"How many" all day

Algebra Level 4

{ x 5 y 3 = 3125 72 3 x + 2 y = 7 \large\begin{cases} x^5y^3=\frac{3125}{72}\\ 3x+2y=7\end{cases} How many real pairs of ( x , y ) (x,y) satisfy the system above?

More than 2 1 2 None

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1 solution

P C
Mar 22, 2016

Use AM-GM for the second equation 7 = 5. 3 5 x + 3. 2 3 y 8 3 2 . 2 3 5 5 x 5 y 3 8 = 8 7= 5.\frac{3}{5}x+3.\frac{2}{3}y\geq 8\sqrt[8]{\frac{3^2.2^3}{5^5}x^5y^3}=8 As can be seen, the inequality above is wrong so N o n e None is the correct answer

*For anyone that questioned: " x , y x,y aren't positive so we can't use AM-GM". In the first equation, notice that x , y x,y have odd power, and R H S > 0 RHS>0 , so x , y x,y must be both positive or both negative. If they're negatives, then the second equation'll be incorrect since L H S < 0 LHS<0 and R H S > 0 RHS>0 . Therefore they must be positive

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