Question

Solve:
$2x-3y+4z=0$
$7x+2y-6z=0$
$4x+3y+z=37$

• A. $x=2$
  $y=9$
  $z=15$
• B. $x=4$
  $y=12$
  $z=10$
• C. $x=2$
  $y=8$
  $z=5$
• D. $x=5$
  $y=2$
  $z=8$

Answer 1

The correct option is C $x=2$

$y=8$
$z=5$
$2x-3y+4z=\mathrm{0....}\left(1\right)$
$7x+2y-6z=\mathrm{0....}\left(2\right)$
$4x+3y+z=\mathrm{37....}\left(3\right)$

From (1) and (2), we get,

$\frac{x}{(-3)(-6)-(2\times 4)}=\frac{y}{4\times 7-(-6)\left(2\right)}=\frac{z}{2\times 2-7.(-3)}$

$\Rightarrow \frac{x}{10}=\frac{y}{40}=\frac{z}{25}$

$\Rightarrow \frac{x}{2}=\frac{y}{8}=\frac{z}{5}$

Let $k$ denote the common value of these fractions.

$\Rightarrow \frac{x}{2}=\frac{y}{8}=\frac{z}{5}=k$

$\Rightarrow x=2k,y=8k,z=5k......\left(A\right)$

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

$\Rightarrow 4x+3y+z=37$
$\Rightarrow 4\left(2k\right)+3\left(8k\right)+\left(5k\right)=37$
$\Rightarrow k(8+24+5)=37$
$\Rightarrow 37k=37$
$\therefore k=1$

Hence, from (A), we get,

$x=2$
$y=8$
$z=5$