關於二元一次聯立方程的數!幫幫忙

{ x^2+xy=6.....(1)
2y+3x=-8.....(2)
By(2) :
y = (-8 - 3x)/2 , sub it to (1) :
x^2 + x(-8 - 3x)/2 = 6
2x^2 - 8x - 3x^2 = 12
x^2 + 8x + 12 = 0
(x + 2)(x + 6) = 0
x = - 2 ,
y = (-8 - 3(-2))/2 = - 1
or
x = - 6
y = (-8 - 3(-6))/2 = 5

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x^2+xy=6----(1)

2y+3x=-8----(2)



From(2)

3x=-2y-8

x=(-2y-8)/3---(3)



Sub (3) into (1)

[(-2y-8)/3]^2+(-2y-8)y/3=6

(-2y-8)^2/9+(-2y-8)y/3=6

(-2y-8)^2+3(-2y-8)y=6x9=54

(4y^2+32y+64)+3(-2y^2-8y)=54

4y^2+32y+64-6y^2-24y=54

-2y^2+8y+64-54=0

-2y^2+8y+10=0

y^2-4y-5=0

(y-5)(y+1)=0

y=5 or -1



when y=5

x=[-2(5)-8]/3

x=(-18)/3

x=-6



when y=-1
x=[-2(-1)-8]/3
x=-6/3
x=-2

The solution is (-2,-1) and (-6,5)

From <2>:
y= -4 -1.5x

Sub into <1>:
x^2 -4x - 1.5x^2 =6
x^2+8x+12=0
(x+2)(x+6)=0
x=-2 or x=-6

y= -1 or 5