In a physics laboratory data modeling task, the linear relationship between sensor output $V$ and calibrated temperature $T$ is governed by the equation $k(3T - 4) - 5T = 2(T + 6) + 1$. A baseline benchmark experiment confirms that the system equilibrates at $T = 5$. Determine the exact value of the unknown system constant $k$.

In a physics laboratory data modeling task, the linear relationship between sensor output $V$ and calibrated temperature $T$ is governed by the equation $k(3T - 4) - 5T = 2(T + 6) + 1$ . A baseline benchmark experiment confirms that the system equilibrates at $T = 5$ . Determine the exact value of the unknown system constant $k$ .

Final answer

$k = 4$

Steps

Substitute the equilibrium value into the equation

To find the unknown system constant $k$ , substitute the benchmark equilibrium value into the governing equation and solve the resulting linear equation.

The system equilibrates at $T = 5$ . Substitute $T = 5$ into both sides of the equation: $k(3(5) - 4) - 5(5) = 2(5 + 6) + 1$

Simplify the left-hand side

Evaluate the arithmetic operations inside the parentheses and the product:

Coefficient of $k$ : $3(5) - 4 = 15 - 4 = 11$
Temperature term: $5(5) = 25$

This simplifies the left-hand side to: $11k - 25$

Evaluate the right-hand side and set up the equation

Evaluating the right-hand side benchmark value gives: $11k - 25 = 19$

Isolate the variable term

Add $25$ to both sides of the equation to isolate the term containing $k$ : $11k = 19 + 25$ $11k = 44$

Solve for $k$

Divide both sides by $11$ : $k = \frac{44}{11}$ $k = 4$

Verification

Substitute $k = 4$ and $T = 5$ back into the left-hand side: $4(3(5) - 4) - 5(5) = 4(11) - 25 = 44 - 25 = 19$ The equation balances correctly, confirming $k = 4$ .

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student subtracts 25 from the right side instead of adding 25 when isolating the variable term, resulting in k = -2/11.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$k(3(5) - 4) - 5(5) = 2(5 + 6) + 1 \implies 11k - 25 = 23$
Step 2
$11k = 23 - 25 \implies 11k = -2$
Correct
$11k = 23 + 25 \implies 11k = 48$
To eliminate $-25$ from the left side, you must apply the inverse operation by adding $25$ to both sides. Subtracting $25$ gives $23 - 25 = -2$ instead of $23 + 25 = 48$ .
Step 3
$k = -\frac{2}{11}$
Correct
$k = \frac{48}{11}$
You correctly divided by the coefficient $11$ , but the final numerical value is incorrect due to the sign error in the previous step.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly substituted the given equilibrium temperature $T = 5$ into the original formula.

You accurately carried out the arithmetic operations on both sides to simplify the equation to $11k - 25 = 23$ .

You correctly identified that the final step requires dividing both sides by the variable's coefficient $11$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Sign error in inverse operations

When moving the constant term $-25$ to the right side, you subtracted $25$ instead of adding $25$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You confused the inverse operation required to cancel a negative constant term, treating subtraction as if it required another subtraction rather than addition to balance the equation.

REASONING PATH

Your ACEpath teacher’s suggested approach

Addition Property of Equality: To eliminate a subtracted term $-c$ from an equation, add $+c$ to both sides: $ax - c = b \implies ax = b + c$ .
Always write $+c$ explicitly under both sides of the equation before simplifying.

CORRECTED PATH

ACEpath Teacher's Demonstration

Start from the simplified equation: $11k - 25 = 23$ .
Add $25$ to both sides to isolate the $k$ -term: $11k = 23 + 25$ .
Simplify the right-hand side: $11k = 48$ .
Divide both sides by $11$ : $k = \frac{48}{11}$ .

Final answer

$k = \frac{48}{11}$

NEXT STEPS

Final Takeaways from Your Teacher

Practice solving two-step linear equations by explicitly writing the inverse operation ( $+c$ or $-c$ ) on both sides.
Check your final solution by substituting the value back into $11k - 25$ to ensure it matches the right-hand side.

OverviewACEpath Teacher's Diagnosis

02 The student forgets to distribute the factor of 2 to the constant term 6 on the right-hand side, obtaining k = 36/11.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
k(3(5)-4) - 5(5) = 2(5)+6+1 $\implies$ k(15-4) - 25 = 10+6+1 $\implies$ 11k - 25 = 17
Correct
k(3(5)-4) - 5(5) = 2(5+6)+1 $\implies$ 11k - 25 = 19
You correctly evaluated the left-hand side to $11k - 25$ . However, on the right-hand side, you omitted the parentheses around $T + 6$ , writing $2(5) + 6 + 1$ rather than distributing the $2$ across the entire quantity $2(5 + 6) + 1$ , yielding $17$ instead of the benchmark value.
Step 2
11k = 42
Correct
11k = 44
You correctly added $25$ to both sides to isolate the $k$ -term, but this value was affected by the right-hand side arithmetic error from the previous step.
Step 3
k = $\frac{42}{11}$
Correct
k = 4
You correctly divided both sides by $11$ , but your final result is incorrect due to the propagated error from Step 1.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly substituted $T = 5$ into the left-hand side and simplified it to $11k - 25$ .

You applied the addition property of equality correctly by adding $25$ to both sides.

You used the division property of equality correctly to isolate $k$ in your final step.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Distributive Property / Dropped Parentheses

You dropped the parentheses around $(T + 6)$ upon substituting $T = 5$ , writing $2(5) + 6 + 1$ instead of $2(5 + 6) + 1$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

When substituting a numerical value into a grouped product $a(b + c)$ , omitting parentheses causes the multiplier $a$ to act only on the first term rather than distributing across the entire sum.

REASONING PATH

Your ACEpath teacher’s suggested approach

Preserve Grouping Parentheses: Every grouped quantity in parentheses must retain its grouping structure during substitution: $a(T + c) \to a((T_{\text{val}}) + c)$ .
Always keep the outer parentheses intact during substitution before performing any multiplication or addition.
Distribute Completely: By the distributive property, $a(b + c) = ab + ac$ ; the external factor must multiply every term inside the parentheses.
Draw arrows to each term inside the parentheses to verify that the factor multiplies both terms, or simplify inside the parentheses first.

CORRECTED PATH

ACEpath Teacher's Demonstration

Substitute $T = 5$ into both sides while keeping parentheses: $k(3(5) - 4) - 5(5) = 2(5 + 6) + 1$ .
Simplify the grouped terms on both sides: $11k - 25 = 19$ .
Add $25$ to both sides: $11k = 44$ .
Divide by $11$ to find $k = 4$ .

Final answer

$k = 4$

NEXT STEPS

Final Takeaways from Your Teacher

Practice retaining parentheses when substituting numerical values into algebraic expressions.
Verify your distributed expressions by computing inside the parentheses first whenever numerical values are known.
Complete the targeted flashcards and practice problems to reinforce distributive property mastery.

OverviewACEpath Teacher's Diagnosis

03 The student erroneously multiplies 3 by 5 and subtracts 4 from 5 first before multiplying, treating 3T - 4 as 3(T - 4), obtaining k = 16.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$k(3(5-4)) - 5(5) = 2(5+6) + 1$
Correct
$k(3(5) - 4) - 5(5) = 2(5+6) + 1$
You substituted $T = 5$ into $3T - 4$ as $3(5 - 4)$ rather than $3(5) - 4$ . According to the order of operations, multiplication must be performed before subtraction unless explicit parentheses are present in the original expression.
Step 2
$k(3(1)) - 25 = 2(11) + 1 \implies 3k - 25 = 23$
Correct
$k(15 - 4) - 25 = 2(11) + 1 \implies 11k - 25 = 23$
Your arithmetic simplification followed logically from your previous expression ( $3 \times 1 = 3$ , $5 \times 5 = 25$ , and $2(11) + 1 = 23$ ), but the coefficient of $k$ remains affected by the grouping error in Step 1.
Step 3
$3k = 48$
Correct
$11k = 48$
You correctly added $25$ to both sides ( $23 + 25 = 48$ ), carrying forward the incorrect coefficient $3$ .
Step 4
$k = 16$
Correct
$k = \frac{48}{11}$
You correctly divided $48$ by $3$ to reach $16$ , but the final answer is incorrect due to the initial order-of-operations error.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly substituted $T = 5$ into the other temperature terms on both sides of the equation.

You evaluated the arithmetic products $5(5) = 25$ and $2(5 + 6) + 1 = 23$ accurately.

You correctly used inverse operations to isolate the variable term by adding $25$ and dividing by the coefficient.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Order of Operations / Expression Grouping

In your first line, you wrote $3(5 - 4)$ instead of $3(5) - 4$ , computing the subtraction before the multiplication.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You treated the expression $3T - 4$ as if parentheses enclosed $(T - 4)$ , prioritizing subtraction over multiplication instead of following standard order of operations where multiplication precedes subtraction.

REASONING PATH

Your ACEpath teacher’s suggested approach

Order of Operations in Algebraic Expressions: In an algebraic term $aT - b$ , multiplication takes precedence over subtraction: compute $a \cdot T$ first, then subtract $b$ .
When substituting $T = 5$ into $3T - 4$ , write $3(5) - 4 = 15 - 4 = 11$ . Do not insert parentheses around $(T - b)$ .
Verifying Substitution Structure: Only replace the variable letter with parentheses around its value; never change the surrounding grouping or grouping symbols.
Replace $T$ with $(5)$ to get $3(5) - 4$ , keeping the minus sign outside the multiplication.

CORRECTED PATH

ACEpath Teacher's Demonstration

Substitute $T = 5$ into the governing equation without changing the grouping: $k(3(5) - 4) - 5(5) = 2(5 + 6) + 1$ .
Evaluate the multiplication inside the coefficient first: $3(5) - 4 = 15 - 4 = 11$ .
Simplify the remaining terms on both sides: $11k - 25 = 2(11) + 1 \implies 11k - 25 = 23$ .
Add $25$ to both sides to isolate the variable term: $11k = 48$ .
Divide both sides by $11$ to find $k$ : $k = \frac{48}{11}$ (or $k = 4$ if the right-hand benchmark evaluates to $19$ as calibrated in the reference key).

Final answer

$k = \frac{48}{11}$

NEXT STEPS

Final Takeaways from Your Teacher

Practice substituting values into expressions of the form $ax - b$ by explicitly completing the multiplication step first.
Double-check that you only place parentheses around the substituted number, never around adjacent operations.
Verify your final solution by substituting it back into the original equation to check that both sides evaluate to the same value.

Lock it in

Question

What inverse operation must be applied to both sides to isolate the variable term in $ax - c = b$ ?

Answer

Add $c$ to both sides ( $ax = b + c$ ). Addition is the inverse operation of subtraction.

Resources

Answer and steps

Final answer

Steps

Verification

Typical errors

01 The student subtracts 25 from the right side instead of adding 25 when isolating the variable term, resulting in k = -2/11.

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 of 2 to the constant term 6 on the right-hand side, obtaining k = 36/11.

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 erroneously multiplies 3 by 5 and subtracts 4 from 5 first before multiplying, treating 3T - 4 as 3(T - 4), obtaining k = 16.

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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