⎩ ⎨ ⎧ 3 x + 3 y = 1 4 x + 4 y = 1
How many real pairs of ( x , y ) satisfy the system above?
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@Gurīdo Cuong It's necessary to specify that x , y must be real in order to claim that t , u ≥ 0
Well,i discovered that for the equations to be true, x and y must be greater than or equal tozero. So by mere observation, (1,0) and (0,1) seem to be the only valid entry.
How can we confirm that those are the only solutions?
A mere observation may be useful in this problem
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Set t = 1 2 x and u = 1 2 y and we'll get the following system { t 4 + u 4 = 1 t 3 + u 3 = 1 ⇔ { t 4 + u 4 = t 3 + u 3 t 3 + u 3 = 1 Considering the first equation t 4 + u 4 = t 3 + u 3 t 3 ( t − 1 ) = u 3 ( 1 − u ) ( ∗ ) Since t , u ≥ 0 and t 3 + u 3 = 1 , we get t , u ∈ [ 0 ; 1 ] . So t 3 ( t − 1 ) ≤ 0 and u 3 ( 1 − u ) ≥ 0 , therefore ( ∗ ) happens when t 3 ( t − 1 ) = u 3 ( 1 − u ) = 0 ⇒ { t 3 ( t − 1 ) = u 3 ( 1 − u ) = 0 t 3 + u 3 = 1 ⇔ ( t , u ) = ( 1 ; 0 ) a n d ( 0 ; 1 ) So finally ( x , y ) = ( 1 ; 0 ) , ( 0 ; 1 ) , 2 is the correct answer