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Question

Solve the following system of linear equations in three variables.
x2y+z=0
9x8y+3z=0
2x+3y+5z=36


Your Answer
A
x=1,y=2,z=3
Correct Answer
B
x=1, y=3, z=5
Your Answer
C
x=6,y=18,7=30
Your Answer
D
x=10,y=30,7=18

Solution

The correct option is B x=1, y=3, z=5
x2y+z=0......(1)
9x8y+3z=0.....(2)
2x+3y+5z=36....(3)

eqn(1)×3eqn(2)

3x6y+3z=0
9x+8y3z=0
__________________
6x+2y=0
3x+y=0.....................(4)

eqn(1)×5eqn(3)

5x10y+5z=0
2x3y5z=36
__________________
3x13y=36
3x+y=0.....................(5)

eqn(4) + eqn(5) gives,

3x+y=0
3x13y=36
_________________
12y=36
y=3......................(6)

Substituting eqn(6) in eqn(4), we get,
3x+3=0
3x=3
x=1.............(7)

Substituting eqn(6) and eqn(7) in eqn(1). we get,

12(3)+z=0
16+z=0
5+z=0
z=5.............(8)

Check:
Substituting the values of x, y and z in eqn(3), we get
LHS = 2x + 3y + 5z 
         = 2(1) + 3(3) + 5(5)
         = 2 + 9 + 25
         = 36
         = RHS

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