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Y nya itu 2
2^(2x+2) + 2^y = 20
2^2 * (2^x)^2 + 2^y = 20
4 (2^x)^2 + 2^y = 20 ..... (1)
2^(x-1) - 2^y = -3 atau
(2^x)/2 - 2^y = -3 .... (2)
elimnasi persamaan (1) dan (2) :
4 (2^x)^2 + 2^y = 20
(2^x)/2 - 2^y = -3
--------------------------- (+)
4(2^x)^2 + (2^x)/2 = 17
kali 2 kedua ruas, get
8 (2^x)^2 + 2^x = 34
8 (2^x)^2 + 2^x - 34 = 0
(8(2^x) + 17)(2^x - 2) = 0
pembuat nol :
2^x - 2 = 0
2^x = 2
x = 1
substitusi x = 1 ke persamaan :
4(x+1) + 2^y = 20
4(1+1) + 2^y = 20
16 + 2^y = 20
2^y = 20 - 16
2^y = 4
y = 2
so, solusi untuk {x,y} = {1,2}