Alternatively, if we multiply the 1st equation, we get x³ + x² y - x y² - y³ =9, Similarly multiplying the 2nd equation , we get x³ - x² y +x y² - y³ = 5 Let these equations be numbered assigned as Eqn 3 and Eqn 4 respectively. Let us add and subtract these two equations. We get 2x³ - 2y³ = 14 or x³ - y³ = 7. And 2x² y - 2x y² = 4 or 2xy (x -y) = 4 or x y(x -y ) =2. Using algebraic identity (x - y)³ = x³ - y³ - 3 xy(x -y), we get (x - y)³ = 7- 3(2) = 1 that is x - y = 1. Using this value in our first 1st equation, which is (x + y)² (x -y) = 9, we get (x + y)² =9/1=9 so that x + y = +/- 3. 1st case, we have x + y = + 3 and x - y = 1, so x = (3+1)/2 =2 and y =( 3-1)/2 = 1. Case 2: we have x + y = - 3 and x - y = 1, so our x = (-3+1)/2 = -1 and y = (-3-1)/2 = - 2. Therefore the two sets of equation are {2,1 and -1, -2}
Two times the second equation minus the first one: x³ - 3x²y + 3xy² - y³ = 1 or (x - y)³ = 1 so x - y = 1 for x, y ∈|R. First equation minus the second one: 2xy (x - y) = 4 and with x - y = 1 we get xy = 2 or y = 2/x. So with x - y = 1 and y = 2/x we get x² - x - 2 = 0 with x = 2 or - 1.
The first equation (...=9) has by itself both final solutions (-1,-2) and (2,1) plus the solutions (4,-5) and (5,-4). The second equation (...=5) has by itself also both final solutions and not more. This I found out via matrices in excel, which I more often use for finding solutions for these equations. It gets easier when you already know the solutions.
I liked your solution, but the frogs in my garden pond insisted on doing like this instead. First, expand the two equations: x³ + x²y - xy² - y³ = 9 x³ - x²y + xy² - y³ = 5 Now add and subtract the two equations: 2x³ - 2y³ = 14 2x²y - 2xy² = 4 Divide by two: (a) x³ - y³ = 7 (b) x²y - xy² = 2 Multiply equation (b) by (-3) and add it to equation (a): x³ - 3x²y + 3xy² - y³ = 7 - 6 = 1 (x - y)³ = 1 (x - y) = 1 y = x - 1 Finally, insert y = x - 1 in equation (a): x³ - (x - 1)³ = 7 x³ - (x³ - 3x² + 3x - 1) = 7 3x² - 3x + 1 = 7 3x² - 3x - 6 = 0 x² - x - 2 = 0 x = 2 or x = -1 Since we have y = x - 1, we get the solution set {(2, 1), (-1, -2)}
The second method probably requires a little Galois theory to be effective; first recognizing the integer solutions and then, from the 6-fold symmetry, showing that the remaining two are complex involving +w and -w where w is a cube root of unity.
The way I approached it was by factoring the first equation into (x-y)(x+y)², then subtracting away the second equation I got (x-y)*2xy=4 or (x-y)*xy = 2. That leads to the obvious answer that x=2 and y=1. I suppose that is cheating! That didn't give me the second solution, but it makes it easy to find or to verify.
I got two solution sets in my head. I factored (x² - y²) to restate the first equation as (x + y)² * (x - y) = 9. I tried (x - y) = 1 and (x + y) = 3 to get x = 2 and y = 1, and verified that they satisfied the second equation. I also tried (x + y) = -3, getting x = -1 and y = -2. This also satisfied the second equation. Since we have cubic equations, I will use pencil and paper to try for a third solution. Much Later: the functions change inconsistent with x and y to have both constant, so the solutions are locally unique. I don't know about other solutions.
I saw the thumbnail and assumed integers, so after factoring we have (x-y) = +/-1 as these are the common factors of 5 and 9. Which gives (X+y)^2 = +/-9. So (x-y)=1 and (X+y) = +/-3. X = 2 or -1, y = 1 or -2. These check with (x^2 + y^2) = 5.
With z=x+y and w=x-y, the system is: z²w=9 and w(x²+y²)=5 If z=x+y and w=x-y implies z²=x²+2xy+y² and w²=x²-2xy+y² De donde: x²+y²=z²-2xy or x²+y²=w²+2xy Ahora 2xy=(1/2)(x²+2xy+y²-(x²-2xy+y²)) Por lo tanto 2xy=(1/2)(z²-w²) Así x²+y²=z²-(1/2)(z²-w²)=(1/2)(z²+w²) The system is z²w=9 and (1/2)(z²w+w³)=5 Reemplazando z²w=9: (1/2)(9+w³)=5 or 9+w³=10 De donde w=1. Reemplazando w=1 en z²w=9 se obtiene z²=9 de donde z=±3. Por lo tanto x+y=±3 and x-y=1 Sumando las ecuaciones: 2x=1±3 de donde 2x=4 or 2x=-2. Por lo tanto x=2 or x=-1. Cuando x=2 como x-y=1 entonces 2-y=1 de donde y=1. Cuando x=-1 como x-y=1 entonces -1-y=1 de donde y=-2. Conjunto solución {(-1,-2),(2,1)}.
Sommo e sottraggo le 2 equazioni e ottengo x^3-y^3=7,xy(x-y)=2..posto d=x-y, p=xy.. pd=2,d(d^2+3p)=7..che dà soluzioni per d=1,p=2...risultano le 2 coppie (x, y) =(2, 1),(-1,-2)
No, really. Do calculate the answer by using a and b instead of 9 and 5. The solutions are simply: x = ∛( (a+b±z)/4 ) and y =∛( (-a-b±z)/4 ), where z = (w+b/w)√a and w = √(2b-a).
The factors x-y and x+y must be integers, because we see that the products (x+y)(x^2-y^2) and (x-y)(x^2+y^2) result in integers (9 and 5). (x+y)(x^2-y^2)=9*1=1*9=3*3=-9*-1=-1*-9=-3*-3 9*1 -> x+y=9,(x+y)(x-y)=1 -> x-y=1/9 -> rejected, 1/9 is a fraction. 1*9 -> x+y=1,(x+y)(x-y)=9 -> x-y=9 -> y=-4, x=5, rejected, does not satisfy eq. 2 3*3 -> x+y=3,(x+y)(x-y)=3 -> x-y=1 -> y=1, x=2, accepted, satisfies both eqs -9*-1 -> x+y=-9,(x+y)(x-y)=-1 -> x-y=1/9 -> rejected, 1/9 is a fraction -1*-9 -> x+y=-1,(x+y)(x-y)=-9 -> x-y=9 -> y=-5, x=4 -> rejected, does not satisfy eqs -3*-3 -> x+y=-3,(x+y)(x-y)=-3 -> x-y=1 -> y=-2, x=-1 -> accepted, satisfies both eqs So (x,y)=(2,1) and (x,y)=(-1,-2)
Nice! My method is : z=x+y w=x-y.then you can solve for w and then for z.
And what would x²+y² be?
x2 +y2 +2xy and 2xy =( x +y) 2 - (x-y) 2
Divide 2, sorry for my clumsy
Alternatively, if we multiply the 1st equation, we get x³ + x² y - x y² - y³ =9, Similarly multiplying the 2nd equation , we get x³ - x² y +x y² - y³ = 5 Let these equations be numbered assigned as Eqn 3 and Eqn 4 respectively. Let us add and subtract these two equations. We get 2x³ - 2y³ = 14 or x³ - y³ = 7. And 2x² y - 2x y² = 4 or 2xy (x -y) = 4 or x y(x -y ) =2. Using algebraic identity (x - y)³ = x³ - y³ - 3 xy(x -y), we get (x - y)³ = 7- 3(2) = 1 that is x - y = 1. Using this value in our first 1st equation, which is (x + y)² (x -y) = 9, we get (x + y)² =9/1=9 so that x + y = +/- 3. 1st case, we have x + y = + 3 and x - y = 1, so x = (3+1)/2 =2 and y =( 3-1)/2 = 1. Case 2: we have x + y = - 3 and x - y = 1, so our x = (-3+1)/2 = -1 and y = (-3-1)/2 = - 2. Therefore the two sets of equation are {2,1 and -1, -2}
Two times the second equation minus the first one: x³ - 3x²y + 3xy² - y³ = 1 or (x - y)³ = 1 so x - y = 1 for x, y ∈|R.
First equation minus the second one: 2xy (x - y) = 4 and with x - y = 1 we get xy = 2 or y = 2/x.
So with x - y = 1 and y = 2/x we get x² - x - 2 = 0 with x = 2 or - 1.
THANKS FOR SHARING YOUR WORK
The first equation (...=9) has by itself both final solutions (-1,-2) and (2,1) plus the solutions (4,-5) and (5,-4). The second equation (...=5) has by itself also both final solutions and not more. This I found out via matrices in excel, which I more often use for finding solutions for these equations. It gets easier when you already know the solutions.
I liked your solution, but the frogs in my garden pond insisted on doing like this instead.
First, expand the two equations:
x³ + x²y - xy² - y³ = 9
x³ - x²y + xy² - y³ = 5
Now add and subtract the two equations:
2x³ - 2y³ = 14
2x²y - 2xy² = 4
Divide by two:
(a) x³ - y³ = 7
(b) x²y - xy² = 2
Multiply equation (b) by (-3) and add it to equation (a):
x³ - 3x²y + 3xy² - y³ = 7 - 6 = 1
(x - y)³ = 1
(x - y) = 1
y = x - 1
Finally, insert y = x - 1 in equation (a):
x³ - (x - 1)³ = 7
x³ - (x³ - 3x² + 3x - 1) = 7
3x² - 3x + 1 = 7
3x² - 3x - 6 = 0
x² - x - 2 = 0
x = 2 or x = -1
Since we have y = x - 1, we get the solution set
{(2, 1), (-1, -2)}
The second method probably requires a little Galois theory to be effective; first recognizing the integer solutions and then, from the 6-fold symmetry, showing that the remaining two are complex involving +w and -w where w is a cube root of unity.
Remaining *four* actually, so there are 2+4 solutions in total.
The way I approached it was by factoring the first equation into (x-y)(x+y)², then subtracting away the second equation I got (x-y)*2xy=4 or (x-y)*xy = 2. That leads to the obvious answer that x=2 and y=1. I suppose that is cheating! That didn't give me the second solution, but it makes it easy to find or to verify.
Nice graph... very symmetric 😀
EOT: It looks like some unknown small transparent ocean creature --->
I got two solution sets in my head. I factored (x² - y²) to restate the first equation as (x + y)² * (x - y) = 9. I tried (x - y) = 1 and (x + y) = 3 to get x = 2 and y = 1, and verified that they satisfied the second equation. I also tried (x + y) = -3, getting x = -1 and y = -2. This also satisfied the second equation. Since we have cubic equations, I will use pencil and paper to try for a third solution.
Much Later: the functions change inconsistent with x and y to have both constant, so the solutions are locally unique. I don't know about other solutions.
I saw the thumbnail and assumed integers, so after factoring we have (x-y) = +/-1 as these are the common factors of 5 and 9. Which gives (X+y)^2 = +/-9. So (x-y)=1 and (X+y) = +/-3. X = 2 or -1, y = 1 or -2. These check with (x^2 + y^2) = 5.
without watching:
set s = x + y and d = x - y, then the equations are
s²d = 9
d(s² + d²) = 10
immediately d³ = d = 1 and s = ±3
Got 'em both!
With z=x+y and w=x-y, the system is:
z²w=9 and w(x²+y²)=5
If z=x+y and w=x-y implies
z²=x²+2xy+y² and w²=x²-2xy+y²
De donde:
x²+y²=z²-2xy or x²+y²=w²+2xy
Ahora
2xy=(1/2)(x²+2xy+y²-(x²-2xy+y²))
Por lo tanto
2xy=(1/2)(z²-w²)
Así
x²+y²=z²-(1/2)(z²-w²)=(1/2)(z²+w²)
The system is
z²w=9 and (1/2)(z²w+w³)=5
Reemplazando z²w=9:
(1/2)(9+w³)=5 or 9+w³=10
De donde w=1.
Reemplazando w=1 en z²w=9 se obtiene z²=9 de donde z=±3.
Por lo tanto x+y=±3 and x-y=1
Sumando las ecuaciones:
2x=1±3 de donde 2x=4 or 2x=-2.
Por lo tanto x=2 or x=-1.
Cuando x=2 como x-y=1 entonces 2-y=1 de donde y=1.
Cuando x=-1 como x-y=1
entonces -1-y=1 de donde y=-2.
Conjunto solución {(-1,-2),(2,1)}.
y=kx might work, but depends on the luck with k^3 solution...
Sommo e sottraggo le 2 equazioni e ottengo x^3-y^3=7,xy(x-y)=2..posto d=x-y, p=xy.. pd=2,d(d^2+3p)=7..che dà soluzioni per d=1,p=2...risultano le 2 coppie (x, y) =(2, 1),(-1,-2)
I used a completely different method. got same solutions.
I think your method is better though
Just put y = ax and it drops out *very* fast.
No, really. Do calculate the answer by using a and b instead of 9 and 5. The solutions are simply: x = ∛( (a+b±z)/4 ) and y =∛( (-a-b±z)/4 ), where z = (w+b/w)√a and w = √(2b-a).
Real men solve for a and b instead of 9 and 5 respectively.
x = -1, y = -2
I think double or multiple integration is more interesting than single integration. Try to make some.
x=2, y=1
made it
The factors x-y and x+y must be integers, because we see that the products (x+y)(x^2-y^2) and (x-y)(x^2+y^2) result in integers (9 and 5).
(x+y)(x^2-y^2)=9*1=1*9=3*3=-9*-1=-1*-9=-3*-3
9*1 -> x+y=9,(x+y)(x-y)=1 -> x-y=1/9 -> rejected, 1/9 is a fraction.
1*9 -> x+y=1,(x+y)(x-y)=9 -> x-y=9 -> y=-4, x=5, rejected, does not satisfy eq. 2
3*3 -> x+y=3,(x+y)(x-y)=3 -> x-y=1 -> y=1, x=2, accepted, satisfies both eqs
-9*-1 -> x+y=-9,(x+y)(x-y)=-1 -> x-y=1/9 -> rejected, 1/9 is a fraction
-1*-9 -> x+y=-1,(x+y)(x-y)=-9 -> x-y=9 -> y=-5, x=4 -> rejected, does not satisfy eqs
-3*-3 -> x+y=-3,(x+y)(x-y)=-3 -> x-y=1 -> y=-2, x=-1 -> accepted, satisfies both eqs
So (x,y)=(2,1) and (x,y)=(-1,-2)
Me over here like: "Please substitute with 'u' and sing 'Happy birthday 2 u'!"
First comment