Mathematics - A Rational Inequality and the Sign That Was Ignored
A student is asked to find all x for which (x − 3)/(x + 2) ≤ 1. The student multiplies both sides by (x + 2), cancels, and obtains −3 ≤ 2, then writes 'all real numbers'. The instructor marks it wrong. The student protests that the algebra is valid because the same expression appears on both sides. The problem is part of a unit on domain restrictions and sign analysis, and the class has just finished graphing rational functions. The student needs to understand why the solution set is not all real numbers, and must produce a corrected interval notation answer that a peer can follow without a graphing calculator. The student's calculator graph shows a horizontal asymptote at y = 1 and appears to stay below it on the left, which the student interprets as confirmation.
1. Identify the exact step where the student's method fails and explain why the conclusion '−3 ≤ 2' does not imply all real x.
Read the worked answer
The failure is multiplying both sides by (x + 2) without knowing its sign. If x + 2 is positive, the inequality direction is unchanged; if x + 2 is negative, the direction must reverse. The student treated (x + 2) as if it were a positive number in every case.
The statement −3 ≤ 2 is always true, but it is only the result of the case x + 2 > 0. For x + 2 < 0, multiplying reverses the inequality: x − 3 ≥ x + 2, which gives −3 ≥ 2, a false statement with no solutions. Separately, x = −2 is not in the domain at all. So the valid conclusion is not 'all real numbers' but 'all x greater than −2.' The student's algebra is not wrong because cancellation is illegal in general; it is wrong because the sign of the multiplier was never established.
2. Solve the inequality correctly using a single rational expression and a sign chart. Give the answer in interval notation and explain the role of x = −2.
Read the worked answer
Bring everything to one side:
(x − 3)/(x + 2) − 1 ≤ 0
Combine over the common denominator:
[(x − 3) − (x + 2)]/(x + 2) ≤ 0
(−5)/(x + 2) ≤ 0
Multiply by −1 and reverse the inequality:
5/(x + 2) ≥ 0
The numerator 5 is always positive, so the sign is controlled entirely by the denominator. The expression is positive when x + 2 > 0 and negative when x + 2 < 0. It is zero nowhere, and it is undefined at x = −2.
Sign chart: for x < −2, the expression is negative, so the original ≤ 0 is false. For x > −2, the expression is positive, but after the reversal the condition is ≥ 0, which is true. At x = −2, the expression is undefined and must be excluded.
Answer: x ∈ (−2, ∞). The value x = −2 is not a solution and not a boundary that can be included; it is a vertical asymptote and a domain restriction.
3. The student's graph seems to confirm 'all real numbers'. Explain what the graph actually shows near x = −2 and why the calculator can mislead.
Read the worked answer
Rewrite the function as f(x) = (x − 3)/(x + 2) = 1 − 5/(x + 2). For x > −2, 5/(x + 2) is positive, so f(x) < 1. For x < −2, 5/(x + 2) is negative, so f(x) > 1. Thus the left branch lies above the horizontal asymptote y = 1, not below it. As x approaches −2 from the left, the denominator is a small negative number, so f(x) → +∞; the graph rises steeply, not toward 1.
A calculator can mislead when the window is zoomed out: the left branch may look flat and close to y = 1, and the vertical asymptote may be drawn as a near-vertical line that is easy to miss. The student read the overall impression of the graph instead of testing a point such as x = −3, where f(−3) = (−6)/(−1) = 6, which is clearly greater than 1. The judgement is that a graph is evidence only after the domain and a test point are checked; it does not replace the sign analysis.