A radical system of equations

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This video is about a system of equations.
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xy = 12, x^2 + y^2 = 25
corresponds to the Pythagorean triple (3, 4, 5)
thus (3, 4) (4, 3) (-3, -4) (-4, -3)
are solutions

echandler

Very good... It can also be solved in this way :
xy=12 i.e., 2xy=24...(i)
x²+y²=25... (ii)
Firstly adding equation(i) and(ii), we get :
(x+y)²= 49 i.e., x+y=±(7)... (iii)
Secondly, equation(ii)- equation(i), we get :
(x-y)²=1 i.e., x-y=±(1)...(iv)
Solving equation (iii) and equation(iv), we now get:
{x, y} ={4, 3}, {4, -3}, {-4, 3}, {-4, -3}

asitmahata

You dont get 4 pairs of solutions, just 2. Product is 12 so they are both negative or both positive. Because of this it dors matter which of the two equations of the system you use after solving the square equation :)

arekkrolak

There will be two more solutions interchanging x and y. As there is a symmetry between x and y. Hence, altogether there are grout solutions.

tapandatta

5:49 xy=12 x^2+y^2=25, Shorter solution: 2*xy = 24. (x+y)^2=49;(x-y)^2= 1 then system equations x+y=±7, x-y=±1.
From symmetry only 4 real solution (±3, ±4) and (±4, ±3). But there are complex solution to.

golddddus

Trigonometric Substitution: x = r cos theta, y = r sin theta
Dividing the equations we get the value of sin 2 theta. Using sin 2 theta = 2 tan theta/ 1 + tan^2 theta, we can find tan theta (y/x).
r can be found by substituting in sin 2 theta

ribhuhooja

At 5:30 you could have sped it up by:
x^2 + 2xy + y^2 = 25 + 2*12 = 49 = (x + y)^2. So x + y = ±7
x^2 - 2xy + y^2 = 25 - 2*12 = 1 = (x - y)^2. So x - y = ±1 etc

MrLidless

Hi
If you divide the first equation by the second you have :
x=(3/4)y OR x=(4/3)y

taba

Really liked the problem. But I couldn't follow it after about 4:41 mark, probably due to short cuts that everyone else would get. Anyway, I have to educate myself more to follow your vids. But I really like your channel.

jamesmccamish

Divide both equations, as both of them don't contain any zero, you get x = ±3y/4 and x = ±4y/3. Given the symmetry of the problem, we only have to solve for x = ±3y/4 and pick all permutations of the answer. Make the substitution on the second equation and you get |y^3| = 27 which is the same as y = ±3, hence x = ±4. So the solutions are (-4, -3), (-3, -4), (3, 4), (4, 3).

djvalentedochp

When you see xx+yy=25
You could.guess a 3, 4, 5 triangle and when xy=12
that confirms it.
You just have to remember to take the negative pair -3, -4 and note the symmetry in x&y so that
They can be swapped .

davidseed

I liked your way to solving equations:) thanks

b.a.

Author didn't find two more solutions (3;4) and (-3;-4). Тhis follows from the fact that the original equations are symmetric with respect to x and y

valerykolesnikov