☰ Contents · Mathematics

Approximate solving and rational inequalities

Lessons 13–14 · 2 lessons · M. A. Mirzaahmedov, Sh. N. Ismailov, A. Q. Amanov (algebra and analysis), B. Q. Haydarov (geometry). Mathematics Grade 10, Parts 1 and 2, 1st edition. EXTREMUM PRESS, Tashkent, 2017
14

Simple rational inequalities and systems

Textbook, Part 1: pp. 77–78
GoalSolve rational inequalities by the interval method and find the solution of a system of inequalities as the common part of the solution sets.
New words
solution of an inequality · tengsizlikning yechimiinterval method · oraliqlar usulisystem of inequalities · tengsizliklar sistemasicommon part · umumiy qism
Explanation

Relations of the form A(x) > B(x), A(x) < B(x), A(x) ≥ B(x) or A(x) ≤ B(x) with a variable x are inequalities, and a value of x that turns it into a true numerical inequality is a solution. For a rational inequality, move everything to one side to get the form P(x)/Q(x) > 0 (or <, ≥, ≤) and factor the numerator and denominator. The zeros of the numerator and denominator split the number line into intervals; the expression has a constant sign on each interval, so testing one number is enough. A factor of even multiplicity (for example (x − 2)²) does not change the sign when passing the point. Zeros of the denominator never belong to the solution (they are always excluded), and zeros of the numerator belong to it only for non-strict inequalities (≥, ≤). The solution of a system of inequalities is the common part (intersection) of the solutions of each inequality.

Worked examples
(x + 2)(x − 1)(4 − x) ≥ 0. Since 4 − x = −(x − 4), the inequality is equivalent to (x + 2)(x − 1)(x − 4) ≤ 0. The zeros are −2, 1, 4. At x = 5 the product is positive, and the signs alternate from right to left: + on (4, ∞), − on (1, 4), + on (−2, 1), − on (−∞, −2). Negative or zero: (−∞, −2] ∪ [1, 4].
(x² − x − 6)/(x − 1) ≥ 0. Factor the numerator: (x − 3)(x + 2)/(x − 1) ≥ 0. The zeros are −2 and 3 (included) and 1 (excluded). At x = 4 the expression is positive: + on (3, ∞), − on (1, 3), + on (−2, 1), − on (−∞, −2). Answer: [−2, 1) ∪ [3, ∞).
Class activity

“Sign staircase”: roots are placed on a number line on the board, students take turns writing + or − for each interval; then groups change the inequality sign and rewrite the solution.

Practice
1
Why do zeros of the denominator never belong to the solution of a rational inequality?
2
Solve (x − 5)(x + 1) < 0.
3
Solve (x − 2)/(x + 4) ≤ 0.
4
Solve the system 3x + 1 > 7, 10 − x > 2 and explain why the common part is taken.