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.😊😊❤❤😊😊.
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.
Thanks a lot for your feedback. It's very much appreciated. Yes, there would be more than three solution pairs, seeing as the two equations represent PLANES and we would need a third dimension (z-axis) to plot both planes and find all their points of intersection. I only gave the real number solutions (and notice that y is the golden ratio 🙂).
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 ).
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 )
WOW! This is detailed! Thanks a lot.
The trick/method that you start introducing at 2:15 is quite clever and cool! 😀👍
Thanks! 😃
That was a great explanation keep up the good work!!
Thank you for your very kind words.
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.😊😊❤❤😊😊.
Thanks for your kind words. Have a great day!
great explanation ssir
Thank you for your kind words!
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
Thanks a lot! I'll try it out.
Smart solution
Thanks for your kind words. Happy Holidays.
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.
Thanks a lot for your feedback. It's very much appreciated. Yes, there would be more than three solution pairs, seeing as the two equations represent PLANES and we would need a third dimension (z-axis) to plot both planes and find all their points of intersection. I only gave the real number solutions (and notice that y is the golden ratio 🙂).
X,2×+5=8[n3]
I knew x and y equal 1 in 1 seconds
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 ).
Thanks.
By inspection
X= 1, Y = 1
You are correct. But we also need to find the other roots.
binomial theory
You are right.
Smart solution
Thanks for your kind words. Happy Holidays.