Question

Solve the following pair of equations for \[x\] and \[y\]; \[\dfrac{{15}}{{x - y}} + \dfrac{{22}}{{x + y}} = 5\] & \[\dfrac{{40}}{{x - y}} + \dfrac{{55}}{{x + y}} = 13\] where \[x \ne y{\text{ and }}x \ne - y\].

Answer 1

Hint: Convert the given pair of equations into linear equations in two variables by substituting the suitable variables. Then solve the obtained linear equations in two variables. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
Put \[\dfrac{5}{{x - y}} = h\] and \[\dfrac{{11}}{{x + y}} = k\]
So, the equation \[\dfrac{{15}}{{x - y}} + \dfrac{{22}}{{x + y}} = 5\]becomes as
\[
   \Rightarrow 3\left( {\dfrac{5}{{x - y}}} \right) + 2\left( {\dfrac{{11}}{{x + y}}} \right) = 5 \\
   \Rightarrow 3h + 2k = 5 \\
   \Rightarrow 2k = 5 - 3h \\
   \Rightarrow k = \dfrac{{5 - 3h}}{2}..................................................\left( 1 \right) \\
\]
And the equation \[\dfrac{{40}}{{x - y}} + \dfrac{{55}}{{x + y}} = 13\] becomes as
\[
   \Rightarrow 8\left( {\dfrac{5}{{x - y}}} \right) + 5\left( {\dfrac{{11}}{{x + y}}} \right) = 13 \\
   \Rightarrow 8h + 5k = 13 \\
   \Rightarrow 5k = 13 - 8h \\
   \Rightarrow k = \dfrac{{13 - 8h}}{5}.....................................................\left( 2 \right) \\
\]
From equations (1) and (2), we have
\[
   \Rightarrow \dfrac{{5 - 3h}}{2} = \dfrac{{13 - 8h}}{5} \\
   \Rightarrow 5\left( {5 - 3h} \right) = 2\left( {13 - 8h} \right) \\
   \Rightarrow 25 - 15h = 26 - 16h \\
   \Rightarrow 16h - 15h = 26 - 25 \\
  \therefore h = 1 \\
\]
Substituting \[h = 1\] in the equation \[5k = 13 - 8h\]
\[
   \Rightarrow 5k = 13 - 8\left( 1 \right) \\
   \Rightarrow 5k = 13 - 8 \\
   \Rightarrow 5k = 5 \\
  \therefore k = 1 \\
\]
So, \[h = 1\] & \[k = 1\]
Substituting \[h = 1\]in \[\dfrac{5}{{x - y}} = h\]
\[
   \Rightarrow \dfrac{5}{{x - y}} = 1 \\
   \Rightarrow x - y = 5........................................................\left( 3 \right) \\
\]
Substituting \[k = 1\]in \[\dfrac{{11}}{{x + y}} = k\]
\[
   \Rightarrow \dfrac{{11}}{{x + y}} = 1 \\
   \Rightarrow x + y = 11..................................................\left( 4 \right) \\
\]
Adding equation (3) and (4)
\[
   \Rightarrow \left( {x - y} \right) + \left( {x + y} \right) = 5 + 11 \\
   \Rightarrow 2x = 16 \\
  \therefore x = 8 \\
\]
Substituting \[x = 8\] in equation (4)
\[
   \Rightarrow 8 + y = 11 \\
   \Rightarrow y = 11 - 8 \\
  \therefore y = 3 \\
\]
Thus, \[x = 8\] & \[y = 3\].

Note: We can also solve the pair of equations directly. But it involves quadratic equations in terms of \[x\] & \[y\], which is difficult to solve the equations. In these types of problems always use the substitution method to reach the solution easily.