Răspuns:
a) 2(y+3)-4(x+1)=8 2y+3-4x-4-8=0
-4x+2y-11=0 /(-1) 4x-2y+11=0
b) 3(x+2y)-2(2x+y)=5 3x+6y-4x-2y-5=0
-2x+4y-5=0 /(-1) 2x-4y+5=0
c) 2(x+y+1)-3(y+2x+1)=0 2x+2y+2-3y-6x-3=0
-4x-y-1=0 /(-1) 4x+y+1=0
d) 5(x-y+1)=4(x+y+2) 5x-5y+5=4x+4y+8
x-9y-3=0
e) [tex]\frac{x+3}{2} +\frac{y-4}{3} =\frac{x+3y+1}{6}[/tex] /·6 3(x+3)+2(y-4)=x+3y+1
3x+9+2y-8-x-3y-1=0 2x-y=0
f) [tex]\frac{x+y-1}{3} +\frac{x-y+2}{2}=1\frac{1}{6}[/tex] [tex]\frac{x+y-1}{3} +\frac{x-y+2}{2}=\frac{7}{6}[/tex] /·6
2(x+y-1)+3(x-y+2)-7=0 2x+2y-2+3x-3y+6-7=0
x-y-3=0
Explicație pas cu pas:
