若y1=x-1,y2=2x+1,且y1-3y2=0,則x=_____,y1+y2=_____怎麼算.

若y1=x-1,y2=2x+1,且y1-3y2=0,則x=_____,y1+y2=_____怎麼算.


y1-3y2=0
則x-1-3(2x+1)=0
x-1-6x-3=0
5x=-4
x=-4/5
y1+y2=x-1+2x+1=3x=-12/5



整式化簡1/3(9y-3)+2(y+1)


1/3×(9y-3)+2(y+1)
=3y-1+2y+2
=5y+1



(2x+1)(2x-1)-(3-2x)(2x-3),其中x=2,先化簡再求值


(2x+1)(2x-1)-(3-2x)(2x-3)
=(2x)^2-1^2+(2x-3)^2
=4x^2-1+(4x^2-12x+9)
=8x^2-12x+8
把x=2代入
原式=8*4-24+8=16