Lessons 10 · 1 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
10
Simple rational equations and systems
Textbook, Part 1: pp. 58–63
GoalKnow the algorithm for rational equations; reject extraneous roots by checking denominators; use substitution and simple systems of rational equations.
The ratio of two polynomials P(x)/Q(x) is a rational expression, and an equation A(x) = B(x) between rational expressions is a rational equation. If every solution of the first equation is also a solution of the second, the second is a consequence of the first; if the solution sets coincide, the equations are equivalent. The equation P(x)/Q(x) = 0 holds exactly when P(x) = 0 and Q(x) ≠ 0 simultaneously. To solve a general rational equation: find the common denominator of the fractions, multiply both sides by it, solve the resulting equation, and discard roots that make the common denominator zero (they are extraneous, because multiplying only gives a consequence equation). In harder equations it helps to replace an expression in x by a letter t; for systems we use addition, substitution and new variables, always keeping every denominator nonzero.
Worked examples
(x² − 9)/(x − 3) = 0. Numerator: x² − 9 = 0 ⇒ x = 3 or x = −3. x = 3 makes the denominator zero and is discarded. Answer: x = −3.
(x/(x + 1))² − 3 · x/(x + 1) + 2 = 0. Put t = x/(x + 1): t² − 3t + 2 = 0, t = 1 or t = 2. t = 1: x = x + 1 — no solution. t = 2: x = 2x + 2 ⇒ x = −2; x + 1 = −1 ≠ 0. Answer: x = −2.
Class activity
“Extraneous-root hunt”: each group writes a rational equation designed so that one root makes a denominator zero; another group solves it and finds which root is extraneous.
Practice
1
When does P(x)/Q(x) = 0 have a solution?
When P(x) = 0 and Q(x) ≠ 0 for the same x.
2
Solve (x² − 16)/(x + 4) = 0.
x² = 16 ⇒ x = 4 or −4; x = −4 makes the denominator zero. Answer: x = 4.
3
Solve 3/(x − 1) − 2/(x + 1) = 6/(x² − 1).
Multiplying by (x − 1)(x + 1): 3(x + 1) − 2(x − 1) = 6, x + 5 = 6, x = 1. But x = 1 makes a denominator zero, so there is no solution.
4
Why must roots of a rational equation be checked against the denominators?
Multiplying by the denominator gives a consequence equation: an x that makes the denominator zero may appear as a new “root”, while in the original equation division by 0 is undefined.