Question

Solve:
$x+6y=5z$
$7x+z=6y$
$5x+6y-4z=24$

• A. $x=2$
  $y=8$
  $z=5$
• B. $x=4$
  $y=6$
  $z=8$
• C. $x=14$
  $y=6$
  $z=8$
• D. $x=14$
  $y=26$
  $z=18$

Answer 1

The correct option is B $x=4$

$y=6$
$z=8$
$x+6y=5z.........\left(1\right)$
$7x+z=6y.........\left(2\right)$
$5x+6y-4z=\mathrm{24.....}\left(3\right)$

From (1), $x+6y-5z=0$
From (2), $7x+z-6y=0$

Hence,
$\frac{x}{6\times 1-(-6).(-5)}=\frac{y}{(-5\times 7)-(1\times 1)}=\frac{z}{1.(-6)-7.\left(6\right)}$

$\Rightarrow \frac{x}{6-30}=\frac{y}{-35-1}=\frac{z}{-6-42}$

$\Rightarrow \frac{x}{-24}=\frac{y}{-36}=\frac{z}{-48}$

$\Rightarrow \frac{x}{2}=\frac{y}{3}=\frac{z}{4}$
[Multiplying each fraction by -12]

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

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

$\Rightarrow x=2k,y=3k,z=4k......\left(A\right)$

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

$\Rightarrow 5x+6y-4z=24$
$\Rightarrow 5\left(2k\right)+6\left(3k\right)-4\left(4k\right)=24$
$\Rightarrow k(10+18-16)=24$
$\Rightarrow 12k=24$
$\therefore k=2$

Hence, from (A), we get,

$x=4$
$y=6$
$z=8$