Question

Solve the following system of linear equation in three variables.
5x + 6y + 8z = 0
3x + 4y + 6z = 0
3x + 5y + 16z = 7

• A. x=2 , y= 4 and z=1
• B. x=2 , y= -3 and z=2
• C. x=2 , y= -3 and z=1
• D. x=3 , y= -3 and z=1

Answer 1

The correct option is C x=2 , y= -3 and z=1

We have,
5x + 6y + 8z = 0.............(1)
3x + 4y + 6z = 0..............(2)
3x + 5y + 16z = 7.............(3)

Multiplying equation (1) by 3 and equation (2) by 5, we get

15x+18y+24z=0 ................(4)
15x+20y +30z =0 ..............(5)

Subtracting equation (5) from equation (4) , we get

-2y-6z =0

$\Rightarrow$ y+3z =0 ............(6)

Subtracting equation (3) from equation (2) , we get

-y -10z = -7 ....................(7)

Adding equations (6) and (7), we get

-7z = -7
$\Rightarrow$ z= 1

Putting z=1 in the equation (6), we get

y + 3(1) =0

$\Rightarrow$ y= -3

Putting z=1 and y= -3 in the equation (1), we get

5x+ 6(-3) +8(1) =0

$\Rightarrow$ 5x -18 +8 =0

$\Rightarrow$ 5x=10

$\Rightarrow$ x=2

Thus, we have x=2 , y= -3 and z=1

Check/Verification :

Substituting the values of x,y and z in the equation (3), we get

LHS = 3(2) +5 (-3) +16 (1)
= 6-15+16
= 7
=RHS