Solve and interpret inequalities

Try

A fictional workshop charges 12 dollars to book a room and 5 dollars per participant. You have at most 47 dollars. Write 12 + 5n ≤ 47 and find the allowed nonnegative whole-number participant counts.

Predict

Does “at most” include exactly 47 dollars? Yes. Predict whether 7 and 8 participants are allowed. Their costs are 47 and 52 dollars, so 7 is included and 8 is excluded.

Experiment

Subtract 12 and divide by positive 5 to obtain n ≤ 7. On paper, draw a number line with a filled point at 7 and a ray to the left. This shows the real-number solution. Then impose the story’s restrictions: retain only whole numbers from 0 through 7. The algebraic interval and the feasible counts are different representations of different domains.

Understand

Adding the same quantity preserves order. Multiplication by a positive quantity preserves order; multiplication by a negative quantity reverses it. On a number line, multiplying by −1 reflects left and right. For example, 2 is less than 5, while −2 is greater than −5.

Solve 4 − 2x ≤ 10: subtract 4 to obtain −2x ≤ 6, then divide by −2 to obtain x ≥ −3. Check x = 0: 4 ≤ 10 is true. Check x = −4: 12 ≤ 10 is false. Forgetting reversal would accept the wrong side.

A compound inequality joined by “and” requires both conditions. For 1 ≤ 2x + 3 ≤ 9, subtract 3 everywhere and divide by 2 to get −1 ≤ x ≤ 3. The intersection is the segment between the endpoints. With “or,” either condition suffices, so the solution is a union, possibly two separate rays.

Strict inequalities exclude equality. A boundary test is useful, but testing a few values alone does not prove the entire solution set. The order-preserving algebra gives that justification. Multiplying by an expression of unknown sign requires cases; the simple negative-constant rule cannot be applied blindly.

Apply

Use the exact interval choices and contextual calculation below. In the written question, explain the reversal rather than merely citing a memorized rule. Check a value on each side of the boundary.

Continue

Next, isolate one symbol in a formula containing several symbols. Conditions such as nonzero divisors become essential because a parameter can change the kind of solution set.