Solve the system of equations: { ( x − 1 ) x − y 2...

Blericker74

Blericker74

Answered

2022-07-14

Solve the system of equations:
{ ( x 1 ) x y 2 = y ( x 2 y + 1 ) y x 1 + 3 x y 2 = 2 x + y 1
This is my try
Condition: x 1 ; x y 2
We have:
x 2 y + 1 x 2 | y | + 1 x ( y 2 + 1 ) + 1 = x y 2 0 y 0
Applying AM-GM inequality, we have:
2 ( 2 x + y 1 ) = 2 y x 1 + 6 x y 2 ( y 2 + x 1 ) + 3 ( 1 + x y 2 ) = 4 x 2 y 2 + 2
y 2 + y 2 0 0 y 1
Squaring both side of first equation, we get:
( x 2 y 2 + 2 y 1 ) ( x 2 x + 2 y 2 2 x y ) = 0
If x 2 y 2 + 2 y 1 = 0, we get 1 2 x 1. Thus, we get x = 1 and y = 0 or y = 1. Two pairs of ( x , y ) don't satisfy the system of equations.
If x 2 x + 2 y 2 2 x y = 0...

Answer & Explanation

owerswogsnz

owerswogsnz

Expert

2022-07-15Added 12 answers

since you have
x 2 x + 2 y 2 2 x y = 0 x y 2 = x 2 2 x y + y 2 = ( x y ) 2
Note x > y,so have
x y 2 = x y
take the second equation we have
y x 1 + 3 ( x y ) = 2 x + y 1
so we have
y = x + 1 4 x 1 [ 0 , 1 ] 1 x 2
take
x 2 x + 2 y 2 2 x y = 0 x 2 x + 2 ( x + 1 4 x 1 ) 2 2 x x + 1 4 x 1 = 0 , 1 x 2
x 3 6 x 1 x 2 + 8 x 2 + 10 x 1 x 19 x + 2 ( x 1 4 ) 2 = 0
take x 1 = t,then we have
( t 2 + 1 ) 3 6 t ( t 2 + 1 ) 2 + 8 ( t 2 + 1 ) 2 + 10 t ( t 2 + 1 ) 19 ( t 2 + 1 ) + 2 = 0
( t 1 ) ( t 5 5 t 4 + 6 t 3 + 4 t 2 + 4 t + 8 ) = 0 , 0 t 1
since
f ( t ) = t 5 5 t 4 + 6 t 3 + 4 t 2 + 4 t + 8 = t 5 + 6 t 3 + 4 t 2 + 4 t + ( 8 5 t 4 ) > 0 , 0 t 1
so t = 1 then you have x = 2

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