Question

Solve the following equations:
$$\begin{gathered} \left(x+y\right)\left(x+z\right)=30 \\ \left(y+z\right)\left(y+x\right)=15, \\ \left(z+x\right)\left(z+y\right)=18 \end{gathered}$$
This question has multiple correct answers.
$$\begin{gathered} A.\left(2,4,1\right) \\ B.\left(-2,-4,-1\right) \\ C.\left(3,1,2\right) \\ D.\left(-3,-1,-2\right) \end{gathered}$$

Answer 1

Hint:In this question assume $x+y=a,y+z=b,z+x=c$, find the value of $a,b,c$ by substitution method and the form the sets . Use these steps to find the solution of the pair of linear equations in two variables .

Complete step-by-step answer:
According to the question , the given equations are $\left(x+y\right)\left(x+z\right)=30,\left(y+z\right)\left(y+x\right)=15,\left(z+x\right)\left(z+y\right)=18$
 Put $x+y=a,y+z=b,z+x=c$
We get $ac=30.....\left(i\right)$
$$\begin{gathered} ab=15........\left(ii\right) \\ cb=18.........\left(iii\right) \end{gathered}$$
From $\left(iii\right)$, we have
$b={\displaystyle \frac{18}{c}}$
Substituting $b$ in $\left(ii\right)$, we get
$$\begin{gathered} {\displaystyle \frac{a}{c}}={\displaystyle \frac{15}{18}} \\ \Rightarrow a={\displaystyle \frac{15}{18}}c \end{gathered}$$
Substituting $a$ in $\left(i\right)$, we get
$$\begin{gathered} {\displaystyle \frac{15}{18}}c\times c=30 \\ \Rightarrow c^2=36 \\ \Rightarrow c=\pm 6 \end{gathered}$$
We have $a={\displaystyle \frac{15}{18}}c$
$\Rightarrow a=\pm 5$
Thus $b={\displaystyle \frac{18}{c}}$
$\Rightarrow b=\pm 3$
Now the given equations become
$$\begin{gathered} x+y=6 \\ y+z=5 \\ z+x=3 \end{gathered}$$
and
$$\begin{gathered} x+y=-6 \\ y+z=-5 \\ z+x=-3 \\ . \end{gathered}$$
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 $z=-1$

Note: In such types of questions it is advisable to use either graphical method of pair of linear equations or substitution method of pair of linear equations of two variables to get the required answer.