中四math題=.=

我let上面係 @,下面係#啦
From#,X=3y/2 + 2 ------- !
SUB ! to @
2 (3y/2 +2 )^2 -3y(3y/2 +2) +2y^2 -12 =0
2(9y^2/4+6y+4) - 9y^2/2 -6y +2y^2 -12=0
2y^2 +6y-4 =0
(y+4)(y-1)=0
y=1 or -4
Sub Y= 1 or -4 into #
-2x +3(1)+4=0 and -2x+3(-4)+4=0
x=3.5 and x= -2
希望你睇得明- -

Given -2x + 3y +4 =0
Then x=3y/2 + 2
Substitute x=3y/2 + 2 ino 2x^2 - 3xy - 2y^2 - 12=0
Then 2 (3y/2 +2 )^2 -3(3y/2 +2)y +2y^2 -12 =0
2(9y^2/4+6y+4) - 9y^2/2 -6y +2y^2 -12=0
9y^2/2+12y+8 - 9y^2/2 -6y +2y^2 -12=0
2y^2 +6y-4 =0
y^2+3y-4=0
(y+4)(y-1)=0
y=1 or y=-4
Given x=3y/2 + 2
When y=1, x=3.5
When y=-4, x=-4
The solution is x,y = 3.5, 1 or x,y = -4, -4.
Remark: Actually this is a geometry question. The first equation is a U-shape curve while the second equation is a straight line. The solution (x,y)= (3.5, 1) or (-4, -4) is the the intercept points where the lines cross each other.