x=12-y
y(12-y)=32
12y-y^2-32=0
y^2-12y+32=0
D=12^2-4*32=144-128=16=4^2
y1=(12+4)/(2)=8
y2=(12-4)/(2)=4
x1=12-8=4
x2=12-4=8
Ответ: (4;8) (8;4)
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Answers & Comments
32=16*2
Подходит первое
x=12-y
y(12-y)=32
12y-y^2-32=0
y^2-12y+32=0
D=12^2-4*32=144-128=16=4^2
y1=(12+4)/(2)=8
y2=(12-4)/(2)=4
x1=12-8=4
x2=12-4=8
Ответ: (4;8) (8;4)