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1
2z + 5 - xz
z = -4 x-5 y+2
2nd z=17 1st Z=22 Easy as ABC: Just do the following 39-5=___ ___divide by 2=z z+5=Z
It is: 5(z+5) = 5z+25
6z-4z-z = 6-7-4 z = -5
9z
Z= -1
15z+5+6z=0 21z+5=0 z=-5/21
5 you dummy
int i, x, y, z; x=1; y=5; z=1; for (i=0; i<5; ++i) { printf ("%d %d ", x, y); if (x==3) z=z*(-1); x=x+z; y=y-z; } Output: 1 5 2 4 3 3 4 2 5 1
X(3+) and Z(2-) will give X2Z3. X(3+) and Z(3-) will give XZ. X(5+) and Z(2-) will give X2Z5. X(5+) and Z(3-) will give X3Z5.