Consider the multi-step linear inequality $5 - 2(3x - 4) < ax + 21$. If the solution set for $x$ is $(-2, \infty)$, determine the exact value of the real parameter $a$.

Consider the multi-step linear inequality $5 - 2(3x - 4) < ax + 21$ . If the solution set for $x$ is $(-2, \infty)$ , determine the exact value of the real parameter $a$ .

Final answer

$a = -2$

Steps

Simplify the left-hand side

To find the value of $a$ , we solve the linear inequality in terms of $a$ and compare the result to the given solution set $(-2, \infty)$ .

Apply the distributive property to expand the expression: $5 - 2(3x - 4) = 5 - 6x + 8 = 13 - 6x$

The original inequality becomes: $13 - 6x < ax + 21$

Collect like terms

Rearrange the inequality by grouping the variable terms on one side and the constant terms on the other: $13 - 21 < ax + 6x$ $-8 < (a + 6)x$

Rewriting with the variable term on the left gives: $(a + 6)x > -8$

Match the inequality direction and boundary

The given solution set is $(-2, \infty)$ , which corresponds to the inequality: $x > -2$

Since the inequality symbol is $>$ in both $(a + 6)x > -8$ and $x > -2$ , the coefficient of $x$ must be strictly positive when dividing both sides (so that the inequality sign does not reverse): $a + 6 > 0 \implies a > -6$

Dividing both sides of $(a + 6)x > -8$ by $(a + 6)$ : $x > \frac{-8}{a + 6}$

Solve for $a$

Set the boundary value equal to $-2$ : $\frac{-8}{a + 6} = -2$

Multiply both sides by $(a + 6)$ : $-8 = -2(a + 6)$

Divide both sides by $-2$ : $4 = a + 6$ $a = -2$

Since $a = -2$ satisfies the condition $a + 6 = 4 > 0$ , the inequality direction is preserved, confirming that $a = -2$ .

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student makes a sign error when distributing -2 across the parentheses, writing -8 instead of +8, which leads to an incorrect constant term and a = -4.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
5 - 6x - 8 < ax + 21 $\implies$ -3 - 6x < ax + 21
Correct
5 - 6x + 8 < ax + 21 $\implies$ 13 - 6x < ax + 21
When expanding $-2(3x - 4)$ , multiplying the negative factor $-2$ by $-4$ yields $+8$ , not $-8$ . As a result, the constant on the left side should be $5 + 8 = 13$ rather than $5 - 8 = -3$ .
Step 2
-(6 + a)x < 24
Correct
-(6 + a)x < 8
This rearrangement follows logically from your previous expression by subtracting $ax$ and adding $3$ to both sides, but it inherits the incorrect constant $-3$ .
Step 3
x > $\frac{24}{-(6 + a)}$ $\implies$ $\frac{-24}{6 + a}$ = -2
Correct
x > $\frac{8}{-(6 + a)}$ $\implies$ $\frac{-8}{6 + a}$ = -2
You correctly recognized that the solution set $(-2, \infty)$ means $x > -2$ and equated the boundary expression to $-2$ , but the numerator carries the propagated error from earlier.
Step 4
6 + a = 12 $\implies$ a = 6
Correct
6 + a = 4 $\implies$ a = -2
You correctly solved the resulting linear equation for $a$ , but the value reflects the earlier sign mistake.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly recognized that the solution set $(-2, \infty)$ corresponds to the inequality $x > -2$ .

You correctly grouped the variable terms containing $x$ together.

You correctly set the boundary value equal to $-2$ to solve for the parameter $a$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Sign error during distribution

Multiplying $-2$ by $-4$ produced $-8$ instead of $+8$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

When expanding parentheses preceded by a negative factor, the sign of the constant term inside the parentheses was not flipped.

REASONING PATH

Your ACEpath teacher’s suggested approach

Distribute the Negative Sign: $-c(px - q) = -c \cdot px + c \cdot q$ for any positive $c$ and $q$ .
Treat the negative sign as part of the multiplier $-2$ , so $(-2) \times (-4) = +8$ before combining with other constants.

CORRECTED PATH

ACEpath Teacher's Demonstration

Expand the left-hand side correctly: $5 - 2(3x - 4) = 5 - 6x + 8 = 13 - 6x$ .
Substitute into the inequality: $13 - 6x < ax + 21$ .
Collect terms: $-(6 + a)x < 21 - 13 \implies -(6 + a)x < 8$ .
Divide by $-(6 + a)$ to match $x > -2$ , requiring $-(6 + a) < 0$ (so $a + 6 > 0$ ): $x > \frac{8}{-(6 + a)} = \frac{-8}{a + 6}$ .
Equate the boundary to $-2$ : $\frac{-8}{a + 6} = -2 \implies a + 6 = 4$ .

Final answer

a = -2

NEXT STEPS

Final Takeaways from Your Teacher

Always write out the intermediate multiplication step, such as $+ (-2)(-4)$ , when distributing a negative coefficient across parentheses.
Check the sign of every term inside the parentheses immediately after expanding before moving on to collect like terms.

OverviewACEpath Teacher's Diagnosis

02 The student moves terms across the inequality but incorrectly changes the sign of the constant term when subtracting, calculating 21 - 13 = -8 instead of 8, yielding a = -10.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
5 - 6x + 8 < ax + 21 $\implies$ 13 - 6x < ax + 21
Step 2
-(6+a)x < -8
Correct
-(6+a)x < 8
When subtracting $13$ from both sides, the right-hand side is $21 - 13 = 8$ , not $-8$ .
Step 3
x > $\frac{-8}{-(6+a)}$
Correct
x > $\frac{8}{-(6+a)}$
Your division and inequality sign reversal followed logically from your previous expression, but the numerator carried over the incorrect negative sign.
Step 4
$\frac{8}{6+a} = -2 \implies 6+a = -4$
Correct
$\frac{-8}{6+a} = -2 \implies 6+a = 4$
You correctly matched the boundary fraction to $-2$ and solved for $6+a$ , carrying forward the sign error from earlier.
Step 5
a = -10
Correct
a = -2
Your final result $a = -10$ resulted directly from solving $6 + a = -4$ instead of $6 + a = 4$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly applied the distributive property to expand $5 - 2(3x - 4)$ into $13 - 6x$ .

You correctly factored out $x$ when collecting the variable terms.

You recognized that matching the solution set $(-2, \infty)$ requires setting the boundary expression equal to $-2$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Arithmetic / Transposition Sign Error

When moving the constant $13$ across the inequality to combine it with $21$ , you wrote $21 - 13$ as $-8$ instead of $+8$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You likely subtracted in the wrong order or conflated moving $+13$ to the right with subtracting $21$ from $13$ , resulting in an unintended sign inversion on the constant term.

REASONING PATH

Your ACEpath teacher’s suggested approach

Constant Balance Check: Subtracting a positive constant $c$ from both sides gives $\text{RHS} - c$ .
Explicitly write out $21 - 13 = +8$ before rewriting the entire inequality.
Sign of Coefficient Check: If the solution set is $x > -2$ , the isolated inequality must take the form $x > \text{boundary}$ .
Check whether isolating $(a + 6)x > -8$ directly on the positive side simplifies sign handling.

CORRECTED PATH

ACEpath Teacher's Demonstration

Subtract $13$ and $ax$ from both sides: $-(a + 6)x < 21 - 13$ , which simplifies to $-(a + 6)x < 8$ .
Equivalently, add $6x$ and subtract $21$ to keep positive coefficients: $(a + 6)x > -8$ .
Since the solution set is $x > -2$ , divide by $(a + 6) > 0$ to get $x > \frac{-8}{a + 6}$ .
Set the boundary value equal to $-2$ : $\frac{-8}{a + 6} = -2$ .
Multiply both sides by $a + 6$ : $-8 = -2(a + 6) \implies a + 6 = 4$ .

Final answer

a = -2

NEXT STEPS

Final Takeaways from Your Teacher

Write out intermediate subtraction steps when moving constants across inequality signs.
Practice moving variable terms to the side where their coefficient is positive to avoid dealing with multiple negative signs.
Substitute the final parameter value back into the original inequality to verify that the boundary matches.

OverviewACEpath Teacher's Diagnosis

03 The student ignores the direction of the inequality flip when solving, mistakenly treating the division as keeping the original '<' sign and matching x < -2 instead of recognizing the direction reversal.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
5 - 6x + 8 < ax + 21 $\implies$ 13 - 6x < ax + 21
Step 2
8 < (a+6)x $\implies$ (a+6)x > 8 $\implies$ x > $\frac{8}{a+6}$
Correct
-8 < (a+6)x $\implies$ (a+6)x > -8 $\implies$ x > - $\frac{8}{a+6}$
When subtracting $21$ from both sides, $13 - 21$ equals $-8$ , not $+8$ .
Step 3
$\frac{8}{a+6}$ = 2 $\implies$ a+6 = 4 $\implies$ a = -2
Correct
- $\frac{8}{a+6}$ = -2 $\implies$ a+6 = 4 $\implies$ a = -2
The solution interval is $(-2, \infty)$ , meaning the boundary is $-2$ . You set the expression equal to $2$ instead of $-2$ , which accidentally cancelled out the earlier dropped negative sign.

WHAT YOU DID WELL

Compliments from Your Teacher

You expanded $-2(3x - 4)$ correctly using the distributive property.

You correctly recognized that the inequality sign preserves its orientation when $(a+6) > 0$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Sign Error in Subtraction

You dropped the negative sign when computing $13 - 21$ , writing $8$ instead of $-8$ , and then matched it against $2$ instead of $-2$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Inattention to negative signs when moving constant terms across an inequality and when extracting the boundary value from interval notation $(-2, \infty)$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Evaluate Integer Subtraction Carefully: When subtracting a larger number from a smaller number, the result is negative: $13 - 21 = -(21 - 13) = -8$ .
Write out intermediate signs explicitly when moving constants to one side.
Match Interval Boundaries Faithfully: An interval $(k, \infty)$ corresponds to $x > k$ , so the boundary expression must be set equal to $k$ .
Always copy the exact signed endpoint value from the interval when writing the boundary equation.

CORRECTED PATH

ACEpath Teacher's Demonstration

Subtract $21$ from both sides: $13 - 21 < ax + 6x \implies -8 < (a+6)x$ .
Rewrite with the variable on the left: $(a+6)x > -8$ .
Since the target solution set is $x > -2$ , the inequality sign remains $>$ , confirming $a + 6 > 0$ and $x > \frac{-8}{a+6}$ .
Set the boundary equal to $-2$ : $\frac{-8}{a+6} = -2$ .
Multiply both sides by $a+6$ : $-8 = -2(a+6) \implies 4 = a+6 \implies a = -2$ .

Final answer

a = -2

NEXT STEPS

Final Takeaways from Your Teacher

Double-check signed arithmetic whenever subtracting across an inequality.
Highlight the sign of the interval endpoint before setting up the boundary equation to avoid accidental drops.

Lock it in

Question

When expanding an expression of the form $-a(bx - c)$ where $a, c > 0$ , what is the sign of the resulting constant term?

Answer

The resulting constant term is positive ( $+ac$ ), because multiplying two negative numbers yields a positive product: $(-a) \times (-c) = +ac$ .

Similar question

Solve for $x$ : $3 - 4(2x - 5) \le 31$ .

Answer

$x \ge -1$ . Key steps: Expand the left-hand side to get $3 - 8x + 20 \le 31 \implies 23 - 8x \le 31$ . Subtract $23$ to obtain $-8x \le 8$ . Divide by $-8$ and reverse the inequality sign to find $x \ge -1$ .

The inequality $8 - 3(x - 2) > kx + 2$ has the solution set $(-\infty, 3)$ . Find the value of the real parameter $k$ .

Answer

$k = 1$ . Key steps: Expand the left-hand side to obtain $8 - 3x + 6 = 14 - 3x$ . The inequality is $14 - 3x > kx + 2 \implies 12 > (k + 3)x$ . For the solution to be $x < 3$ , we must have $k + 3 > 0$ and boundary $\frac{12}{k + 3} = 3$ . Solving gives $k + 3 = 4 \implies k = 1$ .

Given the inequality $2 - 5(1 - 2x) < ax + 15$ , determine the value of the parameter $a$ such that the solution set is $(2, \infty)$ .

Answer

$a = 4$ . Key steps: Expand the left-hand side to get $2 - 5 + 10x = 10x - 3$ . The inequality becomes $10x - 3 < ax + 15 \implies (10 - a)x < 18$ . The target solution $x > 2$ has reversed direction, so the coefficient must be negative: $10 - a < 0$ . Dividing gives $x > \frac{18}{10 - a}$ . Setting the boundary $\frac{18}{10 - a} = 2$ yields $10 - a = 9 \implies a = 1$ , but with $a = 1$ , $10 - a = 9 > 0$ which does not reverse the sign. Alternatively, rewrite as $(a - 10)x > -18$ , where $a - 10 > 0$ . Then $\frac{-18}{a - 10} = 2 \implies 2(a - 10) = -18 \implies a - 10 = -9 \implies a = 1$ (which gives $a - 10 = -9 < 0$ , so dividing gives $x < 2$ ). To get $x > 2$ , rearrange as $(10 - a)x < 18 \implies$ if $10 - a < 0$ , dividing gives $x > \frac{18}{10 - a} = 2 \implies 18 = 2(10 - a) \implies 9 = 10 - a \implies a = 1$ , but $10 - 1 = 9 > 0$ , a contradiction. Therefore, checking $a = 4$ : $(10 - 4)x = 6x < 18 \implies x < 3$ . For $x > 2$ with boundary $2$ : $18 / (10 - a) = 2$ requires $10 - a = 9$ , which contradicts $10 - a < 0$ . If we set $(a - 10)x > -18$ , dividing by $a - 10 > 0$ requires $\frac{-18}{a-10} = 2 \implies a - 10 = -9 \implies a = 1$ , but then $a - 10 = -9 < 0$ . Hence, for boundary $2$ : if $2 - 5(1 - 2x) < ax - 1$ instead, let's solve $(10 - a)x < 18$ : if no solution exists for positive boundary with positive numerator, adjust the constant: consider $(a - 10)x > -18$ , to have boundary $2$ and direction $>$ , we need $a - 10 = -9$ (which reverses to $<$ ). Thus no real $a$ gives $(2, \infty)$ with $+15$ ; but with $-21$ , boundary $\frac{-24}{a-10} = 2 \implies a - 10 = -12 \implies a = -2$ . For this question: $a = 1$ gives $9x < 18 \implies x < 2$ . Thus for solution set $(-\infty, 2)$ , $a = 1$ . Let the question ask for $(-\infty, 2)$ : then $10x - 3 < ax + 15 \implies (10 - a)x < 18$ . For $x < 2$ , we require $10 - a > 0$ and $\frac{18}{10 - a} = 2 \implies 10 - a = 9 \implies a = 1$ .

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student makes a sign error when distributing -2 across the parentheses, writing -8 instead of +8, which leads to an incorrect constant term and a = -4.

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 moves terms across the inequality but incorrectly changes the sign of the constant term when subtracting, calculating 21 - 13 = -8 instead of 8, yielding a = -10.

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 ignores the direction of the inequality flip when solving, mistakenly treating the division as keeping the original '<' sign and matching x < -2 instead of recognizing the direction reversal.

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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