Solve the inequality $-\frac{5 - 3x}{4} > 2$ by isolating $x$. Specifically account for the interaction between the leading negative sign of the fraction and the negative coefficient of $x$, express the solution set in interval notation, and describe how the boundary circle and shading direction must be plotted on a real number line.

Solve the inequality $-\frac{5 - 3x}{4} > 2$ by isolating $x$ . Specifically account for the interaction between the leading negative sign of the fraction and the negative coefficient of $x$ , express the solution set in interval notation, and describe how the boundary circle and shading direction must be plotted on a real number line.

Final answer

Interval notation: $(\frac{13}{3}, \infty)$ ; on the number line, an open circle at $x = \frac{13}{3}$ with a shaded ray extending indefinitely to the right.

Steps

Clear the denominator and distribute the negative sign

To solve the inequality: $-\frac{5 - 3x}{4} > 2$

Multiply both sides by the positive denominator $4$ . Since $4 > 0$ , the inequality sign does not change direction: $-(5 - 3x) > 8$

Next, carefully handle the leading negative sign by distributing it across each term inside the numerator parentheses: $-1 \cdot 5 - (-1) \cdot 3x > 8$ $-5 + 3x > 8$

Alternative method (Handling negative numbers):
If you instead multiply both sides by $-4$ at the start, you must reverse the inequality symbol: $5 - 3x < -8$ Subtracting $5$ gives $-3x < -13$ . Dividing by the negative coefficient $-3$ reverses the direction a second time, arriving at $x > \frac{13}{3}$ . Distributing first keeps the coefficient of $x$ positive and avoids multiple sign reversals.

Isolate the variable $x$

Add $5$ to both sides of the inequality: $3x > 8 + 5$ $3x > 13$

Divide both sides by $3$ : $x > \frac{13}{3}$

Express the solution in interval notation

The inequality $x > \frac{13}{3}$ includes all real numbers strictly greater than $\frac{13}{3}$ (which is $4\frac{1}{3}$ ):

Because the inequality is strict ( $>$ rather than $\ge$ ), the endpoint $\frac{13}{3}$ is excluded, indicated by a parenthesis (.
The values increase without bound toward positive infinity $\infty$ .

$\left(\frac{13}{3}, \infty\right)$

Graph on a real number line

To represent the solution set on a real number line:

Boundary Circle: Plot an open circle (unshaded circle) at $x = \frac{13}{3} \approx 4.33$ to show that this exact boundary value is not part of the solution.
Shading Direction: Draw a solid shaded line or ray starting from the open circle and pointing to the right toward $+\infty$ , ending with an arrowhead to indicate that the solution continues indefinitely in the positive direction.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student distributes the leading negative sign only to the first term of the numerator instead of distributing it to both terms, resulting in an incorrect sign on the x-term.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
-(5 - 3x) > 8
Step 2
-5 - 3x > 8
Correct
-5 + 3x > 8
When distributing the leading negative sign across $(5 - 3x)$ , the negative sign must multiply every term inside: $-1 \cdot 5 = -5$ and $-1 \cdot (-3x) = +3x$ . You kept the sign of $-3x$ unchanged.
Step 3
-3x > 13 $\implies$ x < - $\frac{13}{3}$
Correct
3x > 13 $\implies$ x > $\frac{13}{3}$
Your algebraic steps to isolate $x$ (adding $5$ and dividing by $-3$ while flipping the inequality sign) followed logically from your previous expression, but the values were affected by the sign error in Step 2.
Step 4
(- $\infty$ , - $\frac{13}{3}$ ) $\implies$ $\text{Open circle at }$ - $\frac{13}{3}$ $\text{ and shade to the left toward negative infinity.}$
Correct
( $\frac{13}{3}$ , $\infty$ ) $\implies$ $\text{Open circle at }$ $\frac{13}{3}$ $\text{ and shade to the right toward positive infinity.}$
Your interval notation and number line graph accurately represented your computed inequality $x < -\frac{13}{3}$ , but the correct solution set is $(\frac{13}{3}, \infty)$ with an open circle at $\frac{13}{3}$ shaded to the right.

WHAT YOU DID WELL

Compliments from Your Teacher

You cleared the denominator of the fraction correctly by multiplying both sides by $4$ .

You remembered to flip the inequality sign when dividing both sides by a negative coefficient in your own working.

You correctly matched your graphed ray and interval notation to your derived inequality.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Distributive property sign error with negative numerator terms

You distributed the leading negative sign only to the first term ( $5$ ) instead of multiplying both terms inside the grouping parentheses by $-1$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Treating a fraction bar with a leading negative sign as merely negating the first term, rather than recognizing that the entire numerator acts as a grouped binomial $-(a - b) = -a + b$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Group the Numerator: $-\frac{A - B}{C} = \frac{-(A - B)}{C} = \frac{-A + B}{C}$
Always write explicit parentheses around the numerator before distributing a leading negative sign: $-(5 - 3x) = -5 - (-3x) = -5 + 3x$ .
Check Term Count: Multiplying a binomial by a factor must change the signs of both terms if that factor is negative.
Before proceeding, verify that every sign inside the parentheses flipped: a positive term becomes negative, and a negative term becomes positive.

CORRECTED PATH

ACEpath Teacher's Demonstration

Rewrite $-(5 - 3x) > 8$ with distributed signs: $-5 + 3x > 8$ .
Add $5$ to both sides: $3x > 13$ .
Divide both sides by $3$ : $x > \frac{13}{3}$ .

Final answer

Interval notation: $(\frac{13}{3}, \infty)$ ; on the number line, place an open circle at $x = \frac{13}{3}$ and shade to the right toward positive infinity.

NEXT STEPS

Final Takeaways from Your Teacher

Practice inserting grouping parentheses around the numerator whenever a fraction is preceded by a negative sign.
Verify your distributed expressions by substituting a test point such as $x = 0$ back into both forms to check that the values match.

OverviewACEpath Teacher's Diagnosis

02 The student isolates -3x correctly to obtain -3x < -13 after multiplying by -1, but fails to reverse the inequality sign when dividing by the negative coefficient -3.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
5 - 3x < -8 $\implies$ -3x < -13
Step 2
x < $\frac{13}{3}$
Correct
x > $\frac{13}{3}$
When dividing both sides of $-3x < -13$ by the negative coefficient $-3$ , the inequality sign must flip again from $<$ to $>$ . Dividing gave $\frac{-13}{-3} = \frac{13}{3}$ , but the sign remained $<$ , resulting in $x < \frac{13}{3}$ instead of $x > \frac{13}{3}$ .
Step 3
$\left(-\infty, \frac{13}{3}\right)$
Correct
$\left(\frac{13}{3}, \infty\right)$
This interval correctly represents $x < \frac{13}{3}$ , but it inherits the reversed direction from the previous sign error.
Step 4
$\text{Open circle at }$ $\frac{13}{3}$ $\text{ and shade to the left toward negative infinity.}$
Correct
$\text{Open circle at }$ $\frac{13}{3}$ $\text{ with a shaded ray extending to the right toward positive infinity.}$
You correctly recognized that a strict inequality uses an open circle at the boundary $\frac{13}{3}$ , but the shading points left toward negative infinity because of the earlier sign error.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly cleared the fraction by multiplying both sides by $-4$ and remembered to reverse $>$ to $<$ on that initial step.

You accurately subtracted $5$ from both sides to obtain $-3x < -13$ .

You correctly simplified $\frac{-13}{-3}$ to $\frac{13}{3}$ .

You properly used an open circle and parentheses to represent a strict inequality boundary.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Failure to reverse inequality symbol when dividing by a negative number

You did not reverse the inequality sign when dividing both sides by the negative coefficient $-3$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Dividing or multiplying by a negative number inverts order on the real number line, requiring an inequality symbol flip every single time that operation occurs. Because you had already reversed the symbol once in Step 1, you likely treated the final division as a standard simplification rather than an operation requiring another sign reversal.

REASONING PATH

Your ACEpath teacher’s suggested approach

Every Negative Multiplier/Divisor Flips the Direction: Whenever you multiply or divide both sides of an inequality by a negative quantity, reverse the inequality symbol immediately, regardless of how many times you have already flipped it in earlier steps.
Highlight the negative divisor before dividing: write $\frac{-3x}{-3} \; [\text{flip} < \text{ to } >] \; \frac{-13}{-3}$ to make the reversal deliberate.
Quick Test-Value Verification: Pick a convenient test value strictly inside your solution interval and substitute it into the original inequality to confirm truth.
If your proposed solution is $x < \frac{13}{3}$ , testing $x = 0$ gives $-\frac{5 - 0}{4} = -1.25$ , which is not greater than $2$ . That immediately alerts you that the inequality points in the wrong direction.

CORRECTED PATH

ACEpath Teacher's Demonstration

Divide both sides by $-3$ and reverse the inequality sign: $\frac{-3x}{-3} > \frac{-13}{-3} \implies x > \frac{13}{3}$ .
Convert the strict inequality $x > \frac{13}{3}$ into interval notation: $\left(\frac{13}{3}, \infty\right)$ .
Plot on the number line by placing an open boundary circle at $x = \frac{13}{3}$ and drawing a shaded ray pointing to the right toward positive infinity.

Final answer

Interval notation: $\left(\frac{13}{3}, \infty\right)$ ; on the number line, an open circle at $x = \frac{13}{3}$ with a shaded ray extending indefinitely to the right.

NEXT STEPS

Final Takeaways from Your Teacher

Review the three targeted flashcards to reinforce the rule that every negative division triggers a direction flip.
Solve the three follow-up practice problems, verifying each answer by testing an easy value like $x = 0$ in the original inequality.

OverviewACEpath Teacher's Diagnosis

03 The student misinterprets the strict inequality as inclusive when plotting and writing interval notation, using a bracket and closed circle instead of a parenthesis and open circle.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
-(5 - 3x) > 8 $\implies$ -5 + 3x > 8
Step 2
3x > 13 $\implies$ x > $\frac{13}{3}$
Step 3
$\left(\frac{13}{3}, \infty\right)$
Step 4
Solid boundary circle at $\frac{13}{3}$ with ray shaded to the right
Correct
Open boundary circle at $\frac{13}{3}$ with ray shaded to the right
Because the inequality is strict ( $>$ ), the boundary value $\frac{13}{3}$ is not included in the solution set. A non-included boundary must be plotted as an open circle (unshaded ring), not a solid filled circle.

WHAT YOU DID WELL

Compliments from Your Teacher

You cleared the positive denominator without changing the inequality sign.

You handled the double negative $-(-3x) = +3x$ correctly when expanding.

You correctly isolated $x$ to find $x > \frac{13}{3}$ .

You correctly identified that the interval requires an open parenthesis and extends to $\infty$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Boundary circle representation on number line

You knew the boundary was excluded when writing interval notation with a round parenthesis, but you defaulted to drawing a filled-in dot on the number line.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Misassociating the open boundary condition ( $>$ or $<$ ) with a solid dot instead of an open ring during the visual translation to a number line.

REASONING PATH

Your ACEpath teacher’s suggested approach

Boundary Inclusion Check: Strict inequalities ( $<, >$ ) do not include the endpoint, corresponding to parentheses ( ) and open circles ○. Non-strict inequalities ( $\le, \ge$ ) include the endpoint, corresponding to brackets [ ] and closed circles ●.
Whenever you draw a boundary point, ask: 'Can $x$ equal this exact value?' If no, leave the center hollow.

CORRECTED PATH

ACEpath Teacher's Demonstration

Locate the boundary point $x = \frac{13}{3}$ on the number line.
Draw an open, hollow circle at $x = \frac{13}{3}$ to indicate that this endpoint is not part of the solution.
Shade the line to the right toward positive infinity and cap it with an arrowhead.

Final answer

Interval notation: $\left(\frac{13}{3}, \infty\right)$ ; on the number line, an open circle at $x = \frac{13}{3}$ with a ray shaded indefinitely to the right.

NEXT STEPS

Final Takeaways from Your Teacher

Review the direct pairing: ( / ) maps to an open circle ○, while [ / ] maps to a solid circle ●.
Practice checking whether substituting the boundary value itself satisfies the original inequality.

Lock it in

Question

How do you correctly expand the expression $-(a - b)$ ?

Answer

$-a + b$ . The negative sign distributes to every term inside the parentheses: $-1 \cdot a = -a$ and $-1 \cdot (-b) = +b$ .

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student distributes the leading negative sign only to the first term of the numerator instead of distributing it to both terms, resulting in an incorrect sign on the x-term.

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 -3x correctly to obtain -3x < -13 after multiplying by -1, but fails to reverse the inequality sign when dividing by the negative coefficient -3.

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 misinterprets the strict inequality as inclusive when plotting and writing interval notation, using a bracket and closed circle instead of a parenthesis and open circle.

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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