Question

Solve the following equations: $\dfrac{{xy}}{{ay + bx}} = c,\dfrac{{xz}}{{az + cx}} = b,\dfrac{{yz}}{{bz + cy}} = a$.

Answer 1

Hint- Here, we will simplify the given equations in order to evaluate the values of $x$ ,$y$ and $z$.

Given equations are $\dfrac{{xy}}{{ay + bx}} = c,\dfrac{{xz}}{{az + cx}} = b,\dfrac{{yz}}{{bz + cy}} = a$
The above equations can be re-written as
 $
  \dfrac{{ay + bx}}{{xy}} = \dfrac{1}{c},\dfrac{{az + cx}}{{xz}} = \dfrac{1}{b},\dfrac{{bz + cy}}{{yz}} = \dfrac{1}{a} \Rightarrow \dfrac{{ay}}{{xy}} + \dfrac{{bx}}{{xy}} = \dfrac{1}{c},\dfrac{{az}}{{xz}} + \dfrac{{cx}}{{xz}} = \dfrac{1}{b},\dfrac{{bz}}{{yz}} + \dfrac{{cy}}{{yz}} = \dfrac{1}{a} \\
   \Rightarrow \dfrac{a}{x} + \dfrac{b}{y} = \dfrac{1}{c}{\text{ }} \to {\text{(1) }},\dfrac{a}{x} + \dfrac{c}{z} = \dfrac{1}{b}{\text{ }} \to {\text{(2) }},\dfrac{b}{y} + \dfrac{c}{z} = \dfrac{1}{a}{\text{ }} \to {\text{(3)}} \\
    \\
 $
Now adding all the above three equations, we have
$
   \Rightarrow \dfrac{a}{x} + \dfrac{b}{y} + \dfrac{a}{x} + \dfrac{c}{z} + \dfrac{b}{y} + \dfrac{c}{z} = \dfrac{1}{c} + \dfrac{1}{b} + \dfrac{1}{a} \Rightarrow 2\left[ {\dfrac{a}{x} + \dfrac{b}{y} + \dfrac{c}{z}} \right] = \dfrac{1}{c} + \dfrac{1}{b} + \dfrac{1}{a}{\text{ }} \to {\text{(4)}} \\
   \Rightarrow 2\left[ {\dfrac{a}{x} + \left( {\dfrac{b}{y} + \dfrac{c}{z}} \right)} \right] = \dfrac{1}{c} + \dfrac{1}{b} + \dfrac{1}{a} \\
 $
Since, we have already shown that $\dfrac{b}{y} + \dfrac{c}{z} = \dfrac{1}{a}$ according to equation (3)
$
  \therefore {\text{ }}2\left[ {\dfrac{a}{x} + \dfrac{1}{a}} \right] = \dfrac{1}{c} + \dfrac{1}{b} + \dfrac{1}{a} \Rightarrow \dfrac{{2a}}{x} + \dfrac{2}{a} = \dfrac{1}{c} + \dfrac{1}{b} + \dfrac{1}{a} \Rightarrow \dfrac{{2a}}{x} + \dfrac{1}{a} = \dfrac{1}{c} + \dfrac{1}{b} \\
   \Rightarrow \dfrac{{2a}}{x} = \dfrac{1}{c} + \dfrac{1}{b} - \dfrac{1}{a} \Rightarrow \dfrac{{2a}}{x} = \dfrac{{ab + ac - bc}}{{abc}} \Rightarrow \dfrac{x}{{2a}} = \dfrac{{abc}}{{ab + ac - bc}} \\
   \Rightarrow x = \dfrac{{2{a^2}bc}}{{ab + ac - bc}} \\
 $
 Considering equation (4) again, we have
$ \Rightarrow 2\left[ {\dfrac{a}{x} + \dfrac{b}{y} + \dfrac{c}{z}} \right] = \dfrac{1}{c} + \dfrac{1}{b} + \dfrac{1}{a}{\text{ }} \Rightarrow 2\left[ {\dfrac{b}{y} + \left( {\dfrac{a}{x} + \dfrac{c}{z}} \right)} \right] = \dfrac{1}{c} + \dfrac{1}{b} + \dfrac{1}{a}{\text{ }}$
Since, we have already shown that $\dfrac{a}{x} + \dfrac{c}{z} = \dfrac{1}{b}$ according to equation (2)
$
  \therefore {\text{ }}2\left[ {\dfrac{b}{y} + \dfrac{1}{b}} \right] = \dfrac{1}{c} + \dfrac{1}{b} + \dfrac{1}{a} \Rightarrow \dfrac{{2b}}{y} + \dfrac{2}{b} = \dfrac{1}{c} + \dfrac{1}{b} + \dfrac{1}{a} \Rightarrow \dfrac{{2b}}{y} + \dfrac{1}{b} = \dfrac{1}{c} + \dfrac{1}{a} \\
   \Rightarrow \dfrac{{2b}}{y} = \dfrac{1}{c} + \dfrac{1}{a} - \dfrac{1}{b} \Rightarrow \dfrac{{2b}}{y} = \dfrac{{ab + bc - ac}}{{abc}} \Rightarrow \dfrac{y}{{2b}} = \dfrac{{abc}}{{ab + bc - ac}} \\
   \Rightarrow y = \dfrac{{2a{b^2}c}}{{ab + bc - ac}} \\
 $
Considering equation (4) again, we have
\[ \Rightarrow 2\left[ {\dfrac{a}{x} + \dfrac{b}{y} + \dfrac{c}{z}} \right] = \dfrac{1}{c} + \dfrac{1}{b} + \dfrac{1}{a}{\text{ }} \Rightarrow 2\left[ {\left( {\dfrac{a}{x} + \dfrac{b}{y}} \right) + \dfrac{c}{z}} \right] = \dfrac{1}{c} + \dfrac{1}{b} + \dfrac{1}{a}{\text{ }}\]
Since, we have already shown that $\dfrac{a}{x} + \dfrac{b}{y} = \dfrac{1}{c}$ according to equation (1)
$
  \therefore {\text{ }}2\left[ {\dfrac{1}{c} + \dfrac{c}{z}} \right] = \dfrac{1}{c} + \dfrac{1}{b} + \dfrac{1}{a} \Rightarrow \dfrac{2}{c} + \dfrac{{2c}}{z} + = \dfrac{1}{c} + \dfrac{1}{b} + \dfrac{1}{a} \Rightarrow \dfrac{{2c}}{z} + \dfrac{1}{c} = \dfrac{1}{b} + \dfrac{1}{a} \\
   \Rightarrow \dfrac{{2c}}{z} = \dfrac{1}{a} + \dfrac{1}{b} - \dfrac{1}{c} \Rightarrow \dfrac{{2c}}{z} = \dfrac{{bc + ac - ab}}{{abc}} \Rightarrow \dfrac{z}{{2c}} = \dfrac{{abc}}{{bc + ac - ab}} \\
   \Rightarrow z = \dfrac{{2ab{c^2}}}{{bc + ac - ab}} \\
 $
Therefore, after solving the given equations we get
$x = \dfrac{{2{a^2}bc}}{{ab + ac - bc}}$, $y = \dfrac{{2a{b^2}c}}{{ab + bc - ac}}$ and $z = \dfrac{{2ab{c^2}}}{{bc + ac - ab}}$.

Note- In this particular problem, equations (1), (2) and (3) are used in equation (4) one by one in order to eliminate the other two variables and solve for the remaining one variable only.