Question

Solve the following systems of equations:
$
  \dfrac{{10}}{{x + y}} + \dfrac{2}{{x - y}} = 4 \\
  \dfrac{{15}}{{x + y}} - \dfrac{9}{{x - y}} = - 2 \\
 $

Answer 1

Hint – In this question let $\dfrac{1}{{x + y}} = a{\text{ and }}\dfrac{1}{{x - y}} = b$ and simplify this to get two linear equation in variables of a and b, then solve to get the value of a and later substitute it back into one of the equation obtained between a and b to get the value of b. Then since x+y is earlier computed in terms of a and x-y is computed in terms of b only, so when a is known and b is known hence x and y can be taken out.

Complete Step-by-Step solution:
Let,
$\dfrac{1}{{x + y}} = a$.................. (1)
And
$\dfrac{1}{{x - y}} = b$..................... (2)
So the system of equation becomes
$ \Rightarrow 10a + 2b = 4$................... (3)
$ \Rightarrow 15a - 9b = - 2$................... (4)
Now multiply by 9 in equation (3) and multiply by 2 in equation (4) and add them we have,
$ \Rightarrow 9\left( {10a + 2b} \right) + 2\left( {15a - 9b} \right) = 9\left( 4 \right) + 2\left( { - 2} \right)$
Now simplify the above equation we have,
$ \Rightarrow 90a + 18b + 30a - 18b = 36 - 4$
$ \Rightarrow 120a = 32$
$ \Rightarrow a = \dfrac{{32}}{{120}} = \dfrac{4}{{15}}$
Now substitute the value of (a) in equation (4) we have,
$ \Rightarrow 15\left( {\dfrac{4}{{15}}} \right) - 9b = - 2$
$ \Rightarrow 9b = 4 + 2 = 6$
$ \Rightarrow b = \dfrac{6}{9} = \dfrac{2}{3}$
Now substitute the value of (a) and (b) in equation (1) and (2) we have,
$ \Rightarrow \dfrac{1}{{x + y}} = \dfrac{4}{{15}}{\text{ and }}\dfrac{1}{{x - y}} = \dfrac{2}{3}$
$ \Rightarrow x + y = \dfrac{{15}}{4}$................... (5)
And
$x - y = \dfrac{3}{2}$................. (6)
Now add equation (5) and (6) we have,
$ \Rightarrow x + y + x - y = \dfrac{{15}}{4} + \dfrac{3}{2}$
$ \Rightarrow 2x = \dfrac{{15 + 6}}{4}$
Now divide by 2 we have,
$ \Rightarrow x = \dfrac{{21}}{8}$
Now substitute this value in equation (5) we have,
$ \Rightarrow \dfrac{{21}}{8} + y = \dfrac{{15}}{4}$
$ \Rightarrow y = \dfrac{{15}}{4} - \dfrac{{21}}{8} = \dfrac{{30 - 21}}{8} = \dfrac{9}{8}$
So the solution of the given system of equations is
$ \Rightarrow \left( {x,y} \right) = \left( {\dfrac{{21}}{8},\dfrac{9}{8}} \right)$
So this is the required answer.

Note – The trick point here was the substitution in the starting phase of the solution the reason behind it was if this would not have done we would be getting a two different equations in terms of ${x^2},{y^2}{\text{ and some xy}}$ terms, which can’t be solved to get the values of x and y, that is why why the aim was to break the problem into smaller subdivision.