First of all, No condition was given!! There should be infinite solutions by plotting a curve!! And the first answer appeared in my mind was also (6,0)!!
Without parameters this is also a correct answer. Using his method of selecting the two numbers multiply together to equal 36, using 6 x 6 also fits his method. 0 is not a natural number but there is no rule expressly stated that one must use natural numbers or repeated numbers.
There is a problem with the set N. Older texts considered that 0 was not a natural number but almost all modern texts consider 0 to be a natural number so the solution using modern texts would be x=6 and y = 0.
x^2 - y^2 = 36 x^2 - y^2 = 6^2 x^2 + 6^2 = y^2 according to Pythagoras a^2 + b^2 = c^2 The value of b is 2 times greater than the value of the Egyptian triple: 4^2 + 3^2 = 5^2 Accordingly 8^2 + 6^2 = 10^2 If you rearrange everything in its place: 10^2 - 8^2 = 6^2 100 - 64 = 36
Excellent professor/teacher. He explains very slowly so anyone can learn mathematics. Really good job. Please 🙏 make as many videos as you can in your lifetime. GOD BLESS YOU WITH GOOD HEALTH. AAMEEN
@@ramasaamynachimuthu7965 Wrong. You mistakenly took the factors of 6 * 6 = 36 as x and y themself when you falsely concluded (x+y)=12 and (x-y)=0. because of (x+y)(x-y) = 36 = 6 * 6: x + y = 6 and x - y = 6. Add them together 2x = 12 and x = 12/2 = 6. Now calculate y: x + y = 6 means 6 + y = 6 therefore y = 6 - 6 = 0 or x - y = 6 means 6 - y = 6 therefore -y = 6 - 6 = 0 which is the same as y = 0 (6 + 0)(6 - 0) = 6 * 6 = 36 or back to the original question 6^2 + 0^2 = 36 + 0 = 36
x X x = 36 + y X y so x = 6 if y= 0 when y is 1, 2, 3, 4,5,6,7 then we do not get a number which will have a root which is a whole number when y = 8 , then y X y = 64 x X x = 36 + 64 x X x = 100 so x = 10 , y= 8 or x = 6 , y= 0
X^2-y^2=36 X^2-y^2=6^2 X^2=6^2+y^2 We know 3,4,&5 are Pythagorean triplets. 6 can be converted to 3x2. so to eliminate 2 from 6, Divide both sides by 4 (X/2)^2= 3^2+(y/2)^2 It means x/2= 5 & y/2=4 => X=10 Y=8
In principle at least one of the factors must be an even number. But if one of the factors is even then (X,Y) is (even, even) or (odd, odd) and in both cases both factors would be even.
@@fernandoreescolar418 Or we could say that the difference between X+Y and X-Y is 2Y which is even. So if X+Y is even minus even is even, therefore X-Y is even
I have also got this idea. The problem can be solved in just two steps. A triangle having sides with 3:4:5 ratio is referred to as Egyptian Triangle The thing that any math teacher must tell their students in Grade 5. .
I have no specific formula for solving the problem. But by common-sense inspection and intuition, 100 - 64 = 36. 100 happens to be 10 square, and 64 is 8 square... So "x" must therefore be 10, and "y" equals 8. Or, "x" could be 6, and "y" equals 0. How's that for a quick solution.
Could you kindly provide the source of your claim that this question is from ‘Harvard University Admission Exam’? I am not sure whether they have a written format of entrance exam.
@@DespejalaxconJorge x=6 & y=0. It is called "trivial' solution. On further inspection, 4 (=2 x 2) is a factor too & indicates that both factors (x-y) & (x+y) are each even. x^2 - y^2 = 4 x 9 = (2 x 9) x (2 x 1); (x+y) (x-y)= (2 x 9) x (2 x 1). Hence, ½(x+y) = f odd ½(x-y) = g odd & the product ¼(x+y)(x-y) = ¼(36) = 9. 9 has factors (9, 1) or (3. 3). Factoring out 4 (from 36), gives 9, which is a product of 9 & 1, as 3 & 3 are ruled out being equal (it gives y=0, the trivial solution) - as we seek unequal factors. Thus f=9 & g=1. It can also be proved that swapping the pair of solutions (f=1 & g=9) is acceptable (or admissible) too. adding them gives, (f+g) even, since the sum of two odd numbers is even. Subtraction gives (f-g) even, since the difference of two numbers is even. ½(x+y) = 9 ½(x-y) =1 ___________ added ½(2x) = 10 x = 10. __________subtracted ½(2y) = 8 y = 8 Proof: x^2 - y^2 = 10^2 - 8^2 = 100 -64 x^2 - y^2 = 36
No, because weather if "0" is ∈ ℕ or not isn't well defined, its ambiguous. x,y ∈ ℕ₀ or x,y ∈ ℕ⁺ would be clear in order to avoid confusion especially in a context like this.
I thought of it simply as Pythagorean triple. Move y2 to the other side, think of 36 as 6 squared. Then we can quickly see that the resulting expression is just twice the basic 3, 4 ,5 Pythagorean triple, namely 6, 8 and 10.
@@PandithGautha To indicate an element belongs to the set of natural numbers, you would use the "∈" symbol (like a stylized epsilon) alongside "N". That symbol is on his paper at the top.
@@trouts4444 The symbol ℕ doesn't include or exclude "0". Its not well defined what ℕ means. Several textbooks use it differently. In order to avoid confusion its highly recommended to write ℕ₀ or ℕ⁺. (does it make sense to exclude the neutral element of addition from natural numbers?)
Just to add: Ideally, you should have explained to the students that. When you pick 6 x6, at the point when you will have x as 6 and y as 6, the product will be zero which means that the choice is invalid. I hope I helped a student understand. Also, zero is not a natural number. It should have been easier.
Solutions are : S=[(x,y)=(6,0);(10,8)] X=6 or 10 and Y=0 or 8 36 is = 6*6 or 18*2. The solutions would be if two pairs numbers and the first number must be more or equal than the second to be valid X^2-Y^2 = 36 Case 1 : (X+Y)(X-Y) = 18*2 ===> X+Y = 18 X-Y = 2 === > X=10, Y=8 S1={(X,Y)=(10,8)} Case 2 (X+Y)(X-Y) = 6*6 ===> X+Y = 6 X-Y = 6 === > X=6, Y=0 S2={(X,Y)=(6,0)} S={(X,Y)=(10,8) and X,Y)=(6,0)}
Another solution would be 6 and 0. However, since the Solution must be natural Numbers, 0 is rejected. I got the Solution 0 and 6 using derivatives, however 0 is not a natural Number and the 6,0 are not correct.
36 = 36 x 1 = 6 x 6 = 12 x 3 = 18 x 2; X^2 _ Y^2 = (X+Y) (X-Y) If X+Y = 36 and X-Y = 1 will give nonwhole number. Same with 6×6, 12 x 3. If X+Y = 18 and X-Y = 2, you get x=10 y = 8. 100-64= 36. Qed.
To indicate an element belongs to the set of natural numbers, you would use the "∈" symbol (like a stylized epsilon) alongside "N". That symbol is on his paper at the top.
Just saw your problem on the screen and solved it within 2 minutes in my head only, I'm a 43 year old chef and this is not a Harvard math problem but primary school Pitagorian variant.
That was just a convenient expression but I’m sure the institution setting the question was looking for calculus as the method for solving this. Had the expression been more complicated you couldn’t solve it by using this method.
Look like you miss 6*6. x2 + y2 = 36, y = x - 6. Summary: The x-coordinates of the solutions to this system of equations x2 + y2 = 36, y = x - 6. are 0 and 6.
Because you are dealing with squares, there are actually 4 solutions: x, y can equal 10, 8 10, -8 -10, 8 -10, -8 When the negative numbers are squared, the result becomes positive so any of the 4 sets yields 100 - 64 = 36.
What of -10 and -8? Will they not result in the same solution? Given the way the question was posed, safe an error on my part, i did not see the restriction that the values of X and Y must belong all to the set of Naturall Numbers
Please graph X values that you rejected…. You only find one solution not moving solution … think like a circle and each degree different (r) descend or ascend in one loop of (timplace) If your math doesn’t describe a movement so your at constant level of knowledge Spin your equation
It took me 15 seconds of logical mental gymnastics to obtain the solution: What perfect square numeral less another perfect square numeral = 36? No notes on paper required. Let’s keep it simple and work SMART and CONCISE 👍
It's obviously 100-36. Duh. If you can see that you have a shot at Harvard. Your chances are better if your father went there. If you get it after some some wandering around, you're qualified to apply for your local community college.
x=6 and y =0 is the simplest answer !
true....ye idea to youtuber ke dimaag me bhi nai aaya hoga..
No brother. 0 is not a natural number
First of all, No condition was given!! There should be infinite solutions by plotting a curve!! And the first answer appeared in my mind was also (6,0)!!
Without parameters this is also a correct answer. Using his method of selecting the two numbers multiply together to equal 36, using 6 x 6 also fits his method. 0 is not a natural number but there is no rule expressly stated that one must use natural numbers or repeated numbers.
That was my immediate thought when I saw the question.
There is a problem with the set N. Older texts considered that 0 was not a natural number but almost all modern texts consider 0 to be a natural number so the solution using modern texts would be x=6 and y = 0.
Simply by inspection:
Which square minus which square gives 36?
100 - 64 = 36.
Therefore x = 10
And y = 8
X = 6 & y = 0 also work
@@paulhardy3746 x and y belongs to natural numbers so 0 cant exist
@@paulhardy3746 Dude, x and y belong to N (Natural number). 0 aint a natural number.
@paulhardy3746 "0" is not natural number.
Yeah. I did it in my head in about ten seconds. Harvard, here I come! ;-)
You're making a very easy math problem complicated
As in algebra, you show your work, instead of simply coming up with the correct answer without being able to sow how it was done.
Yes
If this is the best Harvard can come up with, God help us!! No one should need 9 minutes and/or a pencil to solve this one.
The trick is you need to have a pair of even numbers in order to attain an outcome of natural numbers as the solution.
It's true
x^2 - y^2 = 36
x^2 - y^2 = 6^2
x^2 + 6^2 = y^2 according to Pythagoras a^2 + b^2 = c^2
The value of b is 2 times greater than the value of the Egyptian triple: 4^2 + 3^2 = 5^2
Accordingly 8^2 + 6^2 = 10^2
If you rearrange everything in its place:
10^2 - 8^2 = 6^2
100 - 64 = 36
Please elaborate please
If x^2 - y^2 = 6^2
then x^2 + 6^2 = y^2 ==> How?
Since 6٨2 + y٨2 = x٨2, you can list all the pythagorean triple containing 6 where 6 is not the hypotenuse.
😮
Elemental. Es el triángulo rectángulo 3, 4, y 5, multiplicado por 4.
You can get 10 and 8 in your head through substitution in less than a minute.
It is immediately clear that both x^2 and y^2 should be sguare numbers , then we have 100- 64=36, so x=10, and y=8
Да слишком легко)
@paulhardy3746
7 hours ago
X = 6 & y = 0 also work
So basically since y can equal to zero, it doesn’t need to be square number.
If. Solution. Could be decimal. Numbers, then. X=6.5. and Y=2.5 Could be solutions
Good observation
Excellent illustration for lay man understanding & inspirational demonstration.
Stay blessed.
Excellent professor/teacher. He explains very slowly so anyone can learn mathematics. Really good job. Please 🙏 make as many videos as you can in your lifetime. GOD BLESS YOU WITH GOOD HEALTH. AAMEEN
Thank you so much for your kind words! 🙏❤️🙏
6x6 is also 36 you need to present it and omitting it as x-y will be zero, but 6 and 0 is also a solution for x and y respectively
6,6 can be considered as another pair.
So x+y=12 and x-y =0.
But (x+y)(x-y)=36
12X0=36
0=36.
12X0=36
0=36.
So 6 ,6 is rejected.
@ramasaamynachimuthu7965 That's what I said, bro,
@@ramasaamynachimuthu7965
Wrong. You mistakenly took the factors of 6 * 6 = 36 as x and y themself when you falsely concluded (x+y)=12 and (x-y)=0.
because of (x+y)(x-y) = 36 = 6 * 6:
x + y = 6 and
x - y = 6.
Add them together
2x = 12 and
x = 12/2 = 6.
Now calculate y:
x + y = 6 means 6 + y = 6 therefore y = 6 - 6 = 0
or
x - y = 6 means 6 - y = 6 therefore -y = 6 - 6 = 0 which is the same as y = 0
(6 + 0)(6 - 0) = 6 * 6 = 36
or back to the original question
6^2 + 0^2 = 36 + 0 = 36
x X x = 36 + y X y
so x = 6 if y= 0
when y is 1, 2, 3, 4,5,6,7 then we do not get a number
which will have a root which is a whole number
when y = 8 , then y X y = 64
x X x = 36 + 64
x X x = 100
so
x = 10 , y= 8
or
x = 6 , y= 0
6 x 0 = 0
X^2-y^2=36
X^2-y^2=6^2
X^2=6^2+y^2
We know 3,4,&5 are Pythagorean triplets. 6 can be converted to 3x2.
so to eliminate 2 from 6, Divide both sides by 4
(X/2)^2= 3^2+(y/2)^2
It means x/2= 5 & y/2=4
=> X=10 Y=8
X=6 & Y=0
Suberb
Analyzing the expression (X+Y)(X-Y), we come to the conclusion that both terms can only be even numbers. So only the pair 18-2 can be a solution.
can you elaborate?
In principle at least one of the factors must be an even number. But if one of the factors is even then (X,Y) is (even, even) or (odd, odd) and in both cases both factors would be even.
Wrong sir @@fernandoreescolar418
@@fernandoreescolar418
X is 10
Y is 8
@@fernandoreescolar418 Or we could say that the difference between X+Y and X-Y is 2Y which is even. So if X+Y is even minus even is even, therefore X-Y is even
x*x - y*y= 36
x*x = 36 + y*y
x*x = 36 + ( and here I ran basic times table to 8.... 8 squared is 64)
x*x = 36 + 64 = 100
x = 10
you're the genius! coz i did the same and just squared 1-10 coz its always one of the answers.
Using Pythagoras law
In right angled triangle sides 6,8,10
x² = y²+ 6²
10² = 8² +6²
I have also got this idea. The problem can be solved in just two steps.
A triangle having sides with 3:4:5 ratio is referred to as Egyptian Triangle
The thing that any math teacher must tell their students in Grade 5. .
Should be 10^2=8^2+6^2
I have no specific formula for solving the problem. But by common-sense inspection and intuition, 100 - 64 = 36. 100 happens to be 10 square, and 64 is 8 square... So "x" must therefore be 10, and "y" equals 8. Or, "x" could be 6, and "y" equals 0. How's that for a quick solution.
Call it guesswork
Fantastic and simple way of teaching.
Thank you very much always giving a good analysis and a step by step solution!!!
This detailed explanation is for slow learners. Appreciate please 🎉🎉
I’m glad you found it helpful! 😊
You left out 6x6 which would solve that (x,y) =(6,0).
However there seems to be a debate whether 0 is a natural number or not.
But given the requirement that x+y > x-y, 6x6 cannot be an option.
Then again, (x+y)(x-y) = 6x6 when 0 ∈ N, so I think you are correct.
@attiylanen The requirement should be that x+y >= x-y if 0 is a natural number
Okay but (6+6)(6-6) doesn't that just give you 0?
@ohmy9261 you misunderstood my response. What I said is that x=6, y=0 solves the equation if 0 is considered a natural number
XX - YY = 36
.
*. 36 = 10x10 - 8x8 = 100 - 64
**. X = 10, Y = 8.
Could you kindly provide the source of your claim that this question is from ‘Harvard University Admission Exam’? I am not sure whether they have a written format of entrance exam.
Abaixo, considerar: x=(a+b) e y =(a-b). Sempre haverá uma relação, onde 4ab=x^2-y^2.
What about 6 x 6 = 36?
If x+y=x-y,y=0
@@i_am_sb but x=6,y=0 is a correct solution.
@@Merilix2 I didn't say it is wrong
@ I didn't say you said it is wrong.
(X x X) - (Y x Y) =36
(10 x10) - (8 x 8) = 36
(100) - (64) = 36
X = 10 and Y=8
X=10, Y=8. 100 minus 64 = 36. Simples
Yes partial parts of circle ⭕️ one central moving with different r =R which is effected from outside
The condition is that the sum of factors should be even (if x, y ∊ N).
Only (18, 2) satisfy this condition, that yields x=10 and y=8.
Para muchos estaría también la solución x=6 e y=0 debido a que la discusión sobre si el 0 es un número natural o no sigue abierta.
@@DespejalaxconJorge
x=6 & y=0. It is called "trivial' solution.
On further inspection, 4 (=2 x 2) is a factor too & indicates that both factors (x-y) & (x+y) are each even.
x^2 - y^2 = 4 x 9 = (2 x 9) x (2 x 1);
(x+y) (x-y)= (2 x 9) x (2 x 1).
Hence,
½(x+y) = f odd
½(x-y) = g odd
& the product ¼(x+y)(x-y) = ¼(36) = 9.
9 has factors (9, 1) or (3. 3).
Factoring out 4 (from 36), gives 9, which is a product of 9 & 1, as 3 & 3 are ruled out being equal (it gives y=0, the trivial solution) - as we seek unequal factors.
Thus f=9 & g=1. It can also be proved that swapping the pair of solutions (f=1 & g=9) is acceptable (or admissible) too.
adding them gives, (f+g) even, since the sum of two odd numbers is even.
Subtraction gives (f-g) even, since the difference of two numbers is even.
½(x+y) = 9
½(x-y) =1
___________ added
½(2x) = 10
x = 10.
__________subtracted
½(2y) = 8
y = 8
Proof:
x^2 - y^2 = 10^2 - 8^2 = 100 -64
x^2 - y^2 = 36
No, because weather if "0" is ∈ ℕ or not isn't well defined, its ambiguous.
x,y ∈ ℕ₀ or x,y ∈ ℕ⁺ would be clear in order to avoid confusion especially in a context like this.
I thought of it simply as Pythagorean triple. Move y2 to the other side, think of 36 as 6 squared. Then we can quickly see that the resulting expression is just twice the basic 3, 4 ,5 Pythagorean triple, namely 6, 8 and 10.
Let X=Y+ n, plug into the equation, you can quick immediately get the answers
Excellent work!
Solved the problem with easy method
You're welcome!
حسب النسب المثلثية المعروفة
١٠٠ - ٦٤ = ٣٦
(١٠×١٠) - (٨×٨) = ٣٦
مع التحية لكم
x²-y²=36; solution; x²-y²-36=0;x²-y²-9-11-16=0;x²-9-y²-16-11=0; x²-3²-y²-4²-11=0;(x-3)(x+3)-(y-4)(y+4)-11=0; (x-3)(x+3)-(y-4)(y+4)=11; x+3=11; x=11-3; x=8; x-3=11; x=11+3; x=14; -(y²+4y-4y-16)=11; -y²-4y+4y+16=11; -y²+16=11; -y²=11-16; -y²=-5(-1); y²=5; y²-5=0; y²-(√5)²=0; (y-5)(y+5)=0; y-5=0;y=5;y+5=0; y=-5; S={-5;5;8;14}.
x2 = 36 + y^2 => x = SQRT(36+y2) Now if y2 = 64 => y = 8 since y belongs to N => x = 10
(10x10=100) - (8x8=64) = 100-64 =36
you are missing ONE obvious solution : x=6 and y=0 !!
Zero is not a natural number. x and y are natural numbers
@@nagalakshmidevi7022 there is no such restriction imposed in the question.So anyhting that logically satisfies the equation , will do the JOB !
@@PandithGautha To indicate an element belongs to the set of natural numbers, you would use the "∈" symbol (like a stylized epsilon) alongside "N". That symbol is on his paper at the top.
@@trouts4444 The symbol ℕ doesn't include or exclude "0". Its not well defined what ℕ means. Several textbooks use it differently.
In order to avoid confusion its highly recommended to write ℕ₀ or ℕ⁺.
(does it make sense to exclude the neutral element of addition from natural numbers?)
X=10 Y=8 10x10=100 8x8=64 100-64=36 (6x6)
4ab=x^2-y^2. 4ab=36.ab=1×9.
(9+1)=x , y=(9-1).
Just to add: Ideally, you should have explained to the students that. When you pick 6 x6, at the point when you will have x as 6 and y as 6, the product will be zero which means that the choice is invalid. I hope I helped a student understand. Also, zero is not a natural number. It should have been easier.
To eliminate cases, in the condition that result shall be natural number, then one should look for factors that when added shall be even
You are doing a great work, keep it on.
Since x,y are naturals integers
Why you forgot the first case
X=6 and Y= 0
(6+0)*(6-0)= 36
10 squared - 8 squared = 36 (100-64)
(X-Y).(X+Y)=36 , (10-8).(10+8)=36
X=13/2. Y=5/2
X=10 y=8. Or just in a second x=6 and Y 0. Can do it.
You'r an excellent professor teacher
Thank you! 😃
Solutions are : S=[(x,y)=(6,0);(10,8)]
X=6 or 10 and Y=0 or 8
36 is = 6*6 or 18*2.
The solutions would be if two pairs numbers and the first number must be more or equal than the second to be valid
X^2-Y^2 = 36
Case 1 :
(X+Y)(X-Y) = 18*2 ===>
X+Y = 18
X-Y = 2 === > X=10, Y=8 S1={(X,Y)=(10,8)}
Case 2
(X+Y)(X-Y) = 6*6 ===>
X+Y = 6
X-Y = 6 === > X=6, Y=0 S2={(X,Y)=(6,0)}
S={(X,Y)=(10,8) and X,Y)=(6,0)}
1:44 you missed the 9th choice 36 = 6 x 6 which would lead to x = 12/2 = 6 and y = 6 - x = 6 - 6 = 0.
But is brute force really the way to go?
This guy is insane, period!
This is just one specific solution to the equation. There are infinitely many solutions, as the equation represents a hyperbola.
And where exactly is the "solution" here? Simply permutation of the possibilities which are not that much and can be done w/o need to write at all.
X*x-y*y= 100 - 64
X= +-100 , y= +-6
Or, x=+-6, y=0
No need to check every case. (x+y)+(x-y) = 2x must be an even number. SUm of 2 factor must be even so only (18,2) works.
Use the Pythagorean Theorem
Cos ϴ = 0.8
Therefore X=8/0.8 =10
And y^2 = 100- 64 =36
Then y = 6
Another solution would be 6 and 0. However, since the Solution must be natural Numbers, 0 is rejected.
I got the Solution 0 and 6 using derivatives, however 0 is not a natural Number and the 6,0 are not correct.
I think a simple way was elongated to confuse, if you do math like this everything is going like a big problem.
Amazing explanation!
Thank you so much!🙏❤️🙏
It is a best calculation thank you
Amazing 😍
36 = 36 x 1 = 6 x 6 = 12 x 3 = 18 x 2; X^2 _ Y^2 = (X+Y) (X-Y)
If X+Y = 36 and X-Y = 1 will give nonwhole number. Same with 6×6, 12 x 3.
If X+Y = 18 and X-Y = 2, you get x=10 y = 8. 100-64= 36. Qed.
İ am from Azerbaydjan.Veru useful.Thanks
Perfect . Thank you for your good explanation and teaching 👌🏻👌🏻
He explained the math so well that I'm only 11 but I can still understand the math.👍
A common error in reasoning is that if we make the assumption at the beginning that we are considering only natural numbers, then the cases x
To indicate an element belongs to the set of natural numbers, you would use the "∈" symbol (like a stylized epsilon) alongside "N". That symbol is on his paper at the top.
@@trouts4444 but ℕ isn't well defined. better to use ℕ₀ or ℕ⁺ if inclusion/exclusion of "0" matters.
x = 15/2 =7.5
y = 12 - 7.5 =4.5
7.5² - 4.5² = 56.25 - 20.25
= 36
Why do you reject these numbers ?
x=10 y=8 is the answer
X² - y² =10² -8²
100 - 64 = 36.
x=10, y=8 6, 8 and 10 are pyrhagorean triplets.
Вот именно что там решать если знаешь квадраты чисел от двух до десяти
You didn't apply the rule of multiple two negative numbers that are positive and you didn't put it in the probabilities
• x = 10, y = 8
• x = 8, y = 10
• x = -10, y = -8
• x = -8, y = -10
36/2=18
1st no is (18+2)÷2 =10
and
2nd no is ( 18- 2)÷ 2=8
Если мне не изменяет память, такие сочетания называют пифагоровыми тройками (3²+4²=5² ; 12²+5²=13² и т.д.)
Just saw your problem on the screen and solved it within 2 minutes in my head only, I'm a 43 year old chef and this is not a Harvard math problem but primary school Pitagorian variant.
That was just a convenient expression but I’m sure the institution setting the question was looking for calculus as the method for solving this. Had the expression been more complicated you couldn’t solve it by using this method.
(x+y)(x-y) = 36 = 18*2
so x+y = 18 and x-y = 2
so 2x=20
so x=10 and y = 8
Check
100-64=36
Look like you miss 6*6.
x2 + y2 = 36, y = x - 6. Summary: The x-coordinates of the solutions to this system of equations x2 + y2 = 36, y = x - 6. are 0 and 6.
Nicely explained
Thank you 🙏❤️🙏
Please do not solve this question like that. There are no conditions saying x and y are integer.
بما ان الاعداد الثلاث صحيحة
نكتب المربعات المحتملة وهي
من 1 ، 4 الى 400
ونحصل على مربعين هما 100 و 64 تحققان الحل
(6x6) - (0×0)
It is Pythagoras. 25=16+9, 100=64+36, 225=144+81, 400=256+144...
Case 5: (x+y)=(x-y)=6 -------------> x=6, y=0
Explêndido 👏👏👏
Because you are dealing with squares, there are actually 4 solutions: x, y can equal
10, 8
10, -8
-10, 8
-10, -8
When the negative numbers are squared, the result becomes positive so any of the 4 sets yields 100 - 64 = 36.
A^2_B^2=(A+B)(A-B)=(10-8)(10+8)=36
: A=10 AND B=8
OR
A^2_B^2=36
10^2-8^2=36
:A=10 AND B=8
Analyzing the expression (X+Y)(X-Y),
6 X 6 = 36
X =6
Y = 0
x=6, y=0 gives the right solution
x,y=(10,8)
What of -10 and -8? Will they not result in the same solution?
Given the way the question was posed, safe an error on my part, i did not see the restriction that the values of X and Y must belong all to the set of Naturall Numbers
Please graph X values that you rejected…. You only find one solution not moving solution … think like a circle and each degree different (r) descend or ascend in one loop of (timplace)
If your math doesn’t describe a movement so your at constant level of knowledge
Spin your equation
It took me 15 seconds of logical mental gymnastics to obtain the solution: What perfect square numeral less another perfect square numeral = 36? No notes on paper required. Let’s keep it simple and work SMART and CONCISE 👍
Amazing ❤
Thanks 🙏❤️🙏
Deux solutions possibles:
✓ x=6 & y=0
✓ x=10 & y=8
very good, detail
Thank you! Cheers!
0:42 Without to much thoughs
I found that 100-64=36 so X=10 and Y=8
Binomial. Then you can do it in your head.
X = 10 and Y = 8, resulte in 5 seconds or less, why 9,30 minutes?
It's obviously 100-36. Duh.
If you can see that you have a shot at Harvard. Your chances are better if your father went there.
If you get it after some some wandering around, you're qualified to apply for your local community college.