solve the system of questions.

1/x-1/y+1/z=4

2/x+2/y+1/z=2

3/x+1/y-2/z=-2

Let a= 1/x , b= 1/y, c=1/z.

then,

a-b+c = 4 ...[1]

2a+2b+c=2......[2]

3a+b-2c = -2......................[3]

Solve these eqs. and u will get the values of a,b,c.

Then x= 1/a y=1/b z=1/c.

put 1/x = a , 1/y = b and 1/z = c

the eqns are :1) a-b+c= 4 , 2)2a+2b+c=2, 3)3a+b-2c= -2

consider 1) and 2)             a-b+c = 4

                                     2a+2b+c = 2

                                       subtract the two :    a + 3b = -2     --------- 4)

Consider 1) and 3)                 (   a-b+c=4   ) * 2                   2a-2b+2c = 8

                                              3a+b-2c = -2                        3a+b-2c = -2

                                                                            add the two :     5a-b = 6 --------- 5)

consider 4) and 5)             a + 3b = -2

                                     ( 5a - b = 6) * 3 

                     add the two : 16a = 16 which implies that a =1  but 1/x = a which implies x = 1/a or x=1/1 or x=1

a + 3b = -2  put a =1 in this eqn and you get b = -1  but 1/y = b which implies y = 1/b or y = 1/-1 or y=-1

consider the eqn 1) ie; a-b+c=4 put a=1 and b=-1   you get:  1+1+c=4 which implies c=2 or z=1/2

hence the values of x,y,and z are : 1,-1 and 1/2

alternative: you can also solve this by using cramer's rule.(if u know it)

Take, 1/x = p, 1/y = q, 1/z= r, so it bms,

p - q + r = 4

2p + 2q + r = 2

3p + q - 2r = -2

By solving these eqns. u'll get values of p.q.r & then put x =1/p, y = 1/q, z = 1/r.