Question

Solve the following systems of equations:
$
  \dfrac{2}{{3x + 2y}} + \dfrac{3}{{2x - 2y}} = \dfrac{{17}}{5} \\
  \dfrac{5}{{3x + 2y}} + \dfrac{1}{{3x - 2y}} = 2 \\
 $

Answer 1

Hint – In this question let $\dfrac{1}{{3x + 2y}} = a$ and $\dfrac{1}{{3x - 2y}} = b$, simplify this to get two linear equation in variables of a and b, 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,
$\dfrac{1}{{3x + 2y}} = a$.................. (1)
And
$\dfrac{1}{{3x - 2y}} = b$..................... (2)
So the system of equation becomes
$ \Rightarrow 2a + 3b = \dfrac{{17}}{5}$................... (3)
$ \Rightarrow 5a + b = 2$................... (4)
Now multiply by 3 in equation (4) we have,
$ \Rightarrow 3\left( {5a + b} \right) = 2\left( 3 \right)$
$ \Rightarrow 15a + 3b = 6$...................... (5)
Now subtract equation (3) from equation (5) we have,
$ \Rightarrow 15a + 3b - \left( {2a + 3b} \right) = 6 - \dfrac{{17}}{5}$
Now simplify the above equation we have,
$ \Rightarrow 13a = \dfrac{{30 - 17}}{5} = \dfrac{{13}}{5}$
$ \Rightarrow a = \dfrac{1}{5}$
Now substitute the value of (a) in equation (4) we have,
$ \Rightarrow 5\left( {\dfrac{1}{5}} \right) + b = 2$
$ \Rightarrow b = 2 - 1 = 1$
Now substitute the value of (a) and (b) in equation (1) and (2) we have,
$ \Rightarrow \dfrac{1}{{3x + 2y}} = \dfrac{1}{5}{\text{ and }}\dfrac{1}{{3x - 2y}} = 1$
$ \Rightarrow 3x + 2y = 5$................... (6)
And
$3x - 2y = 1$................. (7)
Now add equation (6) and (7) we have,
$ \Rightarrow 3x + 2y + 3x - 2y = 5 + 1$
$ \Rightarrow 6x = 6$
Now divide by 6 we have,
$ \Rightarrow x = 1$
Now substitute this value in equation (6) we have,
$ \Rightarrow 3\left( 1 \right) + 2y = 5$
$ \Rightarrow 2y = 5 - 3 = 2$
Now divide by 2 we have,
$ \Rightarrow y = 1$
So the solution of the given system of equations is
$ \Rightarrow \left( {x,y} \right) = \left( {1,1} \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 be solved to get the values of x and y, that’s why the aim was to break the problem into smaller subdivision.