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If x(x+y+z)=9,y(x+y+z)=16 and z(x+y+z)=144, then x=?
Read Solution (Total 10)
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- x=9/13,-9/13.
sol:-
adding all equations after that we will get
(x+y+z)^2=169
x+y+z=13,-13
after putting these values in eq-n number one we get
x=-9/13,9/13 ans. - 12 years agoHelpfull: Yes(20) No(2)
- let x+y+z=k
x(x+y+z)=9 can be written as xk=9
similarly yk=16
also zk=144
---------------------------------------
add these k(x+y+z)=169
or k.k=169 (bcoz x+y+z=k assumed)
ie k^2=13^2
k=+13,-13
hence from the first eqn x(x+y+z)=9
x.k=9
x=9/k
or x=9/13 or -9/13 - 12 years agoHelpfull: Yes(15) No(0)
- x=9/13
y=16/13
z=144/13
reason: x+y+z=9/x=16/y=144/z
x:y:z=9:16:144
x(x+16x/9+144x/9)=9
solving x=9/13
so y=16/13 and z=144/13 - 12 years agoHelpfull: Yes(3) No(9)
- x(x+y+z)=9 equation named as 1
y(x+y+z)=16 equation named as 2
z(x+y+z)=144 equation named as 3
let (x+y+z) as A
then sum of this equtions 1,2 and 3 becomes
(A)*A=169
A^2=169 then A=13,
from equation 1
x*A=9
therfore x=(9/A)
ans is x=(9/13).
- 12 years agoHelpfull: Yes(3) No(0)
- x=9/x+y+z..............>(1)
y=16/x+y+z
2=144/x+y+z
x+y+z=(9/x+y+z)+(16/x+y+z)+(144/x+y+z)
x+y+z=169/x+y+z
(x+y+z)2=13*13
x+y+z=13
from 1
x=9/13 - 11 years agoHelpfull: Yes(2) No(0)
- Answer is 9/13.
because if we add all these equations then,
x(x+y+z)+y(x+y+z)+z(x+y+z)=9+16+144
& (x+y+z)(x+y+z) = 169
(x+y+z) = 13
therefore x=9/13 - 11 years agoHelpfull: Yes(0) No(0)
- x=9/13
y=16/13
z=144/13
adding all three equation, we get,
(x+y+z)(x+y+z)=169
(x+y+z)sq 2=13 sq 2
(x+y+z)=13
- 11 years agoHelpfull: Yes(0) No(0)
- x(x+y+z)+y(x+y+z)+z(x+y+z)=9+16+144
now, taking common (x+y+z) in L.H.S
(x+y+z)(x+y+z)=169
(x+y+z)=+-13
x=+9/13,-9/13 - 11 years agoHelpfull: Yes(0) No(0)
- add all three equations
we get
(x+y+z)^2=169
x+y+z=13
put it in equation 1
x=9/13
- 10 years agoHelpfull: Yes(0) No(0)
- x= 9/13, just divide then replace the value of y and z in terms of x
- 8 years agoHelpfull: Yes(0) No(0)
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