Question

Solve the following systems of equations:
$
  \dfrac{4}{x} + 3y = 14 \\
  \dfrac{3}{x} - 4y = 23 \\
 $

Answer 1

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

Complete step-by-step answer:

Let, $\dfrac{1}{x} = a$.................... (1)
So the system of equation becomes
$ \Rightarrow 4a + 3y = 14$..................... (2)
And
$ \Rightarrow 3a - 4y = 23$..................... (3)
Now from equation (2) calculate the value of a in terms of y we have,
$ \Rightarrow a = \dfrac{{14 - 3y}}{4}$.................. (4)
Now substitute this value in equation (3) we have,
$ \Rightarrow 3\left( {\dfrac{{14 - 3y}}{4}} \right) - 4y = 23$
Now simplify this equation by multiply by 4 throughout we have,
$ \Rightarrow 42 - 9y - 16y = 92$
$ \Rightarrow 25y = 42 - 92 = - 50$
Now divide by 25 we have,
$ \Rightarrow y = \dfrac{{ - 50}}{{25}} = - 2$
Now substitute this value in equation (4) we have,
$ \Rightarrow a = \dfrac{{14 - 3\left( { - 2} \right)}}{4} = \dfrac{{14 + 6}}{4} = \dfrac{{20}}{4} = 5$
Now substitute this value in equation (1) we have,
$ \Rightarrow \dfrac{1}{x} = 5$
$ \Rightarrow x = \dfrac{1}{5}$
So the required solution of given system of equation is
$ \Rightarrow \left( {x,y} \right) = \left( {\dfrac{1}{5}, - 2} \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 some xy terms, which can’t be solved to get the values of x and y, that’s why the aim was to break the problem into smaller subdivision.