In a college analytical chemistry laboratory, two students set up a spectrophotometric calibration model comparing absorbance $A$ at concentration $c\text{ (in mM)}$ across two different cuvette path lengths. Student A models the baseline as $3(2c - 1.5) + 4.5$, while Student B formulates the response as $6c + k$. During experimental validation, the spectrophotometer records an absorbance response where no single concentration $c$ can satisfy the equality when a cuvette offset $k = 1.2$ is introduced. Determine the number of solutions to the calibration balance $3(2c - 1.5) + 4.5 = 6c + 1.2$, and find the required offset $k^*$ that would yield infinitely many solutions (a fully calibrated identical instrument response).

In a college analytical chemistry laboratory, two students set up a spectrophotometric calibration model comparing absorbance $A$ at concentration $c\text{ (in mM)}$ across two different cuvette path lengths. Student A models the baseline as $3(2c - 1.5) + 4.5$ , while Student B formulates the response as $6c + k$ . During experimental validation, the spectrophotometer records an absorbance response where no single concentration $c$ can satisfy the equality when a cuvette offset $k = 1.2$ is introduced. Determine the number of solutions to the calibration balance $3(2c - 1.5) + 4.5 = 6c + 1.2$ , and find the required offset $k^*$ that would yield infinitely many solutions (a fully calibrated identical instrument response).

Final answer

The equation $6c = 6c + 1.2$ has no solution (inconsistent, $0 = 1.2$ ). An offset of $k^* = 0$ is required to achieve infinitely many solutions (an identity $6c = 6c$ ).

Steps

Simplify the Left-Hand Side (Student A's Expression)

Begin by applying the distributive property to expand and simplify Student A's model:

$3(2c - 1.5) + 4.5 = 3(2c) - 3(1.5) + 4.5$

$= 6c - 4.5 + 4.5$

Combining the constant terms:

$= 6c$

Determine the Number of Solutions when $k = 1.2$

Substitute the simplified expression for Student A and set $k = 1.2$ for Student B:

$6c = 6c + 1.2$

Subtract $6c$ from both sides to isolate the variable terms:

$6c - 6c = 1.2$

$0 = 1.2$

Because $0 = 1.2$ is a mathematical contradiction (a false statement that holds for no value of $c$ ), the equation is inconsistent. Therefore, there is no solution.

Find the Offset $k^*$ for Infinitely Many Solutions

For a linear equation of the form $ac + b_1 = ac + b_2$ to have infinitely many solutions, it must be an identity (true for every real value of $c$ ). This occurs when the constant terms on both sides are identical:

$6c = 6c + k^*$

Subtract $6c$ from both sides:

$0 = k^*$

Thus, setting the offset to $k^* = 0$ yields the identity $6c = 6c$ (or $0 = 0$ ), producing infinitely many solutions.

Common Pitfall to Avoid

When variable terms cancel completely (e.g., $6c - 6c = 0$ ), do not assume $c = 0$ . Instead, look at the remaining numerical statement:

If it is false (such as $0 = 1.2$ ), there is no solution.
If it is always true (such as $0 = 0$ ), there are infinitely many solutions.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets to distribute the factor of 3 to the constant term -1.5 inside the parentheses, resulting in an incorrect linear constant.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
3(2c - 1.5) + 4.5 = 6c + 1.2 $\implies$ 6c - 1.5 + 4.5 = 6c + 1.2 $\implies$ 6c + 3 = 6c + 1.2
Correct
3(2c - 1.5) + 4.5 = 6c + 1.2 $\implies$ 6c - 4.5 + 4.5 = 6c + 1.2 $\implies$ 6c = 6c + 1.2
When expanding $3(2c - 1.5)$ , the factor of $3$ must multiply both terms inside the parentheses. Multiplying $3 \times (-1.5)$ gives $-4.5$ , not $-1.5$ .
Step 2
3 = 1.2 $\implies$ 0 $\text{ solutions}$
Correct
0 = 1.2 $\implies$ $\text{no solution}$
Your reasoning that a false numerical statement implies zero solutions is completely sound; however, the values were carried over from the distribution slip in Step 1.
Step 3
6c + 3 = 6c + k $\implies$ k^* = 3
Correct
6c = 6c + k^* $\implies$ k^* = 0
You applied the correct condition for an identity by matching the constant terms, but using $6c + 3$ instead of $6c$ carried forward the earlier error.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly identified that a contradiction with unequal constants means the equation has no solution.

You matched the constant terms on both sides to set up the condition for infinitely many solutions.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Incomplete distribution of a scalar multiplier across grouped terms

You distributed $3$ to $2c$ to get $6c$ , but left the second term inside parentheses as $-1.5$ instead of calculating $3 \times (-1.5) = -4.5$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Overlooking the second term when applying the distributive property $a(b + c) = ab + ac$ , especially when the second term is a decimal or preceded by a negative sign.

REASONING PATH

Your ACEpath teacher’s suggested approach

Full Distributive Check: $a(bx + c) = a \cdot bx + a \cdot c$
Draw expansion arrows or explicitly write both products before combining like terms.
Identity Constant Comparison: For $mx + b_1 = mx + b_2$ to have infinitely many solutions, $b_1$ must equal $b_2$ .
Ensure each side is fully simplified to slope-intercept form before equating constant terms.

CORRECTED PATH

ACEpath Teacher's Demonstration

Expand the left-hand side completely: $3(2c - 1.5) + 4.5 = 6c - 4.5 + 4.5 = 6c$ .
Set up the equation with $k = 1.2$ : $6c = 6c + 1.2 \implies 0 = 1.2$ , which is a contradiction, so there is no solution ( $0$ solutions).
For infinitely many solutions, equate the simplified models: $6c = 6c + k^* \implies k^* = 0$ .

Final answer

The equation has no solution ( $0$ solutions); $k^* = 0$ .

NEXT STEPS

Final Takeaways from Your Teacher

Practice expanding binomials with non-integer constants to build automatic habits with the distributive property.
Always verify that constant terms cancel cleanly before concluding an offset value.

OverviewACEpath Teacher's Diagnosis

02 The student misinterprets the identity 0 = 0 as having a unique solution c = 0 rather than recognizing that k* must be 0 for infinitely many solutions.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
3(2c - 1.5) + 4.5 = 6c + 1.2 ⟹ 6c - 4.5 + 4.5 = 6c + 1.2 ⟹ 6c = 6c + 1.2 ⟹ 0 = 1.2 ⟹ 0 $\text{ solutions}$
Step 2
6c = 6c + k ⟹ $\text{If }$ k = 0 $\text{, then }$ 6c = 6c ⟹ 6c - 6c = 0
Step 3
c = 0 $\text{, so }$ 1 $\text{ solution, thus no value of }$ k $\text{ gives infinitely many solutions}$
Correct
0 = 0 $\text{ is an identity (always true for all }$ c $\text{), so }$ k^* = 0 $\text{ gives infinitely many solutions}$
The statement $6c - 6c = 0$ simplifies to $0 = 0$ , which is an identity true for every real value of $c$ , not an equation whose only solution is $c = 0$ . Therefore, setting $k^* = 0$ produces infinitely many solutions.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly applied the distributive property to expand $3(2c - 1.5) + 4.5$ into $6c$ .

You correctly recognized that the contradiction $0 = 1.2$ means the equation has zero solutions.

You noticed that setting $k = 0$ aligns both sides to $6c = 6c$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Misinterpreting Identity as Zero Solution

You treated $6c - 6c = 0$ as giving $c = 0$ , mistaking an identity where all variables eliminate into $0 = 0$ for a unique solution.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing an identically true equality ( $0 = 0$ , which holds for all values of the variable) with the equation $c = 0$ (which holds only when the variable equals zero).

REASONING PATH

Your ACEpath teacher’s suggested approach

Variable Elimination Rule: When subtracting identical variable terms from both sides ( $ax - ax = 0$ ), the variable $x$ disappears completely.
Check whether the remaining statement is a true identity (like $0 = 0$ ) or a contradiction (like $0 = 1.2$ ), rather than writing $x = 0$ .
Interpreting 0 = 0: An identity $0 = 0$ means any real value substituted for the variable produces a true statement.
Conclude that the equation has infinitely many solutions whenever both sides simplify to identical expressions.

CORRECTED PATH

ACEpath Teacher's Demonstration

Subtract $6c$ from both sides: $6c - 6c = 6c - 6c \implies 0 = 0$ .
Observe that $0 = 0$ contains no variable $c$ and is always true regardless of the value chosen for $c$ .
Conclude that $k^* = 0$ creates an identity, giving infinitely many solutions.

Final answer

No solution when $k = 1.2$ ; $k^* = 0$ yields infinitely many solutions.

NEXT STEPS

Final Takeaways from Your Teacher

Practice substituting small test values (like $c = 1$ and $c = 5$ ) into $6c = 6c$ to verify that any number satisfies the equation.
Review the three outcomes for linear equations: one unique solution ( $x = a$ ), no solution (contradiction $a = b$ ), and infinitely many solutions (identity $a = a$ ).

OverviewACEpath Teacher's Diagnosis

03 The student makes a sign error when expanding -3(1.5), adding 4.5 instead of subtracting, which leads to a constant term of 9.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
3(2c - 1.5) + 4.5 = 6c + 1.2 $\implies$ 6c + 4.5 + 4.5 = 6c + 1.2 $\implies$ 6c + 9 = 6c + 1.2
Correct
3(2c - 1.5) + 4.5 = 6c + 1.2 $\implies$ 6c - 4.5 + 4.5 = 6c + 1.2 $\implies$ 6c = 6c + 1.2
When distributing $3$ across $(2c - 1.5)$ , the product $3 \times (-1.5)$ is $-4.5$ , not $+4.5$ . This caused the left-hand side constant terms to add to $9$ instead of canceling out to $0$ .
Step 2
9 = 1.2 $\implies$ 0 $\text{ solutions}$
Correct
0 = 1.2 $\implies$ 0 $\text{ solutions (no solution)}$
Your algebraic reasoning here is logically sound: subtracting $6c$ leaves a false numerical statement, which correctly indicates no solution. However, the specific values in the contradiction were affected by the previous sign error.
Step 3
6c + 9 = 6c + k $\implies$ k^* = 9
Correct
6c = 6c + k $\implies$ k^* = 0
You correctly matched the constant terms so both sides would be identical identities. Because the left side should simplify to $6c + 0$ , the correct required offset is $k^* = 0$ rather than $9$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly distributed $3$ to the variable term $2c$ to obtain $6c$ .

You recognized that an impossible equality like $9 = 1.2$ corresponds to zero solutions.

You correctly understood that for an equation to have infinitely many solutions, the expressions on both sides must be identical.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Distributive Sign Error

You wrote $+4.5$ instead of $-4.5$ when multiplying the positive factor $3$ by the negative term $-1.5$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Overlooking the negative sign inside parentheses during distribution, treating $3 \times (-1.5)$ as positive.

REASONING PATH

Your ACEpath teacher’s suggested approach

Sign tracking during distribution: For any real numbers $a$ , $b$ , and $c$ , $a(b - c) = ab - ac$ . Multiplying a positive number by a negative number always yields a negative product.
Circle or highlight minus signs inside grouping symbols before expanding to ensure the resulting term retains its negative sign.

CORRECTED PATH

ACEpath Teacher's Demonstration

Expand the left-hand side: $3(2c) + 3(-1.5) + 4.5 = 6c - 4.5 + 4.5 = 6c$ .
For $k = 1.2$ , set $6c = 6c + 1.2$ . Subtracting $6c$ from both sides gives $0 = 1.2$ , which is a contradiction, so there is no solution (0 solutions).
For infinitely many solutions, equate $6c = 6c + k^*$ , which requires the constant terms to match: $k^* = 0$ .

Final answer

0 solutions (no solution) for $k = 1.2$ ; $k^* = 0$ for infinitely many solutions.

NEXT STEPS

Final Takeaways from Your Teacher

Double-check your signs whenever distributing across subtraction: positive $\times$ negative must remain negative.
Combine constant terms carefully after expansion to confirm whether they cancel completely before solving for parameters.

Lock it in

Question

What does the distributive property require when expanding $a(bx - c)$ ?

Answer

Both terms inside must be multiplied by $a$ : $a(bx - c) = a \cdot bx - a \cdot c$ .

Resources

Answer and steps

Final answer

Steps

Common Pitfall to Avoid

Typical errors

01 The student forgets to distribute the factor of 3 to the constant term -1.5 inside the parentheses, resulting in an incorrect linear constant.

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 misinterprets the identity 0 = 0 as having a unique solution c = 0 rather than recognizing that k* must be 0 for 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

03 The student makes a sign error when expanding -3(1.5), adding 4.5 instead of subtracting, which leads to a constant term of 9.

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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