Rational inequality sign repair: retain both intervals

Rational inequality sign repair starts by checking the step that removed the denominator. Suppose a worksheet asks for every real x satisfying (x − 1)/(x + 2) > 0. A proposed answer multiplies both sides by x + 2, keeps the inequality pointing the same way, and concludes x > 1. That answer contains valid values, but it loses an entire interval. Check whether the multiplier is positive or negative.

This is an invented real-number exercise, with no device reading, financial rate or physical measurement attached. Negative x values are allowed wherever the fraction is defined. The full strict solution is x < −2 or x > 1. We will preserve the proposed working, identify its unjustified assumption, and recover both intervals without changing the original expression. A second version using a nonstrict inequality will make the endpoint conditions especially clear.

Save the domain before changing the formula

The denominator x + 2 is zero at x = −2. The original fraction is undefined there, so exclude that value at the start. Write “x is real and x ≠ −2” beside the expression. This restriction belongs to the original problem. A later algebraic form may be defined at −2 even though the fraction was not; that does not make the value an eligible original input.

The numerator is zero at x = 1. At that point the denominator is three, so the fraction is defined and equals zero. The strict condition asks for a result greater than zero, which rejects that endpoint. The two excluded values have different reasons: −2 is outside the domain, while 1 is inside the domain but fails the strict inequality. Keep those explanations separate rather than recording both merely as unexplained open dots.

These two breakpoints divide the remaining real inputs into three intervals: x < −2, −2 < x < 1, and x > 1. Every allowed input other than the numerator zero belongs to exactly one of them. Their boundaries come from the actual numerator and denominator, not from a convenient collection of test values. Retaining the original domain also makes it possible to check a transformed calculation against the question it was meant to answer.

Expose the lost branch with one exact input

Substitute x = −3 into the original fraction. Its numerator is −4 and its denominator is −1, so the result is 4. It satisfies the strict inequality even though −3 does not satisfy the proposed answer x > 1. This exact counterexample shows that the proposed answer is incomplete. It does not yet supply every missing value, but it tells us which earlier operation needs inspection.

At x = −3, the multiplier x + 2 is negative. Multiplying an inequality by a negative number reverses its direction. The proposed transition retained the direction without establishing that its multiplier was positive. It therefore worked only under an unstated sign condition. The repair needs to cover both permitted signs, rather than changing the problem to allow only positive x or dropping the counterexample as an inconvenient input.

Do not replace the proposed conclusion with “all x except −2.” For instance, x = 0 gives −1/2, which fails the requested positive result. We have one positive-quotient witness on the negative side and one failing input in the middle. The next step explains those signs across complete intervals, so the answer will not depend on guessing how many individual substitutions are enough.

Record signs across all three intervals

If x < −2, then x − 1 is negative and x + 2 is negative. A negative numerator divided by a negative denominator is positive. Every real input in that interval therefore works. This reasoning includes nonintegers: the sign conditions follow from the inequalities defining the interval, without requiring a separate calculation for every fraction or irrational number it contains.

If −2 < x < 1, the numerator x − 1 is negative while the denominator x + 2 is positive. The quotient is negative, so every input in that open interval fails. At x > 1, both expressions are positive and the quotient is positive. That entire interval works. Together, the sign statements cover all three intervals identified by the breakpoints.

Write the quotient-sign record as positive, negative, positive from left to right. Then include the endpoints' separate decisions: undefined at −2 and zero at 1. The strict answer is consequently the union of the two open outer intervals, (−∞, −2) and (1, ∞). An ordinary number line should show open circles at both finite endpoints, with the left and right rays selected and the middle interval unselected.

Substituting x = 2 gives 1/4, an additional direct check of the right interval. The examples −3, 0 and 2 agree with the sign record, but the interval arguments do the proving. A few successful samples alone would not establish that every real value between two chosen points has the same sign. Here the linear numerator and denominator have their signs explicitly fixed by each interval's defining inequalities.

Repair multiplication by splitting the cases

One safe algebraic route begins with x + 2 > 0, which means x > −2. In that case, multiplying the original strict inequality by the denominator preserves its direction and gives x − 1 > 0. Hence x > 1. This result already satisfies the case assumption x > −2, so the positive-denominator case contributes exactly the right-hand solution interval.

The other permitted case is x + 2 < 0, or x < −2. Multiplication now reverses the inequality, giving x − 1 < 0. That condition means x < 1, which every value in the case x < −2 already satisfies. This case therefore contributes the entire left-hand interval. The case x + 2 = 0 remains excluded rather than being multiplied through or assigned a quotient.

Combining the two cases returns x < −2 or x > 1. The originally proposed x > 1 trace can be retained as correct work for its positive-denominator branch once the assumption is written beside it. The error was treating that one branch as the whole problem. A useful correction preserves the valid conditional reasoning and adds the missing negative-denominator analysis.

Use a positive multiplier as another check

There is another route that avoids multiplying by an unknown sign. On the original domain x ≠ −2, the square (x + 2)² is strictly positive. Multiplying both sides by that square therefore preserves the direction everywhere the original fraction is defined. Canceling one nonzero denominator factor gives the equivalent condition (x − 1)(x + 2) > 0, while retaining x ≠ −2.

The product is positive when its two factors are both negative or both positive. Both negative requires x < 1 and x < −2, which reduces to x < −2. Both positive requires x > 1 and x > −2, which reduces to x > 1. Opposite signs give a negative product, and either zero factor fails the strict condition. This agrees with both the fraction-sign record and the case-split multiplication.

The square is not strictly positive at −2. That is why the original domain must remain attached to the transformation. We used a positive multiplier only on the permitted inputs, where cancellation was also defined. Writing a polynomial inequality on a fresh page without the accompanying domain could hide the condition that made the equivalence legitimate. The nonstrict control below shows how that omission changes an answer.

Change only strictness for an endpoint control

Consider the separate question (x − 1)/(x + 2) ≥ 0. Keep the same real domain excluding −2. The three open intervals retain the same signs, but the defined zero at x = 1 now qualifies. Its answer is therefore x < −2 or x ≥ 1. This is a new inequality, with the equality case deliberately added; it is not a correction to the earlier strict answer.

If someone instead solves only (x − 1)(x + 2) ≥ 0 and forgets the original domain, the polynomial accepts x = −2 because its product is zero. The original fraction still cannot accept that input. This exposes a domain error that the strict version could conceal, since its product condition happened to reject both roots anyway. A correct nonstrict answer includes 1 and excludes −2 for their distinct original reasons.

Leave the final record with the fraction, its domain, the sign intervals and the chosen strictness together. Another reader can check why both outer intervals appear and why each endpoint is open or closed. The same written discipline keeps a sign reversal from disappearing inside a crossed-out denominator. No calculator setting, graphing output or numerical approximation is needed to establish this exact result.

Keep the supporting references separate

The 15/16-inch laptop sleeve account provides optional worksheet-device storage context. Four whole galleries show complete zippered textile-looking sleeves and native size/color options. A HAWEEL label appears where no catalog brand is declared. No maker identity, actual selected variant, dimensions, material, fit, water resistance or protection performance is authenticated here.

The Crown-pattern sleeve account shows one complete sleeve with black border and zipper. Crown motifs describe its visible pattern. Catalog Jatoya naming is not independently authenticated by an unmarked photograph. No actual size selection, device fit, water resistance, durability or protection result is verified. Neither sleeve supplies the inequality or its proof.

The markup and margin explanation gives two ratios different positive financial bases. This problem keeps one expression fixed and allows its denominator to change sign across real inputs. Identifying a ratio's meaning remains useful, but a positive-cost percentage example does not settle the missing negative-denominator branch.

The dated payment comparison uses a factor 1 + r under an explicit nonnegative-rate assumption, keeping that denominator positive. Here the domain deliberately permits negative x values apart from the pole. Its two-case sign analysis is therefore a different task, without payment dates, conversion rates or a real saving claim.

Keep any sleeve order independent

If a reader separately considers a sleeve order, keep the item and transaction evidence apart from the inequality. A correct solution interval establishes no device fit, selected color, supplied maker or protective performance. Before ordering through republeq.com, identify the exact listing, size option, quantity and condition, and resolve any product question that affects the intended purchase. A diagonal device-size label or a marketing dimension graphic does not replace confirmation of the actual offered variant and its relevant measurements. If cryptocurrency is being considered, inspect the payment choices actually presented for that specific order before deciding how to proceed. Mentioning USDT in a guide does not establish support for the token, a particular network or a particular wallet. Where a suitable cryptocurrency option is actually offered, verify the requested asset, network, destination, amount and order reference against the current instructions before authorizing a transfer. Include shipping, taxes and any transaction charges when checking the complete amount. Retain the item reference, applicable delivery and return terms, and any written clarification that influenced the purchase. If the item or payment instructions change, review the new information rather than carrying an earlier conclusion across automatically. Keep the resulting order confirmation and payment record together so that a later question can be tied to the specific transaction. An uncertain payment or order status calls for reconciliation against those existing records before considering another payment. None of those steps changes the stipulated fraction, its excluded pole or the difference between strict and nonstrict comparisons. The mathematical exercise can be completed without buying either optional sleeve, while a later purchase depends on its own current product and transaction information.