Question

Solve the following equations:
$x-y-z=2$,
$x^2+y^2-z^2=22$,
$xy=5$.

• A. $(-1,-5,2)$
• B. $(1,2,-4)$
• C. $(2,3,7)$
• D. $(-2,-5,3)$

Answer 1

Answer: (A)

$(-1,-5,2)$

Let $x-y-z=2$ ........(i)

$x^2+y^2-z^2=22$ .......(ii)

$xy=5$ .......(iii)

From (i), we have

$x-y=2+z$

Squaring both sides, we get

$x^2+y^2-2xy=4+z^2+4z{x}^2+y^2-z^2-2xy=4+4z$

Susbtituting (ii) and (iii), we have

$22-2\left(5\right)=4+4z4z=8\Rightarrow z=2$

Substituting $z$ in (i)

$\Rightarrow x-y=4$

From (iii) $y=\frac{5}{x}$

$\Rightarrow x-\frac{5}{x}=4\Rightarrow x^2-5=4x\Rightarrow x^2-4x-5=0\Rightarrow x^2-5x+x-5=0\Rightarrow x(x-5)+1(x-5)=0\Rightarrow (x+1)(x-5)=0\Rightarrow x=-1,5\Rightarrow y=\frac{5}{x}\Rightarrow y=-5,1$

So, the values of $x$ are $-1,5$ , values of $y$ are $-5,1$ and value of $z$ is $2$.