The graph of the solution set of a compound inequality on a real number line consists of the union of two disjoint open rays: $(-\infty, -3) \cup (5, \infty)$. If this graph represents the complete solution to an absolute value inequality of the form $|x - c| > d$ where $c$ and $d$ are real constants with $d > 0$, reverse-engineer the initial parameters $c$ and $d$, and state the midpoint and radius of the excluded segment.

The graph of the solution set of a compound inequality on a real number line consists of the union of two disjoint open rays: $(-\infty, -3) \cup (5, \infty)$ . If this graph represents the complete solution to an absolute value inequality of the form $|x - c| > d$ where $c$ and $d$ are real constants with $d > 0$ , reverse-engineer the initial parameters $c$ and $d$ , and state the midpoint and radius of the excluded segment.

Final answer

$c = 1$ , $d = 4$ ; the midpoint of the excluded interval is $x = 1$ and the distance from the midpoint to either boundary is $4$ units.

Steps

Understand the Geometric Meaning of Absolute Value

An absolute value inequality of the form $|x - c| > d$ (with $d > 0$ ) describes all points $x$ on a real number line whose distance from a central reference point $c$ is strictly greater than $d$ .

Algebraically, this unfolds into two separate linear inequalities: $x - c > d \implies x > c + d$ $\text{or}$ $x - c < -d \implies x < c - d$

In interval notation, the solution set is: $(-\infty, c - d) \cup (c + d, \infty)$

Identify the Boundary Values

The given solution set is $(-\infty, -3) \cup (5, \infty)$ . Matching this to the general form yields the system of equations: $\lbrace c - d = -3,\quad c + d = 5$

Solve for the Center ( $c$ ) and Distance ( $d$ )

Midpoint ( $c$ ): Add the two boundary equations to eliminate $d$ : $(c - d) + (c + d) = -3 + 5$ $2c = 2 \implies c = 1$

Geometrically, $c$ is the midpoint of the excluded interval $[-3, 5]$ : $\text{Midpoint} = \frac{-3 + 5}{2} = 1$
Radius ( $d$ ): Substitute $c = 1$ back into $c + d = 5$ : $1 + d = 5 \implies d = 4$

Geometrically, $d$ is the radius (half-length) of the excluded interval: $\text{Radius} = \frac{5 - (-3)}{2} = \frac{8}{2} = 4$

Conclusion

The parameters for the inequality are $c = 1$ and $d = 4$ , representing the absolute value inequality $|x - 1| > 4$ .
The midpoint of the excluded interval is $x = 1$ , and the radius (distance from the center to either boundary) is $4$ units.

Common Pitfall: Be careful with signs when subtracting a negative endpoint to find the distance: $5 - (-3) = 8$ , not $5 - 3 = 2$ .

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student subtracts rather than averages the endpoints to find the midpoint c, confusing the midpoint formula with the distance formula.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
Boundary points are $-3$ and $5$ .
Step 2
Midpoint $c = \frac{5 - (-3)}{2} = \frac{8}{2} = 4$
Correct
Midpoint $c = \frac{5 + (-3)}{2} = \frac{2}{2} = 1$
You subtracted the endpoints instead of adding them. The midpoint (center) of two points $a$ and $b$ is their arithmetic mean, $\frac{a + b}{2}$ , whereas $\frac{b - a}{2}$ gives the half-length (radius).
Step 3
Radius $d = 5 - c = 5 - 4 = 1$
Correct
Radius $d = 5 - c = 5 - 1 = 4$
Your formula $d = 5 - c$ was logically sound, but substituting your incorrect value $c = 4$ resulted in an incorrect radius.
Step 4
$c = 4,\ d = 1 \implies \text{Midpoint} = 4,\ \text{radius} = 1$
Correct
$c = 1,\ d = 4 \implies \text{Midpoint} = 1,\ \text{radius} = 4$
The final parameters and geometric descriptions are reversed due to the error in Step 2.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly extracted the boundary points $-3$ and $5$ from the interval notation.

You correctly handled double negatives when computing the difference $5 - (-3) = 8$ .

You understood the geometric relationship between the upper boundary, the center, and the radius.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Formula confusion: midpoint vs. radius

You used the difference formula $\frac{b - a}{2}$ when calculating the midpoint $c$ , which actually calculates the radius.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing the midpoint (average: $\frac{a + b}{2}$ ) with the distance or radius (half-difference: $\frac{b - a}{2}$ ) on a 1D coordinate line.

REASONING PATH

Your ACEpath teacher’s suggested approach

Midpoint vs. Distance Formula: Midpoint is an average: $\text{Midpoint} = \frac{a + b}{2}$ . Radius is half of the total spread: $\text{Radius} = \frac{|b - a|}{2}$ .
Always add the coordinates when looking for the center point, and subtract them when finding distance or radius.
Sanity Check on a Number Line: The midpoint must sit symmetrically between the two endpoints.
Plot your calculated center: $4$ is $1$ unit from $5$ but $7$ units from $-3$ , so it cannot be the midpoint. In contrast, $1$ is exactly $4$ units away from both endpoints.

CORRECTED PATH

ACEpath Teacher's Demonstration

Compute the center $c$ using the midpoint formula: $c = \frac{-3 + 5}{2} = \frac{2}{2} = 1$ .
Compute the radius $d$ as the distance from the center to either boundary point: $d = 5 - c = 5 - 1 = 4$ (or $d = \frac{5 - (-3)}{2} = 4$ ).

Final answer

$c = 1$ , $d = 4$ ; midpoint is $1$ and radius is $4$ .

NEXT STEPS

Final Takeaways from Your Teacher

Review the distinction between coordinate averages (addition) and segment lengths (subtraction).
Do a quick symmetry check after finding the midpoint to verify that the distance to both endpoints is equal.

OverviewACEpath Teacher's Diagnosis

02 The student confuses the radius d with the total diameter or length of the excluded interval, failing to divide the distance by 2.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
c $\text{ is the midpoint: }$ c = $\frac{-3+5}{2}$ = $\frac{2}{2}$ = 1
Step 2
d $\text{ is the distance between }$ -3 $\text{ and }$ 5: d = 5 - (-3) = 8
Correct
d = $\frac{5 - (-3)}{2}$ = $\frac{8}{2}$ = 4
In the inequality $|x - c| > d$ , $d$ represents the distance from the center $c$ to each boundary, which is the radius (half the total length), not the full distance between the two endpoints.
Step 3
c = 1,, d = 8 $\implies$ $\text{Midpoint}$ = 1,, $\text{radius}$ = 8
Correct
c = 1,, d = 4 $\implies$ $\text{Midpoint}$ = 1,, $\text{radius}$ = 4
This step followed correctly from your previous calculation of $d$ , but it inherits the doubled value.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly identified that $c$ represents the midpoint of the two boundary points.

You computed the average of $-3$ and $5$ accurately to get $c = 1$ .

You correctly calculated the total distance between $-3$ and $5$ as $8$ units.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Conflating radius with total interval length

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You treated the parameter $d$ in $|x - c| > d$ as the total diameter (the distance between both boundary endpoints) rather than the radius (the distance from the center $c$ to each endpoint).

REASONING PATH

Your ACEpath teacher’s suggested approach

Geometric definition of $|x - c| > d$: $|x - c| > d$ means the distance between $x$ and the center $c$ is strictly greater than $d$ .
Always compute $d$ as the radius: $d = \frac{\text{upper} - \text{lower}}{2}$ or $d = |\text{boundary} - c|$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Find the distance from the center $c = 1$ to the upper boundary $5$ : $d = 5 - 1 = 4$ (or divide the total length by $2$ : $d = \frac{8}{2} = 4$ ).
State the final parameters: $c = 1$ and $d = 4$ .
State the midpoint and radius of the excluded segment $[-3, 5]$ : $\text{midpoint} = 1$ and $\text{radius} = 4$ .

Final answer

$c = 1$ , $d = 4$ ; midpoint $= 1$ , radius $= 4$

NEXT STEPS

Final Takeaways from Your Teacher

When setting up $|x - c| > d$ from endpoints $a$ and $b$ , write $c = \frac{a+b}{2}$ and $d = \frac{b-a}{2}$ .
Check your values by plugging $c$ and $d$ back in: $c - d = 1 - 4 = -3$ and $c + d = 1 + 4 = 5$ .

OverviewACEpath Teacher's Diagnosis

03 The student misinterprets |x - c| > d by taking c as the negative of the midpoint due to the minus sign in the formula.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$\text{Midpoint of }$ [-3, 5] $\text{ is }$ $\frac{-3+5}{2}$ = 1
Step 2
$\text{Because the form is }$ |x - c|, c = -1
Correct
c = 1 $\implies$ |x - 1|
In the standard distance formulation $|x - c|$ , the constant $c$ represents the center point directly. Because the midpoint is $1$ , $c = 1$ , which gives the expression $|x - 1|$ .
Step 3
$\text{Radius }$ d = $\frac{5 - (-3)}{2}$ = $\frac{8}{2}$ = 4
Step 4
c = -1, d = 4 $\implies$ $\text{Midpoint}$ = -1, $\text{radius}$ = 4
Correct
c = 1, d = 4; $\text{Midpoint}$ = 1, $\text{radius}$ = 4
Your final parameters and stated midpoint inherited the sign error from Step 2, even though you initially found the correct midpoint $1$ in Step 1.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly used the midpoint formula $\frac{a + b}{2}$ to find the center of the interval $[-3, 5]$ .

You correctly calculated the half-width (radius) $d = \frac{5 - (-3)}{2} = 4$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Sign confusion in absolute value center parameter

You saw the minus sign inside the standard form $|x - c|$ and incorrectly reversed the sign of your computed center $1$ to $-1$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing the coordinate of the center point $c$ with the negative sign inherent to the distance definition $|x - c| = \text{distance between } x \text{ and } c$ . Because distance from $c$ is $|x - c|$ , a positive center $c = 1$ naturally produces $|x - 1|$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Geometric Distance Definition: The expression $|x - c|$ means 'the distance between $x$ and $c$ on a number line'.
Always set $c$ equal to the center's actual coordinate without changing its sign.
Boundary Verification: Plugging either boundary endpoint into $|x - c| = d$ must yield a true equality.
Test the boundary $x = 5$ : $|5 - 1| = 4$ (true), whereas $|5 - (-1)| = |6| \neq 4$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

The center of the interval is $1$ , so $c = 1$ .
The inequality becomes $|x - 1| > 4$ .
Verify the boundaries: $x - 1 > 4 \implies x > 5$ and $x - 1 < -4 \implies x < -3$ .

Final answer

$c = 1$ , $d = 4$ ; midpoint is $1$ and radius is $4$ .

NEXT STEPS

Final Takeaways from Your Teacher

Review the geometric definition of absolute value as distance: $|a - b|$ is the distance between $a$ and $b$ .
Always substitute your boundary values back into $|x - c| = d$ as a sanity check before writing the final parameters.

Lock it in

Question

On a real number line with endpoints $a$ and $b$ , what are the distinct formulas for the midpoint $c$ and the radius $r$ ?

Answer

Midpoint: $c = \frac{a + b}{2}$ (add endpoints to average them). Radius: $r = \frac{|b - a|}{2}$ (subtract endpoints to find half the total length).

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student subtracts rather than averages the endpoints to find the midpoint c, confusing the midpoint formula with the distance formula.

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 confuses the radius d with the total diameter or length of the excluded interval, failing to divide the distance by 2.

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 |x - c| > d by taking c as the negative of the midpoint due to the minus sign in the formula.

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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