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"
x-3y=1
x2-2xy+9y2=17 solve simalteneously
3 answers
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
if y = -4/3, x = 3(-4/3) + 1 = -3
if y = 1 , x = 3(1) + 1 = 4
I'm confused