二元一次方程 {x+y=5,2y-3(x+y)=-11 五分之3x+1 -四分之3y+2=0 五分之2x—1+四分之3y-2=2

2024-12-24 18:19:23
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回答1:

x+y=5,2y-3(x+y)=-11

x=5-y;2y-3(5-y+y)=-11;2y-15=-11;2y=4;y=2
x=2;y=3.

(3x+1)/5-(3y+2)/4=0,(2x-1)/5+(3y-2)/4=2
4(3x+1)-5(3y+2)=0,4(2x-1)+5(3y-2)=40
12x-15y=6,8x+15y=54
y=2
x=3