Solve inequalities and interpret their full solution sets

Learning goal: Construct and solve simple rational inequalities, graph the solution set and interpret its boundary and permitted values in context. Explain why multiplying or dividing by a negative number reverses the comparison.

An inequality compares quantities without requiring equality. The symbols > and < are strict: the boundary value is excluded. The symbols ≥ and ≤ include equality and may express phrases such as at least or no more than. Read the condition carefully before writing a symbol. A goal of more than fifty is different from a goal of at least fifty.

Suppose a fictional weekly payment is 20 units plus three per completed task, and the goal is more than 50. The inequality is 20 + 3x > 50, where x is the number of completed tasks. Subtract 20 from both sides to obtain 3x > 30, then divide by positive three to get x > 10. On a continuous number line, use an open circle at ten and a ray toward larger numbers.

The context requires whole completed tasks, so possible counts are 11, 12, 13 and so on. The continuous ray represents the algebraic inequality; its whole-number points are the feasible task counts. Saying x = 11 would give the smallest feasible count but not the entire solution set. Distinguish a minimum from all solutions when the question asks for one or the other.

For an at-least goal, 20 + 3x ≥ 50 gives x ≥ 10. The number-line circle at ten is closed because ten tasks exactly meets fifty. A single word in the context changes the boundary inclusion. Testing the boundary in the original inequality helps verify which marker belongs on the graph.

Adding or subtracting the same quantity preserves an inequality's direction. Multiplying or dividing by a positive quantity also preserves it. Multiplying or dividing by a negative quantity reverses the inequality's direction; dividing by a negative number is the same as multiplying by its negative reciprocal. For instance, 2 < 5 but −2 > −5 after multiplying by −1. On a number line, taking opposites reflects points across zero, reversing which point lies farther right. This is why the inequality symbol reverses, not an arbitrary extra step.

Solve a model remaining-balance condition: 18 − 2x > 6. Subtract 18 to obtain −2x > −12. Divide by −2 and reverse the comparison to obtain x < 6. Check a value on each side: at x = 5, the original gives eight > six, true; at x = 7, it gives four > six, false. At x = 6, the amount equals six and fails the strict condition.

A context can restrict this solution further. If x counts whole purchases with no negative counts, the feasible values are 0, 1, 2, 3, 4 and 5. The algebraic ray extends left from six, including negative numbers that solve the inequality mathematically, but those negative values are not purchases in the stated scenario. Interpret the intersection of the algebraic solution and the domain.

Rational coefficients and boundaries work similarly. For (3/4)x + 1/2 < 5, subtract 1/2 to get (3/4)x < 9/2. Divide by positive 3/4 to obtain x < 6. For −(1/2)x + 1 < 5/2, subtract one to get −(1/2)x < 3/2, then divide by −1/2 and reverse to obtain x > −3. The open boundary remains because the original comparison was strict.

Some whole-number minimums arise from fractional boundaries. A goal x > 10.4 requires at least eleven whole items, while x ≥ 10.4 also has eleven as the smallest whole item count. Their real-number solution sets differ even though the same smallest integer works. Do not round a boundary before solving; identify the exact inequality and then interpret the allowed discrete values.

A solution graph should have a labeled scale, correct boundary location, appropriate open or closed marker and a ray in the correct direction. Testing a sample value is a useful check, but one true value does not establish the whole set. The algebraic transformations justify the set; the tests help catch a reversed sign or misplaced boundary.

Worked fee limit: A model cost is 2.5 units per meter plus a fixed four, and the budget permits at most nineteen. Write 2.5m + 4 ≤ 19, leading to m ≤ 6. With m ≥ 0 for length, feasible values are from zero through six meters. The closed endpoint includes exactly six because its cost is nineteen.

Worked negative coefficient: −3x + 2 ≤ 11 gives −3x ≤ 9 and x ≥ −3 after division by −3. A closed circle at negative three and rightward ray represent the algebraic set. Check x = 0: two ≤ eleven is true, agreeing with the ray.

Worked wrong graph: A student solves 18 − 2x > 6 as x > 6 without reversing. Testing x = 7 gives four > six, false, so the graph points the wrong way. The corrected x < 6 agrees with both the negative-division rule and the original inequality.

Graphing rehearsal: Graph x > 10, x ≥ 10, x < 6 and x ≥ −3 on equally spaced number lines. For each, test the boundary and one point on each side. Explain what the marker and ray communicate. Then restrict the first three to nonnegative whole-number counts when a context requires them.

Independent performance: Create four inequality problems, including a strict goal, an inclusive budget, a negative coefficient and a fractional boundary. Define the variable and domain, solve with justified operations and graph each full algebraic solution set. State the feasible contextual values and the smallest or largest value only when it exists and the question asks for it. A partner tests a claimed solution and a claimed nonsolution in the original condition.