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Question

Solve the following simultaneous linear equations by method of reduction.
x+2y+z=8
2x+3yz=11
3xy2z=5

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Solution

x+2y+z=8.....(1)
2x+3yz=11.....(2)
3xy2z=5.....(3)
Adding eqn(1)&(2), we have
(x+2y+z)+(2x+3yz)=8+11
3x+5y=19.....(4)
Subtracting equation (2) from (3), we have
(3xy2z)(2x+3yz)=511
x4yz=6.....(5)
Now adding equation (1)&(5), we have
(x+2y+z)+(x4yz)=86
2x2y=2
xy=1.....(6)
Multiplying eqn(6) by 5, we get
5x5y=5.....(7)
Adding eqn(4)&(7), we have
(3x+5y)+(5x5y)=19+5
8x=24
x=3
Substituting the value of x in eqn(6), we get
3y=1
y=31=2
Substituting the value of x aand y in eqn(1), we get
3+(2×2)+z=8
z=87=1
Hence x=3,y=2,z=1.

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