Question

Solve the following equations :
$x+y-4xy=0,y+z-6yz=0,z+x-8zx=0.$

• A. $X=\frac{1}{3}$,Y=1,Z=$\frac{1}{5}$
• B. $X=\frac{1}{2}$,Y=0,Z=$\frac{1}{2}$
• C. $X=\frac{1}{3}$,Y=0,Z=$\frac{1}{5}$
• D. $X=\frac{1}{3}$,Y=1,Z=$\frac{1}{0}$

Answer 1

The correct option is A $X=\frac{1}{3}$,Y=1,Z=$\frac{1}{5}$

$x+y-4xy=0y+z-6yz=0z+x-8zx=0.$

We can write

$x+y-4xy=0\Rightarrow x+y=4xy$

Now dividing the Equation with $xy$

We get, $\frac{1}{x}+\frac{1}{y}=\mathrm{4...................}\left(1\right)$

Similarly, $\frac{1}{y}+\frac{1}{z}=\mathrm{6...............}\left(2\right)$

and $\frac{1}{z}+\frac{1}{x}=\mathrm{8.................}\left(3\right)$

Now , $\left[\right(1)+(3\left)\right]-\left(2\right)$

we get, $\frac{1}{x}+\frac{1}{y}-\frac{1}{y}-\frac{1}{z}+\frac{1}{z}+\frac{1}{x}=4-6+8\Rightarrow \frac{2}{x}=6\Rightarrow x=\frac{1}{3}$

Putting this in eq (1)

$\frac{1}{y}=4-3\Rightarrow y=1$

From eq (3)

$\frac{1}{z}=8-3\Rightarrow z=\frac{1}{5}$