Question

Solve the following equations:
$x^3{y}^{2}z=12,x^{3}y{z}^3=54,x^7{y}^3{z}^2=72$.

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

Answer 1

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

Given, $x^3{y}^{2}z=12$ and $x^{3}y{z}^3=54$

Dividing both, we get

Therefore, $\frac{y}{z^2}=\frac{2}{9}$

$\Rightarrow y=\frac{2}{9}{z}^2$ ......(i)

$\Rightarrow \frac{y}{z^2}=\frac{2}{9}$

$\Rightarrow y=\frac{2}{9}{z}^2$ ......(i)

$\Rightarrow (x^3{y}^{2}z{)}^2={\left(12\right)}^2{x}^6{y}^2{z}^2=144$

Also given $x^7{y}^3{z}^2=72$

Dividing both, we get

$\frac{y}{x}=2$

$\Rightarrow x=\frac{y}{2}$

Substituting $y$ from (i), we have

$x=\frac{2}{9}{z}^2.\frac{1}{2}$

$\Rightarrow x=\frac{z^2}{9}$ .........(ii)

$x^3{y}^{2}z=12$

Substituting (i) and (ii), we have

${\left(\frac{z^2}{9}\right)}^3{\left(\frac{2}{9}{z}^2\right)}^{2}z=12$

$\Rightarrow z^6.z^4.z=3\left(9^3\right){\left(9\right)}^2$

$\Rightarrow z=3$

Put $z=3$ in (ii), we have

$x=1$

$\Rightarrow y=2x$

$\Rightarrow y=2$

So, $x=1,y=2,z=3$.