallel (no solution).
After matching the variable coefficients, always compare the resulting constant terms to see if they are equal or contradictory.

CORRECTED PATH

ACEpath Teacher's Demonstration

For $m = -4$ , rewrite both equations with matching variable signs: $-2x + y = 2 \implies 2x - y = -2$ , and the second equation is $2x - y = 6$ .
Since $-2 \neq 6$ , the lines have identical slopes but different intercepts, making them parallel with no solution.
For $m = 4$ , simplify the second equation: $\frac{4x + 6y = 12}{2} \implies 2x + 3y = 6$ .
Since this is identical to the first equation $2x + 3y = 6$ , the lines are coincident, giving infinitely many solutions.

Final answer

(a) Infinitely many solutions when $m = 4$ ; (b) No solution when $m = -4$ ; (c) Exactly one unique solution when $m \in \mathbb{R} \setminus \{-4, 4\}$ .

NEXT STEPS

Final Takeaways from Your Teacher

When evaluating critical parameter values, put both linear equations into identical standard form $Ax + By = C$ with matching leading signs.
Double-check whether the constant terms match (coincident $\implies$ infinitely many solutions) or contradict (inconsistent $\implies$ no solution).

Lock it in

Question

When the determinant of a $2 \times 2$ linear system is zero ( $D = 0$ ), what are the two possible outcomes?

Answer

The system either has no solution (parallel distinct lines) or infinitely many solutions (coincident lines). You must check the constant terms to tell them apart.

Resources

Answer and steps

Final answer

Steps

Case 1: Exactly one unique solution (D≠0D \neq 0D=0)

Case 2: Test m=4m = 4m=4

Case 3: Test m=−4m = -4m=−4

Summary

Typical errors

01 The student forgets to test the specific values of m in the full system when D = 0, mistakenly concluding that both critical values lead to infinitely many 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 incorrectly calculates the determinant by adding instead of subtracting the off-diagonal product, yielding m = ±√28 instead of ±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 inverts the conditions between infinitely many solutions and no solution when evaluating the critical values.

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

Case 1: Exactly one unique solution ( $D \neq 0$ )

Case 2: Test $m = 4$

Case 3: Test $m = -4$