Determine the exact value of the parameter $a$ for which the equation $4(ax - 3) + 7x = 3(5x - 4) + 2x$ has infinitely many real solutions for $x$, and identify the boundary behavior of the solution set if $a$ deviates from this critical value.

Determine the exact value of the parameter $a$ for which the equation $4(ax - 3) + 7x = 3(5x - 4) + 2x$ has infinitely many real solutions for $x$ , and identify the boundary behavior of the solution set if $a$ deviates from this critical value.

Final answer

$a = \frac{5}{2}$ ; if $a = \frac{5}{2}$ , the solution set is all real numbers $\mathbb{R}$ (infinitely many solutions); if $a \neq \frac{5}{2}$ , there is exactly one unique solution $x = 0$ .

Steps

Expand and simplify both sides

To determine the value of $a$ and describe the solution set, we simplify both sides of the equation and express it in the standard linear form $Cx = D$ .

Expand the left-hand side (LHS): $4(ax - 3) + 7x = 4ax - 12 + 7x = (4a + 7)x - 12$

Expand the right-hand side (RHS): $3(5x - 4) + 2x = 15x - 12 + 2x = 17x - 12$

Set up the equation in standard form

Equate the simplified expressions: $(4a + 7)x - 12 = 17x - 12$

Add $12$ to both sides: $(4a + 7)x = 17x$

Subtract $17x$ from both sides to group all terms containing $x$ together: $(4a + 7 - 17)x = 0$ $(4a - 10)x = 0$

Condition for infinitely many solutions

A linear equation in the form $Cx = D$ has:

Infinitely many solutions if and only if $C = 0$ and $D = 0$ (producing the identity $0 \cdot x = 0$ ).
No solution if $C = 0$ and $D \neq 0$ (producing a contradiction such as $0 \cdot x = k$ ).
Exactly one unique solution if $C \neq 0$ , given by $x = \frac{D}{C}$ .

Here, the constant term is already $D = 0$ . Therefore, to have infinitely many solutions, the coefficient of $x$ must also equal zero: $4a - 10 = 0$ $4a = 10$ $a = \frac{10}{4} = \frac{5}{2}$

When $a = \frac{5}{2}$ , the equation becomes $0 \cdot x = 0$ , which is satisfied by every real number. Hence, the solution set is all real numbers $\mathbb{R}$ .

Behavior when $a$ deviates from $\frac{5}{2}$

If $a \neq \frac{5}{2}$ , then the coefficient $(4a - 10) \neq 0$ . Dividing both sides of $(4a - 10)x = 0$ by the non-zero quantity $(4a - 10)$ gives: $x = \frac{0}{4a - 10} = 0$

Thus, whenever $a$ deviates from $\frac{5}{2}$ , the equation has exactly one unique solution, $x = 0$ .

Common Pitfall: Do not assume that an equation with variable coefficients automatically has "no solution" when the condition for infinitely many solutions fails. Always check the constant term $D$ : because $D = 0$ here, non-zero values of $C$ yield the single solution $x = 0$ , not "no solution".

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student misapplies the distributive property by multiplying only the first term inside the parentheses, leading to an incorrect value of a.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
4ax - 3 + 7x = 15x - 4 + 2x
Correct
4ax - 12 + 7x = 15x - 12 + 2x
When expanding $4(ax - 3)$ and $3(5x - 4)$ , the factor outside each set of parentheses must multiply every term inside: $4 \cdot (-3) = -12$ and $3 \cdot (-4) = -12$ .
Step 2
(4a + 7)x - 3 = 17x - 4 $\implies$ (4a - 10)x = -1
Correct
(4a + 7)x - 12 = 17x - 12 $\implies$ (4a - 10)x = 0
You correctly combined like terms and moved expressions across the equals sign, but the constant terms were carried over from the incomplete distribution in Step 1.
Step 3
4a - 10 = 0 $\implies$ a = $\frac{5}{2}$
Correct
4a - 10 = 0 $\text{ and }$ 0 = 0 $\implies$ a = $\frac{5}{2}$
Although $a = \frac{5}{2}$ is the correct numerical value, in your reduced equation $(4a - 10)x = -1$ , setting $4a - 10 = 0$ yields $0 \cdot x = -1$ , which produces no solution rather than infinitely many solutions. Infinitely many solutions require the constant side to be $0$ as well.
Step 4
$\text{If }$ a $\neq$ $\frac{5}{2}$ ,\ (4a - 10)x = -1 $\implies$ x = $\frac{-1}{4a - 10}$
Correct
$\text{If }$ a $\neq$ $\frac{5}{2}$ ,\ (4a - 10)x = 0 $\implies$ x = $\frac{0}{4a - 10}$ = 0
Because the correct simplified equation is $(4a - 10)x = 0$ , dividing by the non-zero coefficient $(4a - 10)$ gives $x = 0$ , not a variable fraction.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly grouped like terms in $x$ involving the parameter $a$ .

You recognized that the coefficient of $x$ must be zero to eliminate the variable term.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Distributive Property & Linear System Classification

You multiplied the outer factor only into the first term of each set of parentheses, leaving $-3$ and $-4$ unchanged instead of multiplying them by $4$ and $3$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Incomplete distribution across binomials, paired with checking only the $x$ -coefficient rather than ensuring both sides reduce to the identity $0 = 0$ for infinitely many solutions.

REASONING PATH

Your ACEpath teacher’s suggested approach

Distribute to every term: $k(u + v) = ku + kv$
Draw distribution arrows or write out each individual multiplication before simplifying.
Identity condition: $Cx = D$ has infinitely many solutions if and only if $C = 0$ and $D = 0$ .
Always verify that the constant term $D$ is also zero; if $C = 0$ but $D \neq 0$ , the equation has no solution.

CORRECTED PATH

ACEpath Teacher's Demonstration

Expand both sides completely: $4ax - 12 + 7x = 15x - 12 + 2x$
Combine like terms: $(4a + 7)x - 12 = 17x - 12$
Add $12$ to both sides and subtract $17x$ : $(4a - 10)x = 0$
For infinitely many solutions, set the coefficient to $0$ : $4a - 10 = 0 \implies a = \frac{5}{2}$
For $a \neq \frac{5}{2}$ , divide both sides by $(4a - 10)$ : $x = \frac{0}{4a - 10} = 0$

Final answer

$a = \frac{5}{2}$ ; if $a = \frac{5}{2}$ , there are infinitely many real solutions; if $a \neq \frac{5}{2}$ , there is exactly one unique solution $x = 0$ .

NEXT STEPS

Final Takeaways from Your Teacher

Practice expanding expressions with parentheses by explicitly writing the product for both terms.
Review the three solution cases for standard linear equations $Cx = D$ : one solution ( $C \neq 0$ ), no solution ( $C = 0, D \neq 0$ ), and infinitely many solutions ( $C = 0, D = 0$ ).

OverviewACEpath Teacher's Diagnosis

02 The student assumes that when the parameter deviates from the critical value with infinitely many solutions, the equation necessarily has no solution.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
4ax - 12 + 7x = 15x - 12 + 2x $\implies$ (4a + 7)x - 12 = 17x - 12 $\implies$ (4a - 10)x = 0
Step 2
4a - 10 = 0 $\implies$ a = $\frac{5}{2}$
Step 3
$\text{If }$ a $\neq$ $\frac{5}{2} $\text{, there are no solutions.}$
Correct
$\text{If }$ a $\neq$ $\frac{5}{2}$$\text{, then }$ 4a - 10 $\neq$ 0 $\text{, so }$ x = $\frac{0}{4a - 10}$ = 0$\text{ (exactly one unique solution).}$$
When $a \neq \frac{5}{2}$ , the coefficient $(4a - 10)$ is non-zero. Dividing the right-hand side $0$ by this non-zero number yields $x = 0$ , which is a valid unique solution, not 'no solution'.

WHAT YOU DID WELL

Compliments from Your Teacher

You expanded both sides of the linear equation accurately.

You successfully collected like terms and isolated the variable factor to reach $(4a - 10)x = 0$ .

You correctly solved $4a - 10 = 0$ to obtain $a = \frac{5}{2}$ for the identity condition.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Boundary behavior misinterpretation in $Cx = D$

Assuming that failing the condition for infinitely many solutions automatically implies no solution.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Conflating a linear equation with non-zero constants with the homogeneous form $Cx = 0$ . In $Cx = D$ , 'no solution' requires $C = 0$ and $D \neq 0$ . Here, $D = 0$ , so any non-zero value of $C$ yields the well-defined unique root $x = \frac{0}{C} = 0$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Inspect the Constant Term $D$: In a linear equation reduced to $Cx = D$ , look at both $C$ and $D$ before deciding the number of solutions.
When $C \neq 0$ , the equation always has exactly one solution: $x = \frac{D}{C}$ . Never declare 'no solution' unless you reach a contradiction like $0 \cdot x = D$ where $D \neq 0$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Write the simplified equation: $(4a - 10)x = 0$ .
Assume $a \neq \frac{5}{2}$ , which means $(4a - 10) \neq 0$ .
Divide both sides by the non-zero quantity $(4a - 10)$ : $x = \frac{0}{4a - 10} = 0$ .
Conclude that when $a$ deviates from $\frac{5}{2}$ , there is exactly one unique solution, $x = 0$ .

Final answer

$a = \frac{5}{2}$ ; if $a = \frac{5}{2}$ , the solution set is all real numbers $\mathbb{R}$ (infinitely many solutions); if $a \neq \frac{5}{2}$ , there is exactly one unique solution, $x = 0$ .

NEXT STEPS

Final Takeaways from Your Teacher

Review the three solution cases for standard linear equations $Cx = D$ : unique solution ( $C \neq 0$ ), infinitely many solutions ( $C = 0, D = 0$ ), and no solution ( $C = 0, D \neq 0$ ).
Solve the follow-up practice problems to reinforce checking the constant term $D$ when parameters vary.

OverviewACEpath Teacher's Diagnosis

03 The student forgets to include the additional term 7x when collecting like terms on the left side, resulting in an incorrect parameter value.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
4ax - 12 + 7x = 15x - 12 + 2x
Step 2
4ax - 12 = 17x - 12 ⟹ (4a - 17)x = 0
Correct
(4a + 7)x - 12 = 17x - 12 ⟹ (4a - 10)x = 0
The term $+7x$ on the left-hand side was dropped when simplifying. Combining all $x$ -terms should give $(4a + 7 - 17)x = (4a - 10)x = 0$ rather than $(4a - 17)x = 0$ .
Step 3
4a - 17 = 0 ⟹ a = $\frac{17}{4}$
Correct
4a - 10 = 0 ⟹ a = $\frac{5}{2}$
Your reasoning to set the coefficient of $x$ to $0$ is correct, but this step inherits the numerical error from omitting $+7x$ .
Step 4
$\text{If }$ a $\neq$ $\frac{17}{4}$ ,\ (4a - 17)x = 0 ⟹ x = 0
Correct
$\text{If }$ a $\neq$ $\frac{5}{2}$ ,\ (4a - 10)x = 0 ⟹ x = 0
Your logic for the boundary case is mathematically sound: when the linear coefficient is non-zero, dividing $0$ by that non-zero quantity gives the unique solution $x = 0$ . Only the critical threshold value itself was affected by the earlier omission.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly applied the distributive property to expand both sides of the equation.

You correctly identified that an equation in the form $Cx = 0$ has infinitely many solutions precisely when $C = 0$ .

You correctly reasoned that when $C \neq 0$ , the equation yields the unique solution $x = 0$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Term Omission During Like-Term Collection

You accidentally dropped the $+7x$ term on the left-hand side while combining like terms.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Incomplete accounting of unparenthesized linear terms when collecting like terms after expansion.

REASONING PATH

Your ACEpath teacher’s suggested approach

Inventory terms before combining: Every term present after expansion must be accounted for in the grouped coefficient.
Underline or highlight all $x$ -terms on each side before factoring out $x$ .
Condition for infinitely many solutions: An equation $Cx = D$ has infinitely many solutions if and only if $C = 0$ and $D = 0$ .
Ensure both the combined coefficient of $x$ and the combined constant term reduce to zero.

CORRECTED PATH

ACEpath Teacher's Demonstration

Group all $x$ -terms on the left: $(4a + 7)x - 12 = 17x - 12$
Add $12$ to both sides and subtract $17x$ : $(4a + 7 - 17)x = 0$ , which simplifies to $(4a - 10)x = 0$
Set the coefficient to zero for infinitely many solutions: $4a - 10 = 0 \implies a = \frac{5}{2}$
Analyze the boundary case: if $a \neq \frac{5}{2}$ , then $4a - 10 \neq 0$ , giving the unique solution $x = \frac{0}{4a - 10} = 0$

Final answer

$a = \frac{5}{2}$ ; if $a = \frac{5}{2}$ , there are infinitely many solutions (all real numbers $\mathbb{R}$ ); if $a \neq \frac{5}{2}$ , there is exactly one unique solution, $x = 0$ .

NEXT STEPS

Final Takeaways from Your Teacher

Practice using highlighters or tally marks to verify that every expanded term is gathered into the grouped coefficient.
Substitute your solved parameter value back into the original equation as a quick verification check.

Lock it in

Question

What two conditions must be satisfied for a linear equation in the form $Cx = D$ to have infinitely many solutions?

Answer

Both $C = 0$ and $D = 0$ , which yields the identity $0 \cdot x = 0$ .

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student misapplies the distributive property by multiplying only the first term inside the parentheses, leading to an incorrect value of a.

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 assumes that when the parameter deviates from the critical value with infinitely many solutions, the equation necessarily has no solution.

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 forgets to include the additional term 7x when collecting like terms on the left side, resulting in an incorrect parameter value.

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