Solving a Rational Inequality Without Cross-Multiplying
During a fundraiser planning meeting, a student solves the inequality (x − 1)/(x + 3) ≤ 2, where x represents hundreds of tickets sold and x ≥ 0 in the actual context. The student cross-multiplies and writes x − 1 ≤ 2(x + 3), then x ≥ −7, so the solution is [−7, ∞). A second student tests x = −4 and finds the left side equals 5, which is not ≤ 2, so the first answer cannot be right. The first student replies that x = −4 is 'outside the ticket context' and therefore irrelevant. The tutor points out that the algebra problem is not restricted to x ≥ 0 unless stated, and asks the group to solve the inequality on all real numbers where it is defined, then interpret the result under x ≥ 0.
1. Solve (x − 1)/(x + 3) ≤ 2 on all real numbers where the expression is defined. Show the critical values and a sign analysis.
Read the worked answer
Bring all terms to one side: (x − 1)/(x + 3) − 2 ≤ 0. Combine over the common denominator: (x − 1 − 2x − 6)/(x + 3) ≤ 0, so (−x − 7)/(x + 3) ≤ 0. Multiplying numerator and denominator by −1 reverses the inequality: (x + 7)/(x + 3) ≥ 0. The critical values are x = −7, where the numerator is zero, and x = −3, where the expression is undefined. Test intervals: x = −8 gives (−1)/(−5) = 1/5 ≥ 0, true; x = −5 gives 2/(−2) = −1 ≥ 0, false; x = 0 gives 7/3 ≥ 0, true. Include −7 because the inequality is non-strict, and exclude −3 because the denominator is zero. The solution is (−∞, −7] ∪ (−3, ∞).
2. Explain why the first student's cross-multiplication is invalid, and reconcile the algebraic solution with the context x ≥ 0.
Read the worked answer
Cross-multiplying by x + 3 assumes that x + 3 is positive, so that the inequality direction stays the same. When x < −3, x + 3 is negative, and multiplying by it would reverse the inequality. That is why the student's interval [−7, ∞) contains x = −4, which fails, and omits values such as x = −8, which satisfy the inequality. The correct unrestricted solution is (−∞, −7] ∪ (−3, ∞). Under the ticket context x ≥ 0, every x ≥ 0 lies in (−3, ∞), so all nonnegative values satisfy the inequality. Context does not repair the algebra; it only selects the portion x ≥ 0 from the correct solution set.