Pretty smooth. I took (2), isolated the radical and squared both sides. On the right is just xy, but it's cancelled by half the 2xy on the left. Equation (1) neatly subs into the left side, and we wind up with 133 + 49 = 14(x+y), so x + y = 13. Subbing that into (2) we see xy = 36. Not too many to check, x and y are 4 and 9 WLOG.
Let's u=x+y, v=xy Then u²-v=133 u-√v=7 v= (u-7)²=u²-14u+49 u²-u²+14u-49=133 u=182/14=13 v=(13-7)²=6²=36 By the Viete's Theoreme {x,y} are the roots of the quadratic equation z²-13z+36=0 {x,y}=(13±√(13²-4•36))/2= (13±√(169-144))/2= (13±√(25))/2= (13±5)/2= {9;4} {x,y}={9;4},{4;9}
Pretty smooth. I took (2), isolated the radical and squared both sides. On the right is just xy, but it's cancelled by half the 2xy on the left. Equation (1) neatly subs into the left side, and we wind up with 133 + 49 = 14(x+y), so x + y = 13. Subbing that into (2) we see xy = 36. Not too many to check, x and y are 4 and 9 WLOG.
(X,Y):(4,9),(9,4)
Let's u=x+y, v=xy
Then
u²-v=133
u-√v=7
v= (u-7)²=u²-14u+49
u²-u²+14u-49=133
u=182/14=13
v=(13-7)²=6²=36
By the Viete's Theoreme {x,y} are the roots of the quadratic equation
z²-13z+36=0
{x,y}=(13±√(13²-4•36))/2=
(13±√(169-144))/2=
(13±√(25))/2=
(13±5)/2=
{9;4}
{x,y}={9;4},{4;9}
Please I neee the solution:
Solve:
√(x³+17)=2x-5+√(x³+8)