Question

Solve the following systems of equations:
$
  \dfrac{1}{{3x + y}} + \dfrac{1}{{3x - y}} = \dfrac{3}{4} \\
  \dfrac{2}{{2(3x + y)}} - \dfrac{1}{{2(3x - y)}} = \dfrac{{ - 1}}{8} \\
 $

Answer 1

Hint – In this question let $\dfrac{1}{{3x + y}} = a{\text{ and }}\dfrac{1}{{3x - y}} = b$ 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 3x+y is earlier computed in terms of a and 3x-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,
$\dfrac{1}{{3x + y}} = a$.................. (1)
And
$\dfrac{1}{{3x - y}} = b$..................... (2)
So the system of equation becomes
$ \Rightarrow a + b = \dfrac{3}{4}$................... (3)
$ \Rightarrow \dfrac{1}{2}a - \dfrac{1}{2}b = \dfrac{{ - 1}}{8}$
$ \Rightarrow a - b = \dfrac{{ - 2}}{8} = \dfrac{{ - 1}}{4}$................... (4)
Now add equation (3) and (4) we have,
$ \Rightarrow a + b + a - b = \dfrac{3}{4} + \dfrac{{ - 1}}{4}$
Now simplify the above equation we have,
$ \Rightarrow 2a = \dfrac{{3 - 1}}{4} = \dfrac{2}{4}$
Now divide by 2 we have,
$ \Rightarrow a = \dfrac{1}{4}$
Now substitute the value of (a) in equation (4) we have,
$ \Rightarrow \dfrac{1}{4} - b = \dfrac{{ - 1}}{4}$
$ \Rightarrow b = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{2}{4} = \dfrac{1}{2}$
Now substitute the value of (a) and (b) in equation (1) and (2) we have,
$ \Rightarrow \dfrac{1}{{3x + y}} = \dfrac{1}{4}{\text{ and }}\dfrac{1}{{3x - y}} = \dfrac{1}{2}$
$ \Rightarrow 3x + y = 4$................... (5)
And
$3x - y = 2$................. (6)
Now add equation (5) and (6) we have,
$ \Rightarrow 3x + y + 3x - y = 4 + 2$
$ \Rightarrow 6x = 6$
Now divide by 6 we have,
$ \Rightarrow x = 1$
Now substitute this value in equation (5) we have,
$ \Rightarrow 3\left( 1 \right) + y = 4$
$ \Rightarrow y = 4 - 3 = 1$
$ \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’t be solved to get the values of x and y, that why the aim was to break the problem into smaller subdivision.