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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.
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