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All courses Math Foundations · MF.EQ.12 39 of 60
Interpret algebraic solutions and state them as meaningful contextual answers, including restrictions or whole-number constraints when needed.

Turn a solved inequality into a context range

Problem
The solved inequality is \(x~\le~5\), where \(x\) is a nonnegative whole-number count of extra guests. Apply the contextual domain and report the complete set with units.
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Hint

Start by translating x <= 5 into words: x can be 5 or fewer. Then remember x is counting extra guests.

A solved inequality gives a whole set of possible values, not just one value. Since x counts guests, use only whole numbers that are not negative.

Solution walkthrough

01

Read the algebraic bound

\[x~\le~5\]

The solved inequality permits real values at or below 5 before context is applied.

02

Apply nonnegative count meaning

\[x~\ge~0~\text{and}~x~\text{is}~a~\text{whole}~\text{number}\]

Extra guests cannot be negative or fractional.

03

Enumerate every feasible value

\[0,~1,~2,~3,~4,~5\]

These are all whole numbers satisfying both zero-or-more and at-most-five conditions.

04

Report the contextual set

\[x~\in~{0,~1,~2,~3,~4,~5}~\text{extra}~\text{guests}\]

The set is complete and retains the guest-count unit.

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Another way

  1. Intersect the algebraic interval (-∞, 5] with the nonnegative whole numbers.

!

Common mistake

Do not include negative integers just because they satisfy x ≤ 5 algebraically. Negative guest counts are not realistic.

Solution walkthrough video