math
posted by motlhalefi .
x3y=1
x22xy+9y2=17 solve simalteneously

from the 1st > x = 3y+1
(3y+1)^2  2y(3y+1) + 9y^2 = 17
9y^2 + 6y + 1  6y^2  2y + 9y^2 = 17
12y^2 + 4y 16 = 0
3y^2 + y  4 = 0
(3y + 4)(y  1) = 0
y = 4/3 or y = 1
if x = 4/3, y = 3(4/3) + 1 = 3
if x = 1 , y = 3(1) + 1 = 4
btw, it is "simultaneously" 
You probably figured it out yourself by at the end it should have said ...
if y = 4/3, x = 3(4/3) + 1 = 3
if y = 1 , x = 3(1) + 1 = 4
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