I have another approach . x(x-y) = 14 , y(x+y) = 60 , multiplying , xy(x^2-y^2) = 840 , but x^2-y^2 = 2xy-46 (obtained by subtracting the equations . Therefore , 2x^2y^2 - 46xy - 840 = 0 , xy = 35 , xy = -12 , when xy = 35 , x = +-7 and y = +-5 and when xy = -12 , x = +-sqrt2 and y = +-6sqrt2
I wonder why you didn't consider the following approach. Rewrite the equations as x² = 14 + xy y² = 60 − xy Multiplying both equations this gives x²y² = (14 + xy)(60 − xy) Now let xy = z and we have z² = (14 + z)(60 − z) z² = 840 + 46z − z² 2z² − 46z − 840 = 0 z² − 23z − 420 = 0 (z − 35)(z + 12) = 0 z = 35 ⋁ z = −12 So we have xy = 35, which gives the solutions (x, y) = (7, 5) and (x, y) = (−7, −5) or xy = −12 which gives the solutions (x, y) = (√2, −6√2) and (x, y) = (−√2, 6√2).
If you take each term on the left hand side of each of the equations, the sums of the powers of x and y are constant. In this case, they are always 2. For example, x^2 is x^2 y^0, 2 + 0 = 2. xy = x^1 y^1, 1 + 1 = 2. When that happens, you get surprisingly good results by putting in y = ux.
Very interesting problem and solutions! I love seeing the same problem solved using different methods. Keep it up!
I have another approach . x(x-y) = 14 , y(x+y) = 60 , multiplying , xy(x^2-y^2) = 840 , but x^2-y^2 = 2xy-46 (obtained by subtracting the equations . Therefore , 2x^2y^2 - 46xy - 840 = 0 , xy = 35 , xy = -12 , when xy = 35 , x = +-7 and y = +-5 and when xy = -12 , x = +-sqrt2 and y = +-6sqrt2
My solution (using polar coordinates):
x^2-xy=14
y^2+xy=60
Sum them both:
→x^2+y^2=74
Let x=r cosθ and y=r sinθ:
→r^2=74
→r=±√74
Find θ:
→74 cos^2θ-74 sinθ cosθ=14
→37 cos^2θ-37 sinθ cosθ-7=0
→37 cos^2θ-37 sinθ cosθ-7(sin^2θ+cos^2θ )=0
→30 cos^2θ-42 sinθ cosθ+5 sinθ cosθ-7 sin^2θ=0
→5 cosθ (6 cosθ+sinθ )-7 sinθ (6 cosθ+sinθ )=0
→(6 cosθ+sinθ )(5 cosθ-7 sinθ )=0
Case 1: 6 cosθ+sinθ=0
→tanθ=-6
→θ=arctan(-6)
→x=±√74 cos(arctan(-6) )
→x=±√2
When x=-√2:
→(-√2)^2+y√2=14
→y√2=12
→y=6√2
When x=√2:
→(√2)^2-y√2=14
→y√2=-12
→y=-6√2
→{x,y}={-√2,6√2},{√2,-6√2}
Case 2: 5 cosθ-7 sinθ=0
→7 sinθ=5 cosθ
→tanθ=5/7
→θ=arctan(5/7)
→x=±√74 cos(arctan(5/7) )
→x=±7
When x=-7:
(-7)^2+7y=14
→7y=-35
→y=-5
When x=7:
(7)^2-7y=14
→7y=35
→y=5
→{x,y}={-7,-5},{5,7}
ooops. Not me alone is so clever, as it appears. Removing the comment wit my soution.
Nice!
I thought of yet another method which I did not pursue.
Set the two equations equal to each other...
x^2 - xy = y^2 + xy - 46. (since 14 = 60 - 46)
method 4:
multiply both equations
> xy(x-y)(x+y) = 840
> xy(x^2-y^2) = 840 -- (1)
add both equations:
> x^2-y^2 -2xy = -46
> x^2-y^2 = 2xy - 46 -- (2)
substitute (2) into (1)
xy ( 2xy - 46 ) = 840
let m=xy
> m(2m-46)=840 -> m^2 - 23m -420 = 0
> (m+12)(m-35) = 0
> m = -12 or m = 35
when m=-12, x^2 = 2 and y^2 = 72
when m=35, x^2 = 49 and y^2 = 25
with validation, it's easy to come to the 4 sets of answers :)
Wow! This is nice
Thank you so much for your shareing.... Master
7:00, which one? But at 7:33 its squared, so the sign would not matter this case.
I wonder why you didn't consider the following approach. Rewrite the equations as
x² = 14 + xy
y² = 60 − xy
Multiplying both equations this gives
x²y² = (14 + xy)(60 − xy)
Now let
xy = z
and we have
z² = (14 + z)(60 − z)
z² = 840 + 46z − z²
2z² − 46z − 840 = 0
z² − 23z − 420 = 0
(z − 35)(z + 12) = 0
z = 35 ⋁ z = −12
So we have xy = 35, which gives the solutions (x, y) = (7, 5) and (x, y) = (−7, −5) or xy = −12 which gives the solutions (x, y) = (√2, −6√2) and (x, y) = (−√2, 6√2).
Wow! Nice
X²+y²=14+60.=74=25+49.
{x, y}=(5, 7)
Please explain your comment: this is a homogenous system, therefore y = ux
If you take each term on the left hand side of each of the equations, the sums of the powers of x and y are constant. In this case, they are always 2. For example, x^2 is x^2 y^0, 2 + 0 = 2. xy = x^1 y^1, 1 + 1 = 2.
When that happens, you get surprisingly good results by putting in y = ux.