Can you crack this TOUGH math problem? - 2015 German Math Olympiad.

In this video, we will break down how to solve the cubic system of equations x³ + 9x²y = 10 and y³ + xy² = 2 for x and y. Whether you're struggling with solving cubic equations, working with systems of nonlinear equations, or just need help with algebraic manipulation, this easy-to-follow tutorial will guide you through the process.
I’ll walk you through each step, from simplifying the equations to finding the solution for both variables. Whether you’re preparing for an exam or just looking to improve your algebra skills, this video will help you solve similar systems of equations with confidence.

🔑 Key Concepts Covered:
Solving cubic systems of equations.
Algebraic manipulation and substitution.
Step-by-step breakdown of tricky math problems.
Solving for x and y in nonlinear equations.

Perfect for:
Algebra and calculus students.
Math learners of all levels.
Anyone preparing for exams or looking to improve problem-solving skills.

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#matholympiad #maths #algebra

[yurenchu]

Nine solutions (three real, six complex-valued):
(x, y) = (1, 1) OR
(x, y) = ( (-1+i√3)/2, (-1+i√3)/2 ) OR (x, y) = ( (-1-i√3)/2, (-1-i√3)/2 )
OR
(x, y) = ( (5 + 3√5)/2, (1 - √5)/2 ) OR
(x, y) = ( (-1+i√3)(5+3√5)/4, (-1+i√3)(1-√5)/4 ) OR (x, y) = ( (-1-i√3)(5+3√5)/4, (-1-i√3)(1-√5)/4 )
OR
(x, y) = ( (5 - 3√5)/2, (1 + √5)/2 ) OR
(x, y) = ( (-1+i√3)(5-3√5)/4, (-1+i√3)(1+√5)/4 ) OR (x, y) = ( (-1-i√3)(5-3√5)/4, (-1-i√3)(1+√5)/4 )

_CALCULATION_ :
x³ + 9x²y = 10 [eq. 1]
y³ + xy² = 2 [eq. 2]
Note: x = 0 and y = 0 are not a solutions. Multiply (eq. 2) by 5, and combine with (eq. 1):

x³ + 9x²y = 10
5y³ + 5xy² = 10

x³ + 9x²y = 5y³ + 5xy²
.. Divide by x³ ...
1 + 9(y/x) = 5(y/x)³ + 5(y/x)²
.. Substitute u = y/x ...
1 + 9u = 5u³ + 5u²
1 + 9u - 5u² - 5u³ = 0
... Upon inspection, we see u = 1 is a solution; so we can factor out (1 - u) from LHS ...
(1 - u) * (1 + 10u + 5u²) = 0
(1 - u) * (5 + 50u + 25u²) = 0*5
(1 - u) * (5 + 2*5*5u + (5u)²) = 0
(1 - u) * (25 + 2*5*5u + (5u)² - 20) = 0
(1 - u) * ( (5 + 5u)² - 20 ) = 0
(1 - u) * ( (5 + 5u)² - (2√5)² ) = 0
(1 - u) * (5 + 5u - 2√5) * (5 + 5u + 2√5) = 0
(1 - u) = 0 OR (5 + 5u - 2√5) = 0 OR (5 + 5u + 2√5) = 0
u = 1 OR 5u = -5 + 2√5 OR 5u = -5 - 2√5
u = 1 OR u = (-1 + (2/5)√5) OR u = (-1 - (2/5)√5)
y/x = 1 OR y/x = (-1 + (2/5)√5) OR y/x = (-1 - (2/5)√5)

Case 1 : y = x
Case 2 : y = x*(-1 + (2/5)√5) OR y = x*(-1 - (2/5)√5)

Try each case in (eq. 2): y³ + xy² = 2

Case 1 : y = x :
substitute x into (eq. 2)
y³ + y*y² = 2
2y³ = 2
y³ = 1
(x, y) = (1, 1) OR (x, y) = ( (-1+i√3)/2, (-1+i√3)/2 ) OR (x, y) = ( (-1-i√3)/2, (-1-i√3)/2 )

Case 2 : y = x*(-1 ± (2/5)√5) :
... multiply both sides by [-1 ∓ (2/5)√5] ...
y*[-1 ∓ (2/5)√5] = x*(-1 ± (2/5)√5)*[-1 ∓ (2/5)√5]
y*[-1 ∓ (2/5)√5] = x*[ 1 - 4/5]
y*[-1 ∓ (2/5)√5] = x*[ 1/5 ]
y*[ -5 ∓ 2√5] = x

Substitute x into (eq. 2):
y³ + xy² = 2
y³ + [ -5 ∓ 2√5]y*y² = 2
[ -4 ∓ 2√5]y³ = 2
y³ = 1/[ -2 ∓ √5]
y³ = [ -2 ± √5]/( [ -2 ∓ √5]*[ -2 ± √5] )
y³ = [ -2 ± √5]/( 4 - 5 )
y³ = 2 ∓ √5
y³ = (16 ∓ 8√5)/8
y³ = (1 + 3*5 ∓ (3+5)√5)/8
y³ = (1 ∓ 3√5 + 3*5 ∓ 5√5)/8
y³ = ( 1³ ∓ 3(1²)(√5) + 3(1)*(√5)² ∓ (√5)³ )/2³
y³ = (1 ∓ √5)³ /2³
y³ = [ (1∓√5)/2 ]³
y = m*(1∓√5)/2, where m = 1, (-1+i√3)/2 or (-1-i√3)/2

x = y*[ -5 ∓ 2√5] =
= m*((1∓√5)/2) * [ -5 ∓ 2√5]
= m*(1∓√5)*(-5 ∓ 2√5)/2
= m*( (-5 ∓ 2√5) ∓ (-(√5)*5 ∓ 2(√5)*√5) )/2
= m*( -5 ∓ 2√5 ± 5√5 + 10 )/2
= m*( 5 ± 3√5 )/2

(x, y) = ( (5 ± 3√5)/2, (1∓√5)/2 ) OR
(x, y) = ( (-1+i√3)(5 ± 3√5)/4, (-1+i√3)(1∓√5)/4 ) OR
(x, y) = ( (-1-i√3)(5 ± 3√5)/4, (-1-i√3)(1∓√5)/4 )

[rohanjariwala]

That was a great explanation keep up the good work!!

[yurenchu]

The trick/method that you start introducing at 2:15 is quite clever and cool! 😀👍

[HenriLaporte-kvqq]

2 homogeneous equations in x and y. Put y = kx => with some manipulations 5kkk + 5kk = 9k + 1. 1 is a solution and the rest is straightforward

[freddyalvaradamaranon]

Mi hija y mi persona estamos muy agradecidos por compartir la buena y didáctica explicación 😊😊. No conocíamos este sistema de ecuaciones con exponentes cúbicos y cuadrados, realmente muy interesante.😊😊❤❤😊😊.

[renesperb]

The solution {x=1, y=1} is of course easy to see. For the other of your solutions you used good tricks. I have one problem:
MATHEMATICA gives 9 solution pairs, including complex ones.By the way: a different way to start is to divide the two equations, leading to a third order equation for u = x/y .Since u=1 is a solution one can then apply polynomial division to get a quadratic equation for u. Of course you need more work to get from u back to x and y.You get equations of the form x^3*(1+9u) =10
and y^3*(1+u) =2, where u is a solution determined before.Both equations have one real and two complex solutions.

[yurenchu]

Sorry, but you made a mistake at 13:25 . Instead of writing ±, you should have written ∓ .
When y = (1+√5)/2, then x = (5 *−* 3√5)/2 (and _not_ x = (5 *+* 3√5)/2, because that one is paired with y = (1−√5)/2 ).