(x+i)(3-iy)= 1+13i
(3x+y)+(3-xy)i=1+13i
(1) 3x+y=1,
(2) 3-xy=13
(2) 3-xy=13, xy=-10, y=-10/x
(1) 3x+y=1, 3x+(-10/x)=1, 3x^2-x-10=0, x=2, x=-5/3
then
(x,y)=(2,-5)
or
(x,y)=(-5/3, 6)
(3x+y)+(3-xy)i=1+13i
(1) 3x+y=1,
(2) 3-xy=13
(2) 3-xy=13, xy=-10, y=-10/x
(1) 3x+y=1, 3x+(-10/x)=1, 3x^2-x-10=0, x=2, x=-5/3
then
(x,y)=(2,-5)
or
(x,y)=(-5/3, 6)
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