A System With Radicals

[@hieunguyencong8584]

that's good!😃

[@MrGeorge1896]

Given that we are looking for solutions in the real world x and y can not be negative and so sqrt(x) + sqrt(y + 1) = 1 can only be true if x and y are zero as otherwise the sum would be greater than one.

[@vacuumcarexpo]

＜3rd method＞ Let x=(sinh α)^2 and y=(sinh β)^2(α,β≧0), the given system of equations can be writren as follows: sinh α+cosh β=1 cosh α+sinh β=1 ⇔e^α+e^β=2 and e^(-α)-e^(-β)=0 ⇒α=β and e^α=e^β=1 ⇒α=β=0 ⇒x=y=0

[@DergaZuul]

Easier subtract equations sqrt(x)-sqrt(y) + sqrt(y+1)-sqrt(x+1)=0 or sqrt(x+1)-sqrt(x) = sqrt(y+1) -sqrt(y) or 1/(sqrt(x+1)+sqrt(x)) = 1/(sqrt(y+1)+sqrt(y) sqrt(x+1) +sqrt(x) = sqrt(y+1) + sqrt(y) so add to first one 2sqrt(x+1) = 2sqrt(y+1) hence x=y>=0 multiply both sides by sqrt(x+1) -sqrt(x) so we get x+1-x=sqrt(x+1)-sqrt(x) hence sqrt(x+1)-sqrt(x)=1=sqrt(x+1)+sqrt(x) sqrt(x)=-sqrt(x) so x=0 and y=0. No need to square or use derivatives just few simple steps

[@phill3986]

😎👍😎👍

[@msmbpc24]

X=0 and Y=0

[@SidneiMV]

√x + √(y + 1) = 1 √(x + 1) + √y = 1 x + y + 1 + 2√(xy + x) = 1 x + y + 1 + 2√(xy + y) = 1 x + y + 2√(xy + x) = 0 x + y + 2√(xy + y) = 0 xy + x = xy + y => x = y x + y + 2√(xy + x) = 0 2x + 2√(x² + x) = 0 √(x² + x) = -x => x ≤ 0 x² + x = x² => *x = 0 ∧ y = 0*