Question

Solve the following system of linear equations in three variables.
$x-2y+z=0$
$9x-8y+3z=0$
$2x+3y+5z=36$

• A. $x=1,y=2,z=3$
• B. $x=1,y=3,z=5$
• C. $x=6,y=18,7=30$
• D. $x=10,y=30,7=18$

Answer 1

The correct option is B $x=1,y=3,z=5$

$x-2y+z=0......\left(1\right)$
$9x-8y+3z=0.....\left(2\right)$
$2x+3y+5z=36....\left(3\right)$

$\Rightarrow eqn\left(1\right)\times 3-eqn\left(2\right)$

$3x-6y+3z=0$
$-9x+8y-3z=0$
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$-6x+2y=0$
$-3x+y=\mathrm{0.....................}\left(4\right)$

$\Rightarrow eqn\left(1\right)\times 5-eqn\left(3\right)$

$5x-10y+5z=0$
$-2x-3y-5z=-36$
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$3x-13y=-36$
$-3x+y=\mathrm{0.....................}\left(5\right)$

eqn(4) + eqn(5) gives,

$-3x+y=0$
$3x-13y=-36$
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$-12y=-36$
$y=\mathrm{3......................}\left(6\right)$

Substituting eqn(6) in eqn(4), we get,
$-3x+3=0$
$-3x=-3$
$x=\mathrm{1.............}\left(7\right)$

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

$1-2\left(3\right)+z=0$
$1-6+z=0$
$-5+z=0$
$z=\mathrm{5.............}\left(8\right)$

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