Solve the inequality $-\frac{3}{4}(2x - 6) \ge 5 - \frac{1}{2}(4x + 1)$ for $x$. Express the solution set in interval notation, paying careful attention to sign reversals when dividing or multiplying by negative quantities.

Solve the inequality $-\frac{3}{4}(2x - 6) \ge 5 - \frac{1}{2}(4x + 1)$ for $x$ . Express the solution set in interval notation, paying careful attention to sign reversals when dividing or multiplying by negative quantities.

Final answer

$x \ge 0$ , or $[0, \infty)$ ](streamdown:incomplete-link)

Steps

Clear fractions

To solve the inequality, we can clear the fractions first or expand the expressions on both sides. Clearing fractions simplifies the arithmetic.

The denominators in the inequality are $4$ and $2$ . The least common denominator is $4$ . Multiply both sides of the inequality by $4$ . Since $4$ is positive, the inequality sign does not change:

$4 \cdot \left[-\frac{3}{4}(2x - 6)\right] \ge 4 \cdot \left[5 - \frac{1}{2}(4x + 1)\right]$

$-3(2x - 6) \ge 20 - 2(4x + 1)$

Expand and combine like terms

Apply the distributive property to both sides:

$-6x + 18 \ge 20 - 8x - 2$

Simplify the right-hand side:

$-6x + 18 \ge 18 - 8x$

Isolate the variable term

Add $8x$ to both sides to gather the variable terms on the left:

$-6x + 8x + 18 \ge 18$

$2x + 18 \ge 18$

Subtract $18$ from both sides:

$2x \ge 0$

Divide both sides by $2$ :

$x \ge 0$

Key Rule & Alternative Path: If you had instead gathered the $x$ -terms on the right side by adding $6x$ and subtracting $18$ , you would obtain: $0 \ge -2x \quad \text{or} \quad -2x \le 0$ Dividing both sides by the negative number $-2$ requires reversing the inequality sign: $x \ge \frac{0}{-2} \implies x \ge 0$ Both approaches yield the exact same result.

Write the solution in interval notation

The inequality $x \ge 0$ includes $0$ (indicated by a closed bracket $[$ ) and extends indefinitely in the positive direction toward $\infty$ :

$[0, \infty)$ ](streamdown:incomplete-link)

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 the left-hand side, leading to -9/2 instead of +9/2.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
- $\frac{3}{2}$ x - $\frac{9}{2}$ $\ge$ 5 - 2x - $\frac{1}{2}$
Correct
- $\frac{3}{2}$ x + $\frac{9}{2}$ $\ge$ 5 - 2x - $\frac{1}{2}$
When distributing $-\frac{3}{4}$ across $(2x - 6)$ , multiplying a negative coefficient by a negative constant yields a positive term: $\left(-\frac{3}{4}\right)(-6) = +\frac{18}{4} = +\frac{9}{2}$ . Writing $-\frac{9}{2}$ introduced a negative sign error.
Step 2
- $\frac{3}{2}$ x - $\frac{9}{2}$ $\ge$ $\frac{9}{2}$ - 2x
Correct
- $\frac{3}{2}$ x + $\frac{9}{2}$ $\ge$ $\frac{9}{2}$ - 2x
You correctly simplified $5 - \frac{1}{2} = \frac{9}{2}$ on the right-hand side, but the expression carries over the sign error from the previous line.
Step 3
$\frac{1}{2}x \ge 9 \implies x \ge 18$
Correct
$\frac{1}{2}x \ge 0 \implies x \ge 0$
Your algebraic manipulation to collect $x$ on the left and constants on the right was consistent with your previous line, but subtracting $\frac{9}{2}$ from $\frac{9}{2}$ on the correct path gives $0$ rather than $9$ .
Step 4
$[18, \infty)$ ](streamdown:incomplete-link)
Correct
$[0, \infty)$ ](streamdown:incomplete-link)
You formatted the final inequality into interval notation correctly according to your value, but the lower bound was shifted by the earlier sign error.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly expanded the right-hand side expression $-\frac{1}{2}(4x + 1)$ to get $-2x - \frac{1}{2}$ .

You accurately combined the constant terms $5 - \frac{1}{2}$ to obtain $\frac{9}{2}$ .

You isolated the variable term correctly by adding $2x$ and multiplying by $2$ without altering the direction of the inequality sign.

You correctly translated your linear inequality into interval notation using a closed bracket and infinity.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Distributive Sign Error with Negative Factor

You distributed $-\frac{3}{4}$ to $-6$ and wrote $-\frac{9}{2}$ instead of $+\frac{9}{2}$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Overlooking the double negative when distributing an external negative multiplier over a subtraction inside parentheses.

REASONING PATH

Your ACEpath teacher’s suggested approach

Track the Sign of Every Term: $-a(b - c) = -ab + ac$
Circle or highlight the negative sign in front of the parentheses together with the negative sign inside before multiplying.
Substitute a Test Value: A test value in the original inequality must satisfy the expanded expression.
Substitute $x = 3$ into the original left-hand side: $-\frac{3}{4}(2(3) - 6) = 0$ . Compare it with your expansion $-\frac{3}{2}(3) - \frac{9}{2} = -9 \ne 0$ to catch sign discrepancies immediately.

CORRECTED PATH

ACEpath Teacher's Demonstration

Apply the distributive property with correct signs: $-\frac{3}{4}(2x - 6) = -\frac{3}{2}x + \frac{9}{2}$ .
Equate with the simplified right-hand side: $-\frac{3}{2}x + \frac{9}{2} \ge \frac{9}{2} - 2x$ .
Subtract $\frac{9}{2}$ from both sides: $-\frac{3}{2}x \ge -2x$ .
Add $2x$ to both sides: $\frac{1}{2}x \ge 0$ .
Multiply both sides by $2$ : $x \ge 0$ .

Final answer

$[0, \infty)$ ](streamdown:incomplete-link)

NEXT STEPS

Final Takeaways from Your Teacher

Practice distributing negative constants over expressions containing subtractions.
Consider clearing fractions across the entire inequality as a first step by multiplying by the least common denominator to work exclusively with integers.
Perform a quick mental check on individual product signs before moving to the next line.

OverviewACEpath Teacher's Diagnosis

02 The student isolates the variable on the left as -1/2 x ≤ 0 by subtracting 2x instead of adding 2x, but fails to reverse the inequality sign when dividing by -1/2, concluding x ≤ 0 instead of x ≥ 0.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$-\frac{3}{2}x + \frac{9}{2} \ge 5 - 2x - \frac{1}{2} \implies -\frac{3}{2}x + \frac{9}{2} \ge \frac{9}{2} - 2x$
Step 2
$-\frac{7}{2}x + \frac{9}{2} \ge \frac{9}{2} \implies -\frac{7}{2}x \ge 0$
Correct
$\frac{1}{2}x + \frac{9}{2} \ge \frac{9}{2} \implies \frac{1}{2}x \ge 0$
To eliminate $-2x$ from the right-hand side, you must add $2x$ to both sides. Performing $-\frac{3}{2}x + 2x$ yields $\frac{1}{2}x$ , but you subtracted $2x$ instead to get $-\frac{7}{2}x$ .
Step 3
$x \ge 0$
Correct
$x \ge 0$
Following your intermediate line $-\frac{7}{2}x \ge 0$ , dividing both sides by the negative coefficient $-\frac{7}{2}$ should have reversed the inequality symbol from $\ge$ to $\le$ , yielding $x \le 0$ .
Step 4
$(-\infty, 0]$
Correct
$[0, \infty)$ ](streamdown:incomplete-link)
The condition $x \ge 0$ means all values greater than or equal to $0$ , which corresponds to $[0, \infty)$ . The notation $(-\infty, 0]$ represents $x \le 0$ .

WHAT YOU DID WELL

Compliments from Your Teacher

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

You accurately combined the constant terms $5 - \frac{1}{2}$ into $\frac{9}{2}$ .

You correctly subtracted $\frac{9}{2}$ from both sides to clear the constant term.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Inverse Operation & Inequality Sign Reversal

You applied subtraction instead of addition when transferring the $-2x$ term across the inequality symbol.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You encountered difficulty tracking inverse operations when moving negative variable terms across an inequality, and you overlooked reversing the inequality direction when dividing by a negative number.

REASONING PATH

Your ACEpath teacher’s suggested approach

Canceling Negative Terms: To cancel a negative term like $-ax$ , add $+ax$ to both sides of the inequality.
Always write out the operation explicitly on both sides: $-\frac{3}{2}x + 2x = \frac{1}{2}x$ .
Dividing by a Negative: Multiplying or dividing both sides of an inequality by a negative number reverses the direction of the inequality symbol.
Whenever the coefficient of $x$ is negative, immediately flip the inequality sign as you divide.

CORRECTED PATH

ACEpath Teacher's Demonstration

Starting from $-\frac{3}{2}x + \frac{9}{2} \ge \frac{9}{2} - 2x$ , add $2x$ to both sides: $\frac{1}{2}x + \frac{9}{2} \ge \frac{9}{2}$ .
Subtract $\frac{9}{2}$ from both sides: $\frac{1}{2}x \ge 0$ .
Multiply both sides by $2$ (a positive number, so the sign stays): $x \ge 0$ .

Final answer

$[0, \infty)$ ](streamdown:incomplete-link)

NEXT STEPS

Final Takeaways from Your Teacher

Practice using explicit addition and subtraction steps when transposing variable terms.
Review the rule that dividing or multiplying by a negative number reverses the inequality symbol.
Check that your final interval notation matches the inequality direction ( $x \ge a$ uses $[a, \infty)$ , while $x \le a$ uses $(-\infty, a]$ ).

OverviewACEpath Teacher's Diagnosis

03 The student clears denominators by multiplying the entire inequality by 4, but forgets to multiply the standalone constant 5 by 4, yielding 5 instead of 20.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
-3(2x - 6) $\ge$ 5 - 2(4x + 1)
Correct
-3(2x - 6) $\ge$ 20 - 2(4x + 1)
When multiplying both sides of the inequality by the common denominator $4$ , you must multiply every separate term by $4$ . The term $5$ becomes $4 \cdot 5 = 20$ , not $5$ .
Step 2
-6x + 18 $\ge$ 5 - 8x - 2 $\implies$ -6x + 18 $\ge$ 3 - 8x
Correct
-6x + 18 $\ge$ 20 - 8x - 2 $\implies$ -6x + 18 $\ge$ 18 - 8x
Your expansion of the parentheses and combination of like terms were algebraically correct based on your previous line, but they inherited the missing factor of $4$ from the right-hand side constant.
Step 3
2x $\ge$ -15 $\implies$ x $\ge$ - $\frac{15}{2}$
Correct
2x $\ge$ 0 $\implies$ x $\ge$ 0
You correctly added $8x$ to both sides and divided by $2$ without altering the inequality sign, but the constant on the right was incorrect due to the initial multiplication error.
Step 4
$\left[-\frac{15}{2}, \infty\right)$ ](streamdown:incomplete-link)
Correct
$[0, \infty)$ ](streamdown:incomplete-link)
Your conversion of $x \ge -\frac{15}{2}$ into interval notation was mechanically correct, but the numerical boundary was affected by the earlier mistake.

WHAT YOU DID WELL

Compliments from Your Teacher

You identified $4$ as the correct least common denominator to eliminate both fractions.

You accurately distributed negative numbers across parentheses on both sides.

You maintained the correct inequality direction when dividing by a positive coefficient.

You correctly formatted the final inequality into interval notation with an inclusive bracket.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Distributive property omitted on standalone constant when clearing fractions

You cleared the fractions by multiplying by $4$ , but left the standalone constant $5$ unchanged instead of multiplying it to get $20$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Treating fraction-clearing as simply 'canceling the denominators' rather than applying the multiplication property of inequalities to the entire side, causing terms without denominators to be skipped.

REASONING PATH

Your ACEpath teacher’s suggested approach

Enclose the entire side in brackets: Multiplying an inequality by $c$ means calculating $c \cdot (\text{entire side})$ , distributing $c$ to every additive term.
Write $4 \cdot \left[5 - \frac{1}{2}(4x + 1)\right]$ before simplifying so that every individual term is visibly multiplied.

CORRECTED PATH

ACEpath Teacher's Demonstration

Multiply every term on both sides by $4$ : $4 \cdot \left[-\frac{3}{4}(2x - 6)\right] \ge 4 \cdot 5 - 4 \cdot \left[\frac{1}{2}(4x + 1)\right]$ , giving $-3(2x - 6) \ge 20 - 2(4x + 1)$ .
Distribute the coefficients: $-6x + 18 \ge 20 - 8x - 2$ , which simplifies to $-6x + 18 \ge 18 - 8x$ .
Add $8x$ to both sides and subtract $18$ from both sides: $2x \ge 0$ .
Divide both sides by $2$ : $x \ge 0$ .

Final answer

$[0, \infty)$ ](streamdown:incomplete-link)

NEXT STEPS

Final Takeaways from Your Teacher

Always draw big grouping brackets around the full left and right sides before multiplying by a common denominator.
Count the total number of terms separated by addition or subtraction on both sides, ensuring each term receives the multiplier.
Practice multi-step linear inequalities that mix fractional terms with standalone integer constants.

Lock it in

Question

What is the product when you distribute a negative factor across a subtraction, as in $-a(b - c)$ ?

Answer

$-ab + ac$ . Remember that a negative multiplied by a negative always yields a positive term.

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student forgets to distribute the negative sign to the second term inside the parentheses when expanding the left-hand side, leading to -9/2 instead of +9/2.

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 isolates the variable on the left as -1/2 x ≤ 0 by subtracting 2x instead of adding 2x, but fails to reverse the inequality sign when dividing by -1/2, concluding x ≤ 0 instead of x ≥ 0.

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 clears denominators by multiplying the entire inequality by 4, but forgets to multiply the standalone constant 5 by 4, yielding 5 instead of 20.

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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