Question

Solve the following equations:
$(y-z)(z+x)=22$,
$(z+x)(x-y)=33$,
$(x-y)(y-z)=6$.

• A. $x=\pm 6,y\pm 4,z=\pm 3$
• B. $x=\pm 3,y\pm 5,z=\mp 5$
• C. $x=\pm 8,y\pm 5,z=\pm 3$
• D. $x=\pm 10,y\pm 5,z=\pm 5$

Answer 1

Answer: (C)

$x=\pm 8,y\pm 5,z=\pm 3$

Let $(y-z)(z+x)=22$ ......(i)

$(z+x)(x-y)=33$ ......(ii)

$(x-y)(y-z)=6$ .........(iii)

On dividing (i) by (ii), we get

$\frac{(y-z)(z+x)}{(z+x)(x-y)}=\frac{22}{33}\frac{y-z}{x-y}=\frac{2}{3}y-z=\frac{2}{3}(x-y)$

Substituting the above value in equation (iii), we get

${(y-z)}^2=4\Rightarrow y-z=\pm 2$ .......(1)
Substituting the above value in equation (i), we get

$\Rightarrow z+x=\pm 11$ ......(2)

On substituting (1) in (ii), we have

$\Rightarrow x-y=\pm 3$ ........(3)

Adding $\left(1\right)$ and $\left(2\right)$

$\Rightarrow x+y=\pm 13$ ........(4)

Solving $\left(3\right)$ and $\left(4\right)$

$\Rightarrow x=\pm 8,y\pm 5$

Substituting $y$ in (1), we get

$\Rightarrow z=\pm 3$

So, the complete solution is $x=\pm 8,y\pm 5,z=\pm 3$.