A Nice Maths Problem | X=? & Y=? 

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Fantastic

Beautifull

I love your exercise ❤❤❤❤ I caught on you everything, but I've never thought about it😂😂😂

I cannot solve this problem, however I'm positive this method described is the hardest path to solve it 😢

(3+2)= 5 (1+2)= 3

1-9+10=2 it is not -2 correct the mistake

Very long sleeping it can be solved in four steps

You made a bunch of minor errors. I'm surprised you gave up on finding all the solutions. I list all of them, using a more general approach. *Background Info* For z = x, y as roots of a quadratic equation: (z - x)*(z - y) = 0 z² - (x + y)*z + x*y = 0 { Note sum and product } z = ((x + y) ± √((x + y)² - 4*x*y))/2 { E1 } = ((x + y) ± √(x² + 2*x*y + y² - 4*x*y))/2 = ((x + y) ± √(x² - 2*x*y + y²))/2 = ((x + y) ± √(x - y)²)/2 = ((x + y) ± (x - y))/2 z = x, y { For "±", the "+" yields x, the "-" yields y. } Let s = (x + y) and p = (x*y) in E1: z = (s ± √(s² - 4*p))/2 { E1' } Let z⁺ = (s + √(s² - 4*p))/2 Let z⁻ = (s - √(s² - 4*p))/2 Thus, (x, y) = (z⁺, z⁻) Since x & y are interchangeable, this means: (x, y) = (z⁻, z⁺), (z⁺, z⁻) { E1" } *Problem* I list the squares as E2 and the cubes as E3: x² + y² = 3 { E2 } x³ + y³ = 5 { E3 } Note: x & y are interchangeable. Use E2 in the expansion of the square of the sum: (x + y)² = x² + 2*x*y + y² (x + y)² = (x² + y²) + 2*x*y s² = 3 + 2*p p = (s² - 3)/2 { E2' } Use E3 in the expansion of the cube of the sum: (x + y)³ = x³ + 3*x²*y + 3*x*y² + y³ (x + y)³ = (x³ + y³) + 3*x*y*(x + y) s³ = 5 + 3*p*s { E3' } Use E2' in E3' to eliminate p, and find s: s³ = 5 + 3*((s² - 3)/2)*s s³ = 5 + 3*s³/2 - 9*s/2 0 = 5 + s³/2 - 9*s/2 s³ - 9*s + 10 = 0 s³ - 4*s - 5*s + 10 = 0 s*(s - 2)*(s + 2) - 5*(s - 2) = 0 (s - 2)*(s² + 2*s - 5) = 0 s = 2, (-2 ± √24)/2 = 2, -1 ± √6 = 2, √6-1, -√6-1 Use E2' in E1', and substitute s: z = (s ± √(s² - 4*(s² - 3)/2))/2 = (s ± √(s² - 2*(s² - 3)))/2 = (s ± √(6 - s²))/2 = (2 ± √2)/2, (√6-1 ± √(2*√6-1))/2, (-(√6+1) ± i*√(2*√6+1))/2 From E1", we list the solutions: (x, y) = (z⁻, z⁺), (z⁺, z⁻) = (1 - √2/2, 1 + √2/2), (1 + √2/2, 1 - √2/2), ((√6-1 - √(2*√6-1))/2, (√6-1 + √(2*√6-1))/2), ((√6-1 + √(2*√6-1))/2, (√6-1 - √(2*√6-1))/2), (-(√6+1)/2 - i*√(2*√6+1)/2, -(√6+1)/2 + i*√(2*√6+1)/2), (-(√6+1)/2 + i*√(2*√6+1)/2, -(√6+1)/2 - i*√(2*√6+1)/2)

Faltou destacar as respostas, por esse motivo, não gostei.

Ojo: Utiliza la misma variable, para diferentes funciones.confunde al alumno

Please don't make mistake while writing.

u dont finish it.