Question

Solve the following system of linear equations in three variables.

y + z - 7x = -16
z + x - 7y = -24
x + y - 7z = -40

• A. x = 5, y = 5, z = 7
• B. x = 4, y = 5, z = 7
• C. x = 4, y = 5, z = 9
• D. x = 4, y = 6, z = 7

Answer 1

The correct option is B x = 4, y = 5, z = 7

$y+z-7x=-\mathrm{16.........}\left(1\right)z+x-7y=-\mathrm{24.........}\left(2\right)x+y-7z=-\mathrm{40.........}\left(3\right)$

$eqn\left(1\right)-eqn\left(2\right)z+y-7x=-16-z+7y-x=+24$
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$8y-8x=8y-x=\mathrm{1.............}\left(4\right)$

$eqn\left(1\right)\times 7+eqn\left(3\right)7y+7z-49x=-112y-7z+y=-40$
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$8y-48x=-152y-6x=-\mathrm{19.............}\left(5\right)$

$eqn\left(4\right)-eqn\left(5\right)y-x=1-y+6x=+19$
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$5x=20x=\mathrm{4.............}\left(6\right)$

Substituting eqn(6) in eqn(4), gives,

$y-4=1y=\mathrm{5.............}\left(7\right)$

Substituting eqn(6) and eqn(7) in eqn (1), gives,
$5+z-7\left(4\right)=-165+z-28=-16z-23=-16z=23-16z=7$

Verification:
Substituting the values of x,y and z in eqn(2), we get,

$LHS=7+4-7\left(5\right)=11-35=-24=RHS$