Question

Solve $(x+y)(x+z)=30$,
$(y+z)(y+z)=15$,
$(z+x)(z+y)=18$.

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

Answer 1

Answer: (A), (B)

$(2,4,1)$
$(-2,-4,-1)$

Given equations are $(x+y)(x+z)=30,(y+x)(y+z)=15$ and $(z+x)(z+y)=18$

Put $x+y=a,y+z=b,z+x=c$

we get $ac=30$ .....(i),

$ab=15$ .....(ii),

$cb=18$ .....(iii)

From (iii), we have

$b=\frac{18}{c}$

Substituting $b$ in (ii), we get

$\frac{a}{c}=\frac{15}{18}\Rightarrow a=\frac{15}{18}c$

Substituting $a$ in (i), we get

$\frac{15}{18}c\times c=30\Rightarrow c^2=36\Rightarrow c=\pm 6$

We have $a=\frac{15}{18}c$

$\Rightarrow a=\pm 5$

Thus $b=\frac{18}{c}$

$\Rightarrow b=\pm 3$

Now the given equations become

$\{\begin{array}{c}x+y=6\\ y+z=5\\ z+x=3\end{array}$ and $\{\begin{array}{c}x+y=-6\\ y+z=-5\\ z+x=-3\end{array}$

Solving the first set of equations, we get

$x=2,y=4$ and $z=1$

Solving the second set, we get

$x=-2,y=-4$ and $z=-1$