Question

Solve the following equations:
$x^2{y}^2{z}^{2}u=12,x^2{y}^{2}z{u}^2=8,z^{2}y{x}^2{u}^2=1,3x{y}^2{z}^2{u}^2=4$.

• A. $(2,5,\frac{1}{2},\frac{1}{5})$
• B. $(2,3,4,7)$
• C. $(3,4,\frac{1}{2},\frac{1}{3})$
• D. $(0,1,3,5)$

Answer 1

The correct option is C $(3,4,\frac{1}{2},\frac{1}{3})$

Let $x^2{y}^2{z}^{2}u=12$ .....(i)

and $x^2{y}^{2}z{u}^2=8$ .........(ii)

Dividing (i) by (ii), we get

$\frac{x^2{y}^2{z}^{2}u}{x^2{y}^{2}z{u}^2}=\frac{12}{8}$

$\Rightarrow u=\frac{2}{3}z$ ........(1)

$z^{2}y{x}^2{u}^2=1$ ........(iii)

$3x{y}^2{z}^2{u}^2=4$ ........(iv)

Dividing (ii) by (iv), we get

$\frac{x^2{y}^{2}z{u}^2}{3x{y}^2{z}^2{u}^2}=\frac{8}{4}$

$\Rightarrow x=6z$ .........(2)

Dividing (ii) by (iii), we get

$\frac{x^2{y}^{2}z{u}^2}{z^{2}y{x}^2{u}^2}=\frac{8}{1}$

$\Rightarrow y=8z$ .......(3)

Substituting values of (1), (2) and (3) in (i), we get

${\left(6z\right)}^2{\left(8z\right)}^2{z}^2\frac{2}{3}z=12$

For $z^7=\frac{1}{128}=\frac{1}{2^7}$

$\Rightarrow z=\frac{1}{2}$

For $x=6z$

$\Rightarrow x=3$

For $y=8z$

$\Rightarrow y=4$

For $u=\frac{2}{3}z$

$\Rightarrow u=\frac{1}{3}$

So, the complete solution is $x=3,y=4,z=\frac{1}{2},u=\frac{1}{3}$.