An Incredible System of Equations | Math Olympiad

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In this algebraic video, we'll explore how to solve a challenging system of equations. Whether you're a math enthusiast or a student seeking to improve your problem-solving skills, these strategies will enhance your understanding and boost your confidence. We'll break down solution step-by-step, providing clear explanations and examples to ensure you grasp the concepts. Don't miss out on mastering these valuable techniques!
Topics covered:
System of equations
Algebra
Problem solving
Algebraic identities
Algebraic manipulations
Solving systems of equations
Factorization
Synthetic division
Quadratic Equation
Math enthusiast
Radical equations
Substitution
Rational root theorem
Math tutorial
Math Olympiad
Math Olympiad Preparation
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[@RashmiRay-c1y]

Defining x^2/3=a and y^2/3=b, we get a^2+b^2=17 and a^3+b^3 = 65 > a=4, b=1 or a=1, b=4. If a=4, b=1 > (x,y) = (8,1), (-8,-1), (8, -1), (-8,1). If a=1, b=4, (x,y) = (1,8), (-1,-8), (1,-8), (-1,8).

[@jaggisaram4914]

(8 , 1). , (1 , 8)

[@Fjfurufjdfjd]

Θετω (x^2)^(1/3)=α και (y^2)^(1/3)=β αρα εχω x^2=α^3 και y^2=β^3 α>0 και β>0. Αν (α+β)= t >0 εχωt^3-51t+130=0 t=5 αν t ανηκει στο Z.εχω τελικα α=4 β=1 ,ή α=1 β=4. Αρα x^2=64 y^2=1 ή x^2=1 y^2=64 και αρα x=+ -8 y=+ -1 ή x=+ -1 y=+ -8

[@SidneiMV]

x = u³ y = v³ u³u + v³v = 17 u⁴ + v⁴ = 17 u⁶ + v⁶ = 65 u² + v² = b u²v² = c t² - bt + c = 0 t² = bt - c u⁴ = bu² - c v⁴ = bv² - c u⁴ + v⁴ = b(u² + v²) - 2c 17 = b² - 2c => c = (b² - 17)/2 t³ = bt² - ct = b(bt - c) - ct t³ = (b² - c)t - bc u⁶ = (b² - c)u² - bc v⁶ = (b² - c)v² - bc u⁶ + b⁶ = (b² - c)(u² + v²) - 2bc 65 = (b² - c)b - 2bc 65 = b(b² - 3c) = b(b² - 2c - c) 65 = b(17 - c) 65 = b[17 - (b² - 17)/2] 130 = b(34 - b² + 17) 130 = b(51 - b²) b³ - 51b + 130 = 0 b³ - 125 - 51b + 255 = 0 b³ - 5³ - 51(b - 5) = 0 (b - 5)(b² + 5b - 26) = 0 b - 5 = 0 => b = 5 c = (b² - 17)/2 => c = 4 u² + v² = 5 u²v² = 4 p² - 5p + 4 = 0 (p - 1)(p - 4) = 0 u² = 1 => v² = 4 x = u³ => x = ± 1 y = v³ => y = ± 8 u² = 4 => v² = 1 x = u³ => x = ± 8 y = v³ => y = ± 1 (x, y) = { (1, 8), (1, -8), (-1, 8), (-1, -8), (8, 1), (8, -1), (-8, 1), (-8, -1) }

[@fhffhff]

{x²/³+y²/³=a,x²/³y²/³=b;{b=0,5a²-17/2,a³-51a-130=0; a=(65+√(65²-17³))¹/³+(..-..)..=2√17 cos(1/3arccos(65/√17³)), n=0 b=34cos²(1/3arccos(65/√17³))-8,5 {y=±(√17cos(1/3arccos(65/√17³)) -+√(-8,5cos(2/3arccos(65/√17 ³))))³/², x=±(√17cos(1/3arccos(65/√17³))± √(-8,5cos(2/3arccos(65/√17³))))³/².

[@kassuskassus6263]

(x,y)=(1,8); (-1,-8); (8,1); (-8,-1)

[@TheDavidlloydjones]

By inspection!

[@john-paulderosa7217]

@@TheDavidlloydjones I did it this way also. Inspection should precede long manipulations of symbols. One has to get a sense of a problem before heading into a blizzard of identities.