Question

Solve the following systems of equations:
$
  \dfrac{2}{{\sqrt x }} + \dfrac{3}{{\sqrt y }} = 2 \\
  \dfrac{4}{{\sqrt x }} - \dfrac{9}{{\sqrt y }} = - 1 \\
 $

Answer 1

Hint – In this question let $\dfrac{1}{{\sqrt x }} = a{\text{ and }}\dfrac{1}{{\sqrt y }} = b$ 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 $\dfrac{1}{{\sqrt x }}$ is earlier computed in terms of a and $\dfrac{1}{{\sqrt y }}$ 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,
$\dfrac{1}{{\sqrt x }} = a$.................... (1)
And
$\dfrac{1}{{\sqrt y }} = b$........................... (2)
So the system of equations become
$ \Rightarrow 2a + 3b = 2$..................... (3)
And
$4a - 9b = - 1$.................. (4)
Now multiply by 3 in equation (3) and add in equation (4) we have,
$ \Rightarrow 3\left( {2a + 3b} \right) + 4a - 9b = 3\left( 2 \right) - 1$
Now simplify the above equation we have,
$ \Rightarrow 6a + 9b + 4a - 9b = 6 - 1$
$ \Rightarrow 10a = 5$
Now divide by 10 we have,
$ \Rightarrow a = \dfrac{5}{{10}} = \dfrac{1}{2}$
Now substitute this value in equation (3) we have,
$ \Rightarrow 2\left( {\dfrac{1}{2}} \right) + 3b = 2$
Now simplify the above equation we have,
$ \Rightarrow 3b = 2 - 1 = 1$
Now divide by 3 we have,
$ \Rightarrow b = \dfrac{1}{3}$
Now substitute the value of (a) and (b) in equation (1) and (2) we have,
$ \Rightarrow \dfrac{1}{{\sqrt x }} = \dfrac{1}{2}$
And
$ \Rightarrow \dfrac{1}{{\sqrt y }} = \dfrac{1}{3}$
Therefore,
$\sqrt x = 2$ and $\sqrt y = 3$
Now square on both sides we have,
$ \Rightarrow {\left( {\sqrt x } \right)^2} = {2^2} = 4$
$ \Rightarrow x = 4$
And
$ \Rightarrow {\left( {\sqrt y } \right)^2} = {3^2} = 9$
$ \Rightarrow y = 9$
So the solution of the given system of equations is
$ \Rightarrow \left( {x,y} \right) = \left( {4,9} \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 ${\text{some }}\sqrt x ,\sqrt y $ 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.