Consider the expression $E(x) = 3(4x - 5) - a(2x - 7) + 6$, where $a$ is a real constant. (a) Apply the distributive property to write $E(x)$ in the form $mx + b$. (b) Determine the critical threshold value of $a$ for which the expression becomes completely independent of $x$ (the linear coefficient vanishes), and evaluate the resulting constant value of $E(x)$ at this threshold.

Consider the expression $E(x) = 3(4x - 5) - a(2x - 7) + 6$ , where $a$ is a real constant. (a) Apply the distributive property to write $E(x)$ in the form $mx + b$ . (b) Determine the critical threshold value of $a$ for which the expression becomes completely independent of $x$ (the linear coefficient vanishes), and evaluate the resulting constant value of $E(x)$ at this threshold.

Final answer

(a) $E(x) = (12 - 2a)x + (7a - 9)$ ; (b) Threshold value $a = 6$ , at which $E(x) = 33$ .

Steps

Part (a): Expanding and Simplifying $E(x)$

Step 1: Apply the distributive property to each grouped term. Multiply each term inside the parentheses by the factor in front:

For the first group: $3(4x - 5) = 3 \cdot 4x - 3 \cdot 5 = 12x - 15$
For the second group, distribute the negative factor $-a$ : $-a(2x - 7) = (-a) \cdot (2x) + (-a) \cdot (-7) = -2ax + 7a$

Common Pitfall: When distributing a negative term like $-a$ across $(2x - 7)$ , remember that multiplying two negative signs yields a positive: $(-a) \cdot (-7) = +7a$ .

Step 2: Combine the expanded terms. Substitute these expansions back into the full expression: $E(x) = (12x - 15) + (-2ax + 7a) + 6$

Step 3: Group like terms into the standard form $mx + b$ .

Group terms containing $x$ : $12x - 2ax = (12 - 2a)x$
Group the constant terms: $-15 + 7a + 6 = 7a - 9$

Putting them together gives: $E(x) = (12 - 2a)x + (7a - 9)$

Part (b): Finding the Threshold Value of $a$ and Evaluating $E(x)$

Step 1: Set the coefficient of $x$ to zero. An expression is independent of $x$ when the variable term vanishes completely, meaning its coefficient $m$ must equal $0$ : $12 - 2a = 0$ $2a = 12$ $a = 6$

Step 2: Evaluate the resulting constant value of $E(x)$ . Substitute $a = 6$ into the constant term $b = 7a - 9$ : $E(x) = (12 - 2(6))x + (7(6) - 9)$ $E(x) = 0x + (42 - 9)$ $E(x) = 33$

Thus, the threshold value is $a = 6$ , yielding the constant value $E(x) = 33$ .

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets to distribute the negative sign to the second term inside the parentheses when expanding -a(2x - 7), writing -7a instead of +7a.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
E(x) = 12x - 15 - 2ax - 7a + 6 $\implies$ E(x) = (12 - 2a)x + (-7a - 9)
Correct
E(x) = 12x - 15 - 2ax + 7a + 6 $\implies$ E(x) = (12 - 2a)x + (7a - 9)
When multiplying $-a$ by the second term $-7$ , the product of two negative values is positive: $(-a) \cdot (-7) = +7a$ . Distributing with a negative sign yielded $-7a$ instead of $+7a$ .
Step 2
12 - 2a = 0 $\implies$ a = 6
Step 3
E(x) = -7(6) - 9 = -42 - 9 = -51
Correct
E(x) = 7(6) - 9 = 42 - 9 = 33
Your arithmetic substitution is consistent with your expression from step 1, but the value is affected by the earlier sign error on $7a$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly applied the distributive property to expand $3(4x - 5)$ into $12x - 15$ .

You correctly factored out $x$ from the linear terms to obtain the coefficient $(12 - 2a)$ .

You correctly identified that the linear coefficient must equal zero for the expression to be independent of $x$ and found $a = 6$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Distributive Sign Error with Negative Factor

You wrote $-7a$ instead of $+7a$ when multiplying $-a$ by $-7$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Dropping the negative sign of the subtracted term inside parentheses when multiplying by an outer negative factor, forgetting that $(-a) \cdot (-b) = +ab$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Distributing a Negative Factor: $-k(cx - d) = -k \cdot cx + (-k) \cdot (-d) = -kcx + kd$
Always treat a subtraction inside the parentheses as adding a negative number, or explicitly write out the two minus signs before simplifying.

CORRECTED PATH

ACEpath Teacher's Demonstration

Expand $-a(2x - 7)$ as $(-a)(2x) + (-a)(-7) = -2ax + 7a$ .
Combine the constant terms: $-15 + 7a + 6 = 7a - 9$ , so $E(x) = (12 - 2a)x + (7a - 9)$ .
Substitute $a = 6$ into the constant term: $E(x) = 7(6) - 9 = 42 - 9 = 33$ .

Final answer

(a) $E(x) = (12 - 2a)x + (7a - 9)$ ; (b) Threshold $a = 6$ , yielding $E(x) = 33$ .

NEXT STEPS

Final Takeaways from Your Teacher

Practice distributing negative monomials over binomials with subtraction.
Double-check every negative-times-negative product before combining like terms.

OverviewACEpath Teacher's Diagnosis

02 The student sets the constant term rather than the linear coefficient to zero to make the expression independent of x.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
E(x) = 12x - 15 - 2ax + 7a + 6 $\implies$ E(x) = (12 - 2a)x + (7a - 9)
Step 2
7a - 9 = 0 $\implies$ a = $\frac{9}{7}$
Correct
12 - 2a = 0 $\implies$ a = 6
For an expression to be independent of $x$ , the variable term must vanish completely, which requires setting the coefficient of $x$ to zero ( $12 - 2a = 0$ ). You set the constant term $7a - 9$ to zero instead.
Step 3
E(x) = $\left(12 - 2\left(\frac{9}{7}\right)\right)$ x = $\left(12 - \frac{18}{7}\right)$ x = $\frac{66}{7}$ x
Correct
E(x) = 7(6) - 9 = 33
Because this substitution used the incorrect threshold $a = \frac{9}{7}$ , your resulting expression still contains $x$ , meaning it did not become independent of $x$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly applied the distributive property to expand $3(4x - 5)$ and $-a(2x - 7)$ .

You handled the double negative multiplication correctly when expanding $(-a)(-7) = +7a$ .

You organized and grouped like terms properly into standard linear form $mx + b$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Confusing linear coefficient with constant term for variable independence

You set the constant part $7a - 9 = 0$ instead of the variable's coefficient $12 - 2a = 0$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You confused eliminating the variable term ( $m = 0$ ) with eliminating the constant term ( $b = 0$ ). An expression $mx + b$ is independent of $x$ only when its rate of change with respect to $x$ is zero, meaning $m = 0$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Condition for Independence: An expression $mx + b$ is independent of $x$ if and only if the coefficient of $x$ vanishes ( $m = 0$ ).
Locate the coefficient multiplying $x$ , set that entire factor equal to $0$ , and solve for the unknown parameter.
Sanity Check the Final Value: If an expression is truly independent of $x$ , evaluating it cannot leave any $x$ terms behind.
Check your evaluated answer: if $x$ is still in the expression, re-verify which term you set to zero.

CORRECTED PATH

ACEpath Teacher's Demonstration

Set the coefficient of $x$ equal to zero: $12 - 2a = 0$ .
Solve for $a$ : $2a = 12 \implies a = 6$ .
Substitute $a = 6$ into the constant term $7a - 9$ : $E(x) = 7(6) - 9 = 42 - 9 = 33$ .

Final answer

(a) $E(x) = (12 - 2a)x + (7a - 9)$ ; (b) Threshold value $a = 6$ , at which $E(x) = 33$ .

NEXT STEPS

Final Takeaways from Your Teacher

Remember that 'independent of $x$ ' means $x$ has no effect on the value, which requires setting the coefficient of $x$ to $0$ .
Always verify that your evaluated result at the threshold is a purely numeric constant without any remaining variable.

OverviewACEpath Teacher's Diagnosis

03 The student neglects to multiply the constant 7 by a when distributing across the parentheses, writing + 7 instead of + 7a.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
E(x) = 12x - 15 - 2ax + 7 + 6
Correct
E(x) = 12x - 15 - 2ax + 7a + 6
When expanding $-a(2x - 7)$ , you must multiply every term inside the parentheses by $-a$ . Multiplying $-a \cdot (-7)$ yields $+7a$ , not $+7$ .
Step 2
E(x) = (12 - 2a)x - 2
Correct
E(x) = (12 - 2a)x + (7a - 9)
You grouped the like terms consistently with your previous line, but because $+7$ was missing the factor of $a$ , the constant term became $-15 + 7 + 6 = -2$ instead of $7a - 9$ .
Step 3
12 - 2a = 0 $\implies$ a = 6
Step 4
E(x) = -2
Correct
E(x) = 7(6) - 9 = 33
Because the constant term was missing the $7a$ component from Step 1, evaluating at $a = 6$ produced $-2$ instead of $7(6) - 9 = 33$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly applied the distributive property to the first term $3(4x - 5) = 12x - 15$ .

You correctly factored out $x$ from the linear terms to obtain the coefficient $(12 - 2a)$ .

You correctly set the variable coefficient to zero and solved for the critical value $a = 6$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Incomplete Distribution Over Parentheses

You distributed $-a$ to the first term $2x$ but only applied the negative sign to the second term $-7$ , dropping the factor $a$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Treating the trailing term inside parentheses as an isolated number rather than distributing the exterior factor to every term inside.

REASONING PATH

Your ACEpath teacher’s suggested approach

Check Every Term in Parentheses: For any expression $k(p + q)$ , both terms must be multiplied: $k \cdot p + k \cdot q$ .
Draw distribution arcs from the outside factor to every individual term inside parentheses before writing down the expanded terms.

CORRECTED PATH

ACEpath Teacher's Demonstration

Expand both products: $3(4x - 5) = 12x - 15$ and $-a(2x - 7) = -2ax + 7a$ .
Combine like terms: $E(x) = (12 - 2a)x + (-15 + 7a + 6) = (12 - 2a)x + (7a - 9)$ .
Set the linear coefficient to zero: $12 - 2a = 0 \implies a = 6$ .
Substitute $a = 6$ into the constant term: $E(x) = 7(6) - 9 = 42 - 9 = 33$ .

Final answer

(a) $E(x) = (12 - 2a)x + (7a - 9)$ ; (b) $a = 6$ with $E(x) = 33$

NEXT STEPS

Final Takeaways from Your Teacher

Practice expanding products where the outside factor is a variable or parameter with a negative sign, such as $-c(3y - 4)$ .
Before combining like terms, count the number of terms inside the parentheses to verify that the expansion has the exact same count of multiplied products.

Lock it in

Question

What is the general expansion of $-a(b - c)$ using the distributive property?

Answer

$-ab + ac$ . Multiplying the negative factor $-a$ by the subtracted term $-c$ yields a positive product: $(-a)(-c) = +ac$ .

Resources

Answer and steps

Final answer

Steps

Part (a): Expanding and Simplifying E(x)E(x)E(x)

Part (b): Finding the Threshold Value of aaa and Evaluating E(x)E(x)E(x)

Typical errors

01 The student forgets to distribute the negative sign to the second term inside the parentheses when expanding -a(2x - 7), writing -7a instead of +7a.

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 sets the constant term rather than the linear coefficient to zero to make the expression independent 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 neglects to multiply the constant 7 by a when distributing across the parentheses, writing + 7 instead of + 7a.

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

Part (a): Expanding and Simplifying $E(x)$

Part (b): Finding the Threshold Value of $a$ and Evaluating $E(x)$