Question

Solve the following systems of equations:
$
  \dfrac{{44}}{{x + y}} + \dfrac{{30}}{{x - y}} = 10 \\
  \dfrac{{55}}{{x + y}} + \dfrac{{40}}{{x - y}} = 13 \\
$

Answer 1

Hint – In this question let $\dfrac{1}{{x + y}} = a$ and $\dfrac{1}{{x - y}} = b$, simplify this to get two linear equation in variables of a and b, then multiply by 4 in equation (3) and multiply by equation (3) in equation then solve to get the value of a and b and later substitute it back to form two equations again in terms of x and y, solve them to get the value of x and y.

Complete step by step answer:
Let,
$\dfrac{1}{{x + y}} = a$.................. (1)
And
$\dfrac{1}{{x - y}} = b$..................... (2)
So the system of equation becomes
$ \Rightarrow 44a + 30b = 10$................... (3)
$ \Rightarrow 55a + 40b = 13$................... (4)
Now multiply by 4 in equation (3) and multiply by 3 in equation (4) and subtract them we have,
$ \Rightarrow 4\left( {44a + 30b} \right) - 3\left( {55a + 40b} \right) = 4\left( {10} \right) - 3\left( {13} \right)$
Now simplify the above equation we have,
$ \Rightarrow 176a + 120b - 165a - 120b = 40 - 39$
$ \Rightarrow 11a = 1$
$ \Rightarrow a = \dfrac{1}{{11}}$
Now substitute the value of (a) in equation (4) we have,
$ \Rightarrow 55\left( {\dfrac{1}{{11}}} \right) + 40b = 13$
$ \Rightarrow 40b = 13 - 5 = 8$
$ \Rightarrow b = \dfrac{8}{{40}} = \dfrac{1}{5}$
Now substitute the value of (a) and (b) in equation (1) and (2) we have,
$ \Rightarrow \dfrac{1}{{x + y}} = \dfrac{1}{{11}}{\text{ and }}\dfrac{1}{{x - y}} = \dfrac{1}{5}$
$ \Rightarrow x + y = 11$................... (5)
And
$x - y = 5$................. (6)
Now add equation (5) and (6) we have,
$ \Rightarrow x + y + x - y = 11 + 5$
$ \Rightarrow 2x = 16$
Now divide by 2 we have,
$ \Rightarrow x = \dfrac{{16}}{2} = 8$
Now substitute this value in equation (5) we have,
$ \Rightarrow 8 + y = 11$
$ \Rightarrow y = 11 - 8 = 3$
So the solution of the given system of equations is
$ \Rightarrow \left( {x,y} \right) = \left( {8,3} \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 the aim was to break the problem into smaller subdivision.