Question

# Solve the following system of linear equations in three variables. x−2y+z=0 9x−8y+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=5x−2y+z=0......(1) 9x−8y+3z=0.....(2) 2x+3y+5z=36....(3) ⇒eqn(1)×3−eqn(2) 3x−6y+3z=0 −9x+8y−3z=0 __________________ −6x+2y=0 −3x+y=0.....................(4) ⇒eqn(1)×5−eqn(3) 5x−10y+5z=0 −2x−3y−5z=−36 __________________ 3x−13y=−36 −3x+y=0.....................(5) eqn(4) + eqn(5) gives, −3x+y=0 3x−13y=−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, 1−2(3)+z=0 1−6+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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