Solve inequalities with an unknown coefficient

Try

Consider the family ax+4>0. Here x is the unknown and a is a real parameter identifying which inequality you are solving. Try a=2, a=−2 and a=0. The answers are x>−2, x<2 and every real x, respectively. One formula obtained by dividing by a cannot describe all three without conditions.

Predict

Predict how the solution changes as a passes through zero. Does the direction merely stay fixed while the boundary moves? Does the zero case have the boundary x=0? Substituting a=0 leaves the true statement 4>0, with no restriction on x. A parameter case must be read from the original inequality when the operation you planned becomes invalid.

Experiment

Complete a case table for (a−1)x≤2. Test one accepted and one rejected x in each nonzero case.

Parameter	Reduced inequality	Real solution set
a=3	2x≤2	x≤1
a=−1	−2x≤2	x≥−1
a=1	0≤2	Every real x

For the general result, use a>1, a<1 and a=1, not a>0, a<0 and a=0. The coefficient whose sign matters is the complete expression a−1. Sketch the resulting number-line rays with filled boundary points, since equality is permitted.

Understand

Why does negative multiplication reverse order? If u<v then v−u is positive. For a negative c, the product c(v−u) is negative, so cv−cu is negative and cv<cu. This argument explains the rule rather than inferring it from a few examples. Division by a negative nonzero number is multiplication by its negative reciprocal and follows the same reasoning.

To solve ax≤b, first split into a>0, a<0 and a=0. In the positive case the answer is x≤b/a; in the negative case it is x≥b/a. At a=0 the original inequality becomes 0≤b, which holds for every real x if b≥0 and for none if b<0. Both sides can depend on parameters, so the zero case may require another test.

For instance, (a−1)x≥a−1 gives x≥1 when a>1, x≤1 when a<1, and every real x at a=1. “Cancel a−1” is incomplete even when you remember to exclude zero, because the sign still determines which ray is correct. A strict inequality changes its zero case: 0>0 is false, whereas 0≥0 is true.

Context adds a separate domain filter. A count may be restricted to nonnegative integers. If a business margin is negative, the formal real solution might demand a negative count and therefore provide no feasible plan. First solve with correct sign cases; then intersect with the allowed contextual domain. Do not round a negative or impossible answer into an invented feasible decision.

Apply

In the practice, construct a case table before writing a final answer. For each boundary, identify whether it is included. Check the zero-coefficient case directly, and distinguish “all real numbers” from “all allowed nonnegative integer counts” when context changes the universe.

A useful debugging test is to substitute one positive, one negative and one zero parameter into your general answer, then check representative x values in the original inequality. This can reveal a missing case, although a general sign argument is still needed to justify the whole family.

Continue

The next lesson uses reversible operations for equations rather than inequalities. Completing a square requires a nonzero leading coefficient; the degenerate case must again be separated before division. Carry the habit of stating an operation’s conditions into every formula derivation.

Continue to Choose a quadratic method and derive the formula.