Question

Rezolvati:
28:(x+1)=3 rest 1
79:(2xZ+1)=11 rest 2
368:(4xZ+1)=40 rest 8
(3xZ+33):4=32 rest 1
(7xZ+8):5=29 rest 3

Answer 1

28:(x+1)=3 rest 1
folosim teorema impartirii cu rest
28=3(x+1)+1
28=3x+3+1
-3x=4-28
-3x=-24    inmultim -1
x=24/3
x=8

79:(2xZ+1)=11 rest 2
79=11(2z+1)+2
79=22z+11+2
-22z=13-79
-22z=-66     ·(-1)
x=66/22
z=3

368:(4xZ+1)=40 rest 8
368=40(4z+1)+8
368=160z+40+8
-160z=48-368
-160z=-320    ·(-1)
z=360/160
z=2

(3xZ+33):4=32 rest 1
3z+33=32·4+1
3z=128+1-33
3z=96
z=96/3
z=32

(7xZ+8):5=29 rest 3
7z+8=29·5+3
7z=145+3-8
7z=140
z=140/7
z=20