(x^2/9)+(y^2)=1
x-3y= -3
2007-05-23 17:34:52 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x = -3+3y
(-3+3y)^2/9 +y^2 = 1
9(y-1)^2/9 + y^2 = 1
y^2- 2y +1 +y^2 = 1 => 2y^2 - 2y = 0
y = 0 & (y - 1)= 0 => y= 1
so , sol are (-3, 0) and (0, 1)
2007-05-23 17:44:44 · answer #1 · answered by me_poori 2 · 0⤊ 0⤋
x = 3y - 3
(3y - 3)² / 9 + y² = 1
(3y - 3)² + 9y² = 9
9y² - 18y + 9 + 9y² = 9
18y² - 18 y = 0
18y.(y - 1) = 0
y = 0 , y = 1
x = - 3 , x = 0
(- 3, 0) and (0 , 1) are solutions.
2007-05-24 09:27:07 · answer #2 · answered by Como 7 · 0⤊ 0⤋
(x^2/9) + (y^2) = 1
x - 3y = - 3
x = 3(y - 1)
(y - 1)^2 + y^2 = 1
2y^2 - 2y = 0
y(y - 1) = 0
y = 0, 1
x = - 3, 0
2007-05-24 00:48:10 · answer #3 · answered by Helmut 7 · 0⤊ 0⤋
You have to substitute in "z".
2007-05-24 00:41:05 · answer #4 · answered by Anonymous · 0⤊ 0⤋