Overview
An inequality describes a set of possible values. Solving it means rewriting that same set in a more useful form, which is why direction and endpoint inclusion matter. A parameter adds another layer: a step that is valid for most parameter values may fail when a coefficient becomes zero. We will begin with a simple sign reversal, then combine intervals, and finally separate ordinary and exceptional cases.
Understand why a negative divisor reverses order
For 5 − 2x ≥ 13, subtract 5 to obtain −2x ≥ 8. Dividing by −2 gives x ≤ −4. The direction reverses because multiplication by a negative number reverses positions on the number line: larger inputs produce smaller products. This is a property of order, not a special instruction attached only to one formula.
Check the boundary first. At x = −4, both sides of the original comparison equal 13, so equality belongs in the answer. Then test a value on the accepted side and one on the rejected side if you need to confirm the orientation. The option x ≥ −4 keeps the correct boundary but the wrong direction. By separating boundary calculation from direction checking, you can diagnose exactly where a proposed answer fails. Preserve the equality bar throughout whenever your reversible operations do not remove it.
Intersect conditions before counting integers
The conditions 3x + 1 > −8 and 2x − 5 ≤ 7 give x > −3 and x ≤ 6. Because both must hold, take their intersection: −3 < x ≤ 6. The eligible integers start at −2 and end at 6, so there are nine.
The difference between the two displayed boundaries is not automatically the integer count. One boundary is excluded and the other is included, and the first eligible integer must be identified explicitly. The distractor 10 includes −3 even though it makes the first comparison an equality rather than a strict inequality. When two inequalities point in different directions, draw their overlap or write one combined interval. If they point the same way, the stronger restriction controls. A count is the last step; first establish the actual set that satisfies every condition.
Identify values that make division invalid
Consider (k² − 4)x = k − 2. For a fixed k, this is an equation in x, and the question asks when it has exactly one positive solution. Before dividing by k² − 4, record the values that make this coefficient zero: k = 2 and k = −2. For all other values, factoring gives (k − 2)(k + 2)x = k − 2, so x = 1/(k + 2).
This ordinary branch has a positive solution precisely when k + 2 > 0, or k > −2. However, the canceled value k = 2 still needs a separate examination. Substituting it into the original equation gives 0 = 0, which permits every x and therefore does not give exactly one positive solution. At k = −2 the equation gives 0 = −4 and has no solution. The final parameter set is k > −2 with k ≠ 2.
Read existence, positivity and uniqueness separately
The near-miss k > −2 gets the positivity condition right for the ordinary formula but forgets that the formula was obtained by dividing by something that can vanish. The narrower answer k > 2 avoids the exceptional point by discarding many valid values unnecessarily. Neither repair is a substitute for explicitly checking the excluded cases.
For a new parameter problem, make a small case table before doing long algebra. A nonzero coefficient usually permits one candidate; a zero coefficient can create either an identity or a contradiction. Only after deciding how many solutions exist should you apply a sign, interval or integer restriction. In practice, keep three questions separate: does a solution exist, is it unique, and does it lie in the requested domain? This structure prevents a correct algebraic expression from becoming an incomplete final answer and gives you a clear way to check every boundary or hole in a proposed interval.
Apply the method to a new case
A delivery fee is 4 + 3n dollars for a whole-number n of packages. A customer has at most 25 dollars and needs more than 3 packages. Which package counts are possible? The budget condition is 4 + 3n ≤ 25, so n ≤ 7; the need is n > 3. Intersect them, then apply the whole-number restriction: n is 4, 5, 6 or 7. There are four possibilities. Including 3 ignores the strict condition; including 8 exceeds the budget. Check the extreme permitted counts in the original model: four packages cost 16 dollars and seven cost 25 dollars. The upper boundary is included because 'at most' permits equality. Translating both restrictions before counting prevents a budget answer such as n ≤ 7 from silently including counts that do not meet the customer's need.