Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

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

seo-qna
Last updated date: 24th Jul 2024
Total views: 453.6k
Views today: 7.53k
Answer
VerifiedVerified
453.6k+ views
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.