For anyone who may be confused by the method in the video, asking "why can we assume that y and x are proportional?" The answer is that, technically, there is no assumption built into the substitution. All he did was perform an arbitrary change of variables. If it helps, you can motivate this change of variables by realizing that every nonzero real number is equal to the product of two nonzero real numbers in uncountably infinite many ways. Perhaps it will all be clearer if instead of using the letter m, which is suggestive, the letter z is used.
Excellent. Try with this. A sphere has volume V=(4/3)pi(2+sqrt(5)) = (4/3)pi*r^3. Let r = x + sqrt(5)y so r^3 = 2+sqrt(5) = (x + sqrt(5)y)^3. This gives two cubic equations in x and y with one solution x = y = 1/2 and r = (1 + sqrt(5))/2.
Nice problem !! I was not able to solve the equation, I tried to get a perfect cube but I was not able to do so. Anyway, I liked how you solved the problem. Thanks for the video :)
@@SyberMath two curves intersect in infinty points or product the grades ( multiplicity is important). If a curve F has degree m and the curve G has degree n then the curve F Intersect G in mn points (with multiplicity and points the infinity) or infinity points (in this case gcd(f,g) ot=1).
@@elkincampos3804 Not sure what you mean, but Setting y = m*x, dividing, and simplifying gives m³ - 8m² + 19m - 12 = 0 which has solutions m = 1, 3 and 4. m = 1 gives x³ = 1 which has solutions [x,y] = [1,1], [(-1-i√3)/2,(-1-i√3)/2] and [(-1+i√3)/2,(-1+i√3)/2] m = 3 gives x³ = 1/5 which has solutions [x,y] = [1/cbrt(5), 3/cbrt(5)], [(-1+i√3)/(2cbrt(5)), (-3+3i√3)/(2cbrt(5))], [(-1-i√3)/(2cbrt(5)), (-3-3i√3)/(2cbrt(5))] m = 4 gives x³ = 1/10 which has solutions [x,y] = [1/cbrt(10), 4/cbrt(10)], [(-1+i√3)/(2cbrt(10)), (-2+2i√3)/cbrt(10)], [(-1-i√3)/(2cbrt(10)), (-2-2i√3)/cbrt(10)]] cbrt(x) is meant to be cube root Plugging those points into the original equations gives the correct values.
could u explain the validity of assuming a linear relation between x and y(x=my), as far as i believe there doesnt have to be a relation between x and y in the first place cuz they both are independent in their respective aspect
from subtracting the equations i get : 2x^3 + 5x^2 * y + y^2 * ( y - x ) = 7 then it is pretty obvious that y = 1 and x = 1 is a solution to the system.
This does help, but I find that this type of approach to solving system of equations is completely pointless. "Solve the system" does not mean "find one or two solutions." It means "find all the solutions, and prove that these are all the solutions." That is the general idea behind these videos.
1 + 22·x + 20·y = x·y x·y - 22·x - 20·y = 1 x·y - 22·x - 20·y + 22·20 = 1 + 22·20 = (x - 20)·(y - 22) = 441 = 21^2 = (3·7)^2. The list of divisors of 441 is thus {-441, -147, -63, -49, -21 - 9, -7, -3, -1, 1, 3, 7, 9, 21, 49, 63, 147, 441}. Set y - 22 to be equal to all of these divisors, one at a time, then solve for y and x. This will give you all the solutions that you seek.
@@SyberMath i used different substitution In fact i got 10th degree equation but it was easily factorable One root was obvious other could be easily found by substitution
For anyone who may be confused by the method in the video, asking "why can we assume that y and x are proportional?" The answer is that, technically, there is no assumption built into the substitution. All he did was perform an arbitrary change of variables. If it helps, you can motivate this change of variables by realizing that every nonzero real number is equal to the product of two nonzero real numbers in uncountably infinite many ways. Perhaps it will all be clearer if instead of using the letter m, which is suggestive, the letter z is used.
The fact that this is a homogenous system makes that kind of substitution useful, doesn't it?
Both methods are very interesting and are two excellent strategies for a lot of questions with 2 equations and two variables.
Why was the equal sign so humble?
Because she knew she wasn’t greater than or less than anyone else.
This couldn't be truer! 😁
why does this have only 3 likes? Two are from Syber an me, what about the others?
Excellent. Try with this. A sphere has volume V=(4/3)pi(2+sqrt(5)) = (4/3)pi*r^3. Let r = x + sqrt(5)y so r^3 = 2+sqrt(5) = (x + sqrt(5)y)^3. This gives two cubic equations in x and y with one solution x = y = 1/2 and r = (1 + sqrt(5))/2.
Great as always. I love Math 😊
Thank you! Cheers!
Me too
Me too
much faster is to use Horner's method to try if a number is the root. Immediately u also obtain the result of polynomial division
Nice explanation. Thanks
Glad you liked it!
Excellent solution!!
Glad you think so!
you got 3 values of y in terms of x, wouldn't solving 2nd equation as a quadratic in y give only 2 of them?
Nice problem !! I was not able to solve the equation, I tried to get a perfect cube but I was not able to do so.
Anyway, I liked how you solved the problem. Thanks for the video :)
Sure! Thanks for watching
Very nice exercise and methods.
Glad you like them!
There is 9 points or infinity by Bèzout theorem. We should see to solution in the complex numbers.
What is Bèzout theorem?
@@SyberMath two curves intersect in infinty points or product the grades ( multiplicity is important). If a curve F has degree m and the curve G has degree n then the curve F Intersect G in mn points (with multiplicity and points the infinity) or infinity points (in this case gcd(f,g)
ot=1).
@@elkincampos3804 You have three values of `m`. Each results in a cubic equation in `x`. Each has one real and two complex solutions, totaling 9.
@@MizardXYT Prove it. It's not easy. We take points in the infinity.
@@elkincampos3804 Not sure what you mean, but
Setting y = m*x, dividing, and simplifying gives m³ - 8m² + 19m - 12 = 0 which has solutions m = 1, 3 and 4.
m = 1 gives x³ = 1 which has solutions [x,y] = [1,1], [(-1-i√3)/2,(-1-i√3)/2] and [(-1+i√3)/2,(-1+i√3)/2]
m = 3 gives x³ = 1/5 which has solutions [x,y] = [1/cbrt(5), 3/cbrt(5)], [(-1+i√3)/(2cbrt(5)), (-3+3i√3)/(2cbrt(5))], [(-1-i√3)/(2cbrt(5)), (-3-3i√3)/(2cbrt(5))]
m = 4 gives x³ = 1/10 which has solutions [x,y] = [1/cbrt(10), 4/cbrt(10)], [(-1+i√3)/(2cbrt(10)), (-2+2i√3)/cbrt(10)], [(-1-i√3)/(2cbrt(10)), (-2-2i√3)/cbrt(10)]]
cbrt(x) is meant to be cube root
Plugging those points into the original equations gives the correct values.
Nice system and solution!
Glad you like it!
But what if y ≠ mx ?
Are you excluding possible solutions? Proof?
could u explain the validity of assuming a linear relation between x and y(x=my), as far as i believe there doesnt have to be a relation between x and y in the first place cuz they both are independent in their respective aspect
Observe that the sum of power of x and y equates the same, say 3 in the case, then we can use y = mx.
I like this problem. I present two solutions here. Which method do you think is better?
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Great work! 👍
Thank you! 👍
from subtracting the equations i get : 2x^3 + 5x^2 * y + y^2 * ( y - x ) = 7
then it is pretty obvious that y = 1 and x = 1 is a solution to the system.
Well, yes!
This does help, but I find that this type of approach to solving system of equations is completely pointless. "Solve the system" does not mean "find one or two solutions." It means "find all the solutions, and prove that these are all the solutions." That is the general idea behind these videos.
Thank you!, but could you help me with this?
Solve for positive integers x,y
1+22x+20y=xy
1 + 22·x + 20·y = x·y x·y - 22·x - 20·y = 1 x·y - 22·x - 20·y + 22·20 = 1 + 22·20 = (x - 20)·(y - 22) = 441 = 21^2 = (3·7)^2. The list of divisors of 441 is thus {-441, -147, -63, -49, -21 - 9, -7, -3, -1, 1, 3, 7, 9, 21, 49, 63, 147, 441}. Set y - 22 to be equal to all of these divisors, one at a time, then solve for y and x. This will give you all the solutions that you seek.
Thanks for the response, Angel! This is a really cool trick aka Simon's Favorite Factoring Trick.
Using substitutions i got cubic with one integer root easy to guess
Substitution method may lead us to extraneous solutions
Does it?
@@SyberMath i used different substitution
In fact i got 10th degree equation but it was easily factorable One root was obvious other could be easily found by substitution
excellent
Many many thanks!
Hocam emeğinize ve zihninize sağlık. 👏👏👏
Cok tesekkur ederim hocam! 🥰🥰🥰
you are legend from India in up
Thank you! 🧡
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Cool but i liked diophantine eqns more
More will come. Thanks for sharing!
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