Find the boundary value of the real parameter $m$ for which the multi-step equation $3(2x - 4) + m(x + 1) = 2(m + 3)x

Find the boundary value of the real parameter $m$ for which the multi-step equation $3(2x - 4) + m(x + 1) = 2(m + 3)x - 5$ has no solution for $x$ .

Final answer

$m = 0$

Steps

To find the value of $m$ for which the equation has no solution, we first simplify both sides and write the equation in the standard linear form $Ax = B$ .

Step 1: Expand both sides of the equation

Expand the left-hand side: $3(2x - 4) + m(x + 1) = 6x - 12 + mx + m$

Expand the right-hand side: $2(m + 3)x - 5 = (2m + 6)x - 5$

Setting both sides equal gives: $(6 + m)x + (m - 12) = (2m + 6)x - 5$

Step 2: Rearrange into the standard form $Ax = B$

Move all terms involving $x$ to one side and all constant terms to the other side: $(m - 12) + 5 = (2m + 6)x - (6 + m)x$ $m - 7 = (2m + 6 - 6 - m)x$ $m - 7 = mx$

Rewriting in the standard form $Ax = B$ : $mx = m - 7$

Step 3: Apply the condition for no solution

A linear equation of the form $Ax = B$ has:

A unique solution if $A \neq 0$ ( $x = \frac{B}{A}$ ).
Infinitely many solutions if $A = 0$ and $B = 0$ ( $0x = 0$ ).
No solution if $A = 0$ and $B \neq 0$ ( $0x = B$ , which simplifies to the false statement $0 = B$ ).

For our equation $mx = m - 7$ :

The coefficient of $x$ must be zero: $m = 0$
The constant term must be non-zero when $m = 0$ : $m - 7 = 0 - 7 = -7 \neq 0$

Since $0x = -7$ leads to the contradiction $0 = -7$ , there is no value of $x$ that satisfies the equation.

Therefore, the boundary value is $m = 0$ .

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student confuses the condition for no solution with the condition for infinitely many solutions by setting the constant term to zero instead of the coefficient of x.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
3(2x - 4) + m(x + 1) = 2(m + 3)x - 5 $\implies$ 6x - 12 + mx + m = 2mx + 6x - 5
Step 2
mx - 7 = 0
Correct
mx = m - 7
When gathering terms onto opposite sides, the $+m$ term on the left side was omitted. Subtracting $mx$ , $6x$ , and adding $5$ gives $mx = m - 7$ (or $mx - m + 7 = 0$ ).
Step 3
m - 7 = 0
Correct
m = 0 $\text{and}$ m - 7 $\neq$ 0
For a linear equation $Ax = B$ to have no solution, the coefficient of $x$ must be zero ( $A = 0$ ) while the constant term is non-zero ( $B \neq 0$ ). Setting $m - 7 = 0$ sets the constant term to zero instead of setting the coefficient of $x$ to zero.
Step 4
m = 7
Correct
m = 0
This final value follows from solving $m - 7 = 0$ , but the correct condition requires setting the coefficient of $x$ to zero, yielding $m = 0$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You expanded both sides using the distributive property without any arithmetic errors.

You recognized that terms involving the parameter $m$ needed to be isolated to find the boundary condition.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Condition for No Solution Misapplied

You dropped the constant $+m$ on the left-hand side when simplifying terms.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You confused the condition for an equation to have no solution with setting the constant term to zero. In standard form $Ax = B$ , no solution occurs when the variable vanishes ( $A = 0$ ) but leaves an impossible statement ( $B \neq 0$ ), not when the constant term equals zero.

REASONING PATH

Your ACEpath teacher’s suggested approach

Standard Form Reduction: Collect all $x$ -terms on one side and all constants on the other to write the equation as $Ax = B$ .
Group like terms carefully: $(2m + 6 - 6 - m)x = m - 12 + 5$ , which simplifies to $mx = m - 7$ .
No Solution Condition: The linear equation $Ax = B$ has no solution if and only if $A = 0$ and $B \neq 0$ (producing $0x = B$ , a contradiction).
Set the coefficient of $x$ equal to zero ( $m = 0$ ) and verify that the constant side is non-zero ( $0 - 7 = -7 \neq 0$ ).

CORRECTED PATH

ACEpath Teacher's Demonstration

Rearrange $6x - 12 + mx + m = 2mx + 6x - 5$ by subtracting $6x$ and $mx$ from both sides: $-12 + m = mx - 5$ .
Add $5$ to both sides to isolate the $x$ -term: $mx = m - 7$ .
Identify the coefficient of $x$ as $A = m$ and the constant term as $B = m - 7$ .
Enforce the condition for no solution: set $A = 0$ , giving $m = 0$ .
Check that $B \neq 0$ when $m = 0$ : $B = 0 - 7 = -7 \neq 0$ , which produces the contradiction $0x = -7$ .

Final answer

m = 0

NEXT STEPS

Final Takeaways from Your Teacher

Practice rewriting parameterized linear equations into the clean form $Ax = B$ before making any conclusions.
Review the three structural outcomes for $Ax = B$ : unique solution ( $A \neq 0$ ), infinitely many solutions ( $A = 0, B = 0$ ), and no solution ( $A = 0, B \neq 0$ ).

OverviewACEpath Teacher's Diagnosis

02 The student forgets to distribute the factor 2 to the term 3 inside the parentheses on the right-hand side, leading to an incorrect coefficient of x.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
6x - 12 + mx + m = 2mx + 3x - 5
Correct
6x - 12 + mx + m = 2mx + 6x - 5
When expanding the right-hand side expression $2(m + 3)x$ , you multiplied $2$ by $m$ to get $2mx$ , but forgot to multiply $2$ by $3$ , writing $3x$ instead of $6x$ .
Step 2
(6 + m - 2m - 3)x = 12 - m - 5 $\implies$ (3 - m)x = 7 - m
Correct
(6 + m - 2m - 6)x = 12 - m - 5 $\implies$ -mx = 7 - m $\text{ (or }$ mx = m - 7 $\text{)}$
You grouped the like terms and rearranged the equation correctly based on your previous line, but the resulting coefficient was affected by the earlier distribution mistake.
Step 3
3 - m = 0 $\implies$ m = 3
Correct
m = 0 $\text{ (since }$ -m = 0 $\text{ and }$ 7 - 0 $\neq$ 0 $\text{)}$
You correctly applied the condition for no solution by setting the coefficient of $x$ equal to $0$ , but this yielded $m = 3$ instead of the correct value $m = 0$ due to the earlier distribution error.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly expanded the left-hand side into $6x - 12 + mx + m$ .

You grouped all terms containing $x$ onto one side and constant terms onto the other side.

You correctly recognized that a linear equation $Ax = B$ has no solution when the coefficient $A = 0$ and the constant term $B \neq 0$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Distributive Property Incomplete

You distributed the multiplier $2$ only to the first term inside $(m + 3)x$ , leaving $3x$ instead of doubling it to $6x$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Overlooking the outer multiplier when distributing across a binomial that has a variable attached to the right, such as $2(a + b)x$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Distribute Completely: Every term inside the parentheses must be multiplied by every factor outside: $k(a + b)x = kax + kbx$ .
Draw arrows from the outer factor to each term inside the parentheses before multiplying.
Check the Constant Term for No Solution: For $Ax = B$ to have no solution, we must have $A = 0$ and $B \neq 0$ .
Always substitute your found parameter back into $B$ to confirm it does not evaluate to $0$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Expand the right-hand side: $2(m + 3)x - 5 = 2mx + 6x - 5$ .
Rearrange terms: $(6 + m)x - (2m + 6)x = 12 - m - 5$ .
Simplify both sides: $-mx = 7 - m$ , or equivalently $mx = m - 7$ .
Set the coefficient of $x$ to $0$ : $m = 0$ .
Verify the constant term: $m - 7 = 0 - 7 = -7 \neq 0$ .

Final answer

m = 0

NEXT STEPS

Final Takeaways from Your Teacher

Practice distributing expressions with factors on both sides of the parentheses, like $c(a + b)x$ .
Re-substitute parameter values into both sides of the equation to verify whether they produce an impossible statement like $0 = -7$ .

OverviewACEpath Teacher's Diagnosis

03 The student incorrectly applies the distributive property to m(x + 1) as mx + 1, resulting in an incorrect constant term condition.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
6x - 12 + mx + 1 = 2mx + 6x - 5
Correct
6x - 12 + mx + m = 2mx + 6x - 5
When distributing $m$ across $(x + 1)$ , you must multiply $m$ by both terms, giving $m \cdot x + m \cdot 1 = mx + m$ , not $mx + 1$ .
Step 2
-mx - 11 = -5 $\implies$ -mx = 6
Correct
mx = m - 7
Your algebraic rearrangement into the form $Ax = B$ followed correctly from your previous line, but the constant terms were affected by the earlier distribution mistake.
Step 3
-m = 0 $\implies$ m = 0, $\text{ but }$ 6 $\neq$ 0 $\text{ so }$ m = -6
Correct
m = 0 $\text{ gives }$ 0x = -7 $\implies$ m = 0
An equation $Ax = B$ has no solution when the coefficient of $x$ is zero ( $A = 0$ ) and the constant term is non-zero ( $B \neq 0$ ). The non-zero constant ( $6 \neq 0$ ) confirms that $m = 0$ is indeed the boundary value, rather than ruling it out.

WHAT YOU DID WELL

Compliments from Your Teacher

You successfully expanded $3(2x - 4)$ to $6x - 12$ and $2(m + 3)x$ to $2mx + 6x$ .

You correctly collected like terms and isolated the variable term to form $-mx = 6$ .

You identified that the coefficient of $x$ must equal $0$ to create a no-solution scenario.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Distributive Property & No-Solution Condition

You wrote $m(x + 1)$ as $mx + 1$ instead of distributing $m$ to both terms.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Incomplete distribution across parentheses with a variable coefficient, paired with a misconception that finding a contradiction ( $0 \neq B$ ) means discarding the parameter value rather than keeping it as the valid condition.

REASONING PATH

Your ACEpath teacher’s suggested approach

Full Distribution: Every term inside parentheses must be multiplied by the factor outside: $a(b + c) = ab + ac$ .
Always draw distribution arrows or verify each product individually when distributing a variable parameter.
No-Solution Criterion: For a linear equation $Ax = B$ , no solution exists if and only if $A = 0$ and $B \neq 0$ .
When $A = 0$ produces a contradiction like $0 = -7$ , conclude that this parameter value is the correct answer.

CORRECTED PATH

ACEpath Teacher's Demonstration

Expand both sides fully: $6x - 12 + mx + m = (2m + 6)x - 5$ .
Group terms in the form $Ax = B$ : $(2m + 6 - 6 - m)x = m - 12 + 5$ , which simplifies to $mx = m - 7$ .
Set the coefficient of $x$ to zero: $m = 0$ .
Check the constant term at $m = 0$ : $m - 7 = 0 - 7 = -7 \neq 0$ . Since $0x = -7$ is impossible for any $x$ , $m = 0$ yields no solution.

Final answer

m = 0

NEXT STEPS

Final Takeaways from Your Teacher

Review distributive property flashcards focusing on distributing variables over constants.
Practice identifying the difference between no-solution ( $0x = B, B \neq 0$ ) and infinitely many solutions ( $0x = 0$ ).

Lock it in

Question

For a linear equation in standard form $Ax = B$ , under what exact conditions does the equation have no solution?

Answer

$A = 0$ and $B \neq 0$ .

Principle: The variable disappears ( $0x = 0$ ), leaving the false statement $0 = B$ .

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student confuses the condition for no solution with the condition for infinitely many solutions by setting the constant term to zero instead of the coefficient of x.

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 forgets to distribute the factor 2 to the term 3 inside the parentheses on the right-hand side, leading to an incorrect coefficient of x.

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 incorrectly applies the distributive property to m(x + 1) as mx + 1, resulting in an incorrect constant term condition.

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