Solve the rational inequality $\frac{2x - 8}{x + 3} \le 0$ for $x \in \mathbb{R}$. Identify the critical threshold values that partition the number line, evaluate whether each boundary point is included or excluded based on mathematical domain constraints, express the solution set in interval notation, and describe its number line graph.

Solve the rational inequality $\frac{2x - 8}{x + 3} \le 0$ for $x \in \mathbb{R}$ . Identify the critical threshold values that partition the number line, evaluate whether each boundary point is included or excluded based on mathematical domain constraints, express the solution set in interval notation, and describe its number line graph.

Final answer

Critical points: $x = -3$ (vertical asymptote / undefined denominator) and $x = 4$ (numerator zero). Solution set: $(-3, 4]$ ; on the number line, an open circle at $x = -3$ , a solid closed circle at $x = 4$ , and a shaded line segment connecting them.

Steps

Find the critical threshold values

To solve the rational inequality $\frac{2x - 8}{x + 3} \le 0$ , we find where the expression changes sign, determine whether each boundary point belongs to the solution set, and test the resulting intervals.

A rational expression can only change signs where its numerator is equal to zero or where its denominator is equal to zero:

Numerator zero: $2x - 8 = 0 \implies 2x = 8 \implies x = 4$
Denominator zero (undefined point): $x + 3 = 0 \implies x = -3$

These two critical points partition the real number line into three intervals: $(-\infty, -3)$ , $(-3, 4)$ , and $(4, \infty)$ .

Determine inclusion or exclusion of boundary points

Denominator constraint: Division by zero is undefined. Therefore, $x \neq -3$ . The point $x = -3$ must be excluded from the solution set, represented by an open circle on a number line and a round parenthesis ( in interval notation.
Inequality condition ( $\le 0$ ): The inequality allows the expression to equal zero. Since the numerator equals zero at $x = 4$ and the denominator is non-zero there, $x = 4$ makes the expression equal to $0$ . Thus, $x = 4$ is included, represented by a solid closed circle on a number line and a square bracket ] in interval notation.

Test each interval

Choose a convenient test value in each interval to check the sign of the rational expression $\frac{2x - 8}{x + 3}$ :

Interval $(-\infty, -3)$ : Test $x = -4$ : $\frac{2(-4) - 8}{-4 + 3} = \frac{-16}{-1} = 16 > 0 \quad (\text{Positive})$
Interval $(-3, 4)$ : Test $x = 0$ : $\frac{2(0) - 8}{0 + 3} = \frac{-8}{3} < 0 \quad (\text{Negative})$
Interval $(4, \infty)$ : Test $x = 5$ : $\frac{2(5) - 8}{5 + 3} = \frac{2}{8} = \frac{1}{4} > 0 \quad (\text{Positive})$

We are looking for values where the expression is less than or equal to zero ( $\le 0$ ), which corresponds to the negative interval and the included endpoint.

Express the solution set and describe the number line

Interval notation: $(-3, 4]$
Number line graph: Draw a horizontal number line marked with integers. Place an open circle at $-3$ , a solid closed circle at $4$ , and shade the continuous line segment connecting them.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student includes the denominator root in the solution set because the inequality is non-strict, ignoring domain constraints.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
2x - 8 = 0 $\implies$ x = 4 $\text{and}$ x + 3 = 0 $\implies$ x = -3 $\implies$ $\text{Critical values are }$ x = -3 $\text{ and }$ x = 4
Step 2
$\text{The inequality is }$ $\le$ 0 $\text{, so both boundary points }$ x = -3 $\text{ and }$ x = 4 $\text{ are included.}$
Correct
x = 4 $\text{ gives }$ 0 $\le$ 0 $\text{ (included)}$ , $\text{ but }$ x = -3 $\text{ makes the denominator }$ 0 $\text{ (undefined, must be excluded)}$ .
Even when an inequality is non-strict ( $\le$ or $\ge$ ), any value causing division by zero cannot belong to the domain. Therefore, $x = -3$ must be excluded with an open circle and a round parenthesis.
Step 3
$\text{Test }$ x = 0: $\frac{0 - 8}{0 + 3}$ = - $\frac{8}{3}$ $\le$ 0 $\text{ (true)}$
Step 4
$\text{Solution set is }$ [-3, 4] $\implies$ $\text{Closed solid circles at }$ -3 $\text{ and }$ 4 $\text{ with shading between them.}$
Correct
(-3, 4] $\text{with an open circle at }$ -3 $\text{, a solid closed circle at }$ 4 $\text{, and shading between them.}$
Because $x = -3$ was incorrectly treated as included in Step 2, this error carried forward into using a square bracket and a solid closed circle at $-3$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly identified both critical values by setting the numerator and denominator to zero.

You picked an appropriate test point within the partition interval and evaluated the inequality condition accurately.

You correctly identified that the interval between the two critical points makes the expression negative.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Domain constraint violation / denominator endpoint inclusion

You noticed the non-strict inequality sign $\le 0$ and included both boundary endpoints without checking if either endpoint made the denominator zero.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You associated the non-strict inequality symbol $\le$ globally with solid dots and square brackets, forgetting that domain restrictions take precedence over the inequality relation: division by zero is mathematically undefined.

REASONING PATH

Your ACEpath teacher’s suggested approach

Denominator Zero Check: A denominator can never equal zero in real numbers ( $x \neq c$ ).
Always mark critical values coming from the denominator with an open circle and a parenthesis, regardless of whether the inequality has $\le$ or $\ge$ .
Numerator vs. Denominator Endpoints: Numerator roots equal zero; denominator roots make the fraction undefined.
Separate your endpoint analysis: only include roots of the numerator if the inequality has $\le$ or $\ge$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Evaluate the domain restriction: $x + 3 \neq 0 \implies x \neq -3$ , so $x = -3$ is strictly excluded (open circle, round parenthesis).
Evaluate the numerator root: at $x = 4$ , $\frac{2(4)-8}{4+3} = 0 \le 0$ , so $x = 4$ is included (solid dot, square bracket).
Combine the tested middle interval with the correct endpoint statuses to form the interval notation $(-3, 4]$ .

Final answer

Critical points: $x = -3$ (excluded) and $x = 4$ (included). Solution set: $(-3, 4]$ ; on the number line, an open circle at $x = -3$ , a solid closed circle at $x = 4$ , and a shaded line segment connecting them.

NEXT STEPS

Final Takeaways from Your Teacher

Practice labeling critical points as 'N' (numerator) and 'D' (denominator) before graphing.
Immediately draw an open circle at every denominator root on your number line before shading any region.
Work through rational inequalities with non-strict inequalities to build the habit of checking domain constraints.

OverviewACEpath Teacher's Diagnosis

02 The student clears the denominator by multiplying both sides by x + 3 without considering that x + 3 can be negative, treating it like a linear inequality.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
Multiply both sides by $x + 3$ : $2x - 8 \le 0$
Correct
Find critical values where numerator or denominator equals zero: $2x - 8 = 0 \implies x = 4$ and $x + 3 = 0 \implies x = -3$
You cannot multiply both sides of an inequality by an expression with a variable whose sign depends on $x$ . When $x + 3 < 0$ (meaning $x < -3$ ), multiplying by $x + 3$ would flip the inequality sign. Multiplying it away also removes the denominator's critical point at $x = -3$ entirely.
Step 2
Critical value is $x = 4 \implies x = 4$ satisfies the inequality, so $x = 4$ is included.
Correct
Critical values are $x = 4$ (included, closed circle) and $x = -3$ (excluded due to division by zero, open circle).
You correctly concluded that $x = 4$ is included because it makes the numerator zero and satisfies the non-strict inequality $\le 0$ . However, because the denominator was eliminated in the previous step, you missed the critical value at $x = -3$ where the expression is undefined.
Step 3
$2x \le 8$ gives $x \le 4$
Correct
Test intervals $(-\infty, -3)$ , $(-3, 4)$ , and $(4, \infty)$ to determine where $\frac{2x - 8}{x + 3} \le 0$ .
You solved your linear inequality $2x \le 8$ accurately to obtain $x \le 4$ , but rational inequalities must be solved by testing sign changes across all critical intervals rather than solving linearly.
Step 4
Solution set is $(-\infty, 4] \implies$ A closed circle at 4 with shading extending infinitely to the left.
Correct
Solution set is $(-3, 4]$ , graphed with an open circle at $-3$ , a solid closed circle at $4$ , and a shaded segment between them.
Because the interval $(-\infty, -3)$ actually yields a positive quotient (for example, at $x = -4$ , $\frac{-16}{-1} = 16 > 0$ ), that region cannot be part of the solution set. The true solution is restricted to $(-3, 4]$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly identified $x = 4$ as a boundary value that makes the numerator equal to zero.

You correctly determined that $x = 4$ should be included with a closed circle because the inequality includes equality ( $\le 0$ ).

You accurately solved the linear inequality $2x - 8 \le 0$ to get $x \le 4$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Multiplying by variable denominator

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Treating a rational inequality as a linear inequality by clearing a variable denominator, overlooking that the variable expression can change sign and that division by zero defines an essential critical partition on the number line.

REASONING PATH

Your ACEpath teacher’s suggested approach

Never multiply by a variable denominator: Do not multiply both sides of an inequality by an expression containing $x$ unless its sign is definitively known to be strictly positive or strictly negative for all real numbers.
Keep zero on one side and find the critical values by setting both the numerator and the denominator equal to zero separately.
Domain constraints are always excluded: Values that make any denominator zero must always be excluded from the solution set with an open circle and a parenthesis.
Mark $x = -3$ with an open circle immediately before analyzing test intervals.

CORRECTED PATH

ACEpath Teacher's Demonstration

Set the numerator equal to zero: $2x - 8 = 0 \implies x = 4$ .
Set the denominator equal to zero: $x + 3 = 0 \implies x = -3$ .
Plot $x = -3$ (open circle, undefined) and $x = 4$ (solid circle, makes fraction $0$ ) on a number line, creating intervals $(-\infty, -3)$ , $(-3, 4)$ , and $(4, \infty)$ .
Test signs: on $(-\infty, -3)$ , test $x = -4 \implies \frac{-16}{-1} = +16 > 0$ ; on $(-3, 4)$ , test $x = 0 \implies \frac{-8}{3} < 0$ ; on $(4, \infty)$ , test $x = 5 \implies \frac{2}{8} > 0$ .
Select the negative interval and included endpoint: $(-3, 4]$ .

Final answer

Critical points are $x = -3$ (excluded) and $x = 4$ (included); solution set is $(-3, 4]$ , represented on the number line by an open circle at $-3$ , a solid circle at $4$ , and shading between them.

NEXT STEPS

Final Takeaways from Your Teacher

Practice identifying all critical points by setting both numerator and denominator equal to zero before taking any algebraic steps.
Construct sign charts or test-point tables across all partitioned intervals instead of clearing algebraic fractions in inequalities.

OverviewACEpath Teacher's Diagnosis

03 The student tests the sign of intervals incorrectly by assuming the middle region is positive and outer regions are negative, selecting the union of the outside intervals.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
2x - 8 = 0 $\implies$ x = 4 $\implies$ x + 3 = 0 $\implies$ x = -3 $\implies$ $\text{Critical values are }$ x = -3 $\text{ and }$ x = 4
Step 2
x = -3 $\text{ makes denominator zero so it is excluded}$ ; x = 4 $\text{ makes numerator zero so it is included.}$
Step 3
$\text{For }$ x < -3: $\text{positive}$ $\implies$ $\text{For }$ -3 < x < 4: $\text{positive}$ $\implies$ $\text{For }$ x > 4: $\text{negative}$
Correct
$\text{For }$ x < -3: $\frac{-}{-}$ > 0 $\text{ (positive)}$ ; $\text{For }$ -3 < x < 4: $\frac{-}{+}$ < 0 $\text{ (negative)}$ ; $\text{For }$ x > 4: $\frac{+}{+}$ > 0 $\text{ (positive)}$
Testing specific sample points shows that for $-3 < x < 4$ (such as $x = 0$ ), the expression is $\frac{-8}{3} < 0$ (negative), and for $x > 4$ (such as $x = 5$ ), the expression is $\frac{2}{8} > 0$ (positive). Your interval signs for the middle and right regions were incorrect.
Step 4
$\text{Solution set is }$ (- $\infty$ , -3) $\cup$ 4, $\infty$ ) $\implies$ $\text{Open circle at }$ -3 $\text{ shaded to the left, closed circle at }$ 4 $\text{ shaded to the right.}$
Correct
(-3, 4] $\implies$ $\text{Open circle at }$ -3, $\text{ solid circle at }$ 4, $\text{ shaded line segment connecting them.}$
Because the middle interval $(-3, 4)$ is where the expression is negative, the solution to $\frac{2x - 8}{x + 3} \le 0$ is $(-3, 4]$ , representing the segment between $-3$ and $4$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You found the exact zeros of the numerator and denominator.

You correctly identified that the denominator cannot equal zero, excluding $x = -3$ .

You correctly included $x = 4$ because the inequality allows non-strict comparison.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Sign analysis error

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You did not systematically substitute test numbers into both the numerator and denominator for each region, leading to false sign assignments across the test intervals.

REASONING PATH

Your ACEpath teacher’s suggested approach

Pick a concrete test value: Every interval between critical values must be evaluated using a specific number inside that interval.
Select $x = 0$ for $(-3, 4)$ to see $\frac{2(0)-8}{0+3} = -\frac{8}{3} < 0$ , and $x = 5$ for $(4, \infty)$ to see $\frac{2(5)-8}{5+3} = \frac{2}{8} > 0$ .
Match the inequality symbol: For an expression $f(x) \le 0$ , select intervals where $f(x)$ is negative along with zero-valued endpoints.
Highlight only the negative region $(-3, 4)$ and include the valid endpoint $4$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Substitute $x = -4 \in (-\infty, -3)$ : $\frac{2(-4)-8}{-4+3} = \frac{-16}{-1} = 16 > 0$ .
Substitute $x = 0 \in (-3, 4)$ : $\frac{2(0)-8}{0+3} = -\frac{8}{3} < 0$ .
Substitute $x = 5 \in (4, \infty)$ : $\frac{2(5)-8}{5+3} = \frac{2}{8} = \frac{1}{4} > 0$ .
Since the inequality requires $\le 0$ , select the interval $(-3, 4)$ and include $x = 4$ .

Final answer

(-3, 4]

NEXT STEPS

Final Takeaways from Your Teacher

Create a sign chart with separate rows for the numerator $(2x - 8)$ and denominator $(x + 3)$ to avoid sign errors.
Always plug in a quick test value like $0$ whenever the interval spans across zero.
Verify your shaded interval on the number line matches the requested sign $(\le 0$ means negative or zero).

Lock it in

Question

In a rational inequality with $\le$ or $\ge$ , can a root of the denominator ever be included in the solution set?

Answer

No. Division by zero is undefined, so roots of the denominator are always excluded with an open circle and a round parenthesis ( or ).

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student includes the denominator root in the solution set because the inequality is non-strict, ignoring domain constraints.

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 clears the denominator by multiplying both sides by x + 3 without considering that x + 3 can be negative, treating it like a linear inequality.

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 tests the sign of intervals incorrectly by assuming the middle region is positive and outer regions are negative, selecting the union of the outside intervals.

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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