maths(解方程 二 : 代數方法)

{ y = x2 + x - 1 ... (1)
{ y = 2 - x ... (2)
代(2)入(1) => 2 - x = x2 + x -1
x2 + 2x - 3 = 0
(x + 3)(x - 1) = 0
{x = 1 => y = 1
{x = -3 => y = 5
{ 2/x - y = 1 ... (1)
{ y - 4x = 1 ... (2)
(1) + (2) => 2/x - 4x = 2
2 - 4x2 = 2x
4x2 + 2x - 2 = 0
2x2 + x - 1 = 0
(2x - 1)(x + 1) = 0
x = 1/2或x = -1
利用(2) x = 1/2 => y = 1 + 4(1/2) = 3
x = -1 => y = 1 + 4(-1) = -3
{ y = x2 + x - 1 ... (1)
{ y = 3x - 1 ... (2)
代(2)入(1) => 3x - 1 = x2 + x - 1
x2 - 2x = 0
x(x - 2) = 0
x = 0 或x = 2
利用(2) x= 0 => y = -1
x = 2 => y = 5
{ x + 2y = -2 ... (1)
{ x2 + 4y2 = 2 ... (2)
(1) => x = -2 - 2y
代入(2) (-2 - 2y)2 + 4y2 = 2
4 + 8y + 4y2 + 4y2 = 2
8y2 + 8y + 2 = 0
4y2 + 4y + 1 = 0
(2y + 1)2 = 0
y = -1/2 => x = -2 - 2(-1/2) = -1
3. - xy = x - 2y = 1
x - 2y = 1 ... (1)
- xy = 1 => y = -1/x ...(2)
代入(1), x - 2(-1/x) = 1
x2 + 2 = x
x2 - x + 2 = 0
x = [ 1 +/- √(1 - 8) ] / 2
x = 1/2 + (√7/2)i 或x = 1/2 - (√7/2)i
利用(1) => x = 1/2 + (√7/2)i => 1/2 + (√7/2)i - 2y = 1 => y = -1/4 + (√7/4)i
x = 1/2 - (√7/2)i => 1/2 - (√7/2)i - 2y = 1 => y = -1/4 - (√7/4)i
4. { x + 4y = 2/y ... (1)
{ x + 2y = 3 ... (2)
(1) - (2) => 2y = 2/y - 3
2y2 = 2 - 3y
2y2 + 3y - 2 = 0
(2y - 1)(y + 2) = 0
y = 1/2 y = -2
利用(2)
y = 1/2 => x + 2(1/2) = 3 => x = 2
y = -2 => x + 2(-2) = 3 => x = 7