Question

Solve the following pair of equations by reducing them to a pair of linear equations (by any algebraic method):

$\begin{array}{l} \frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2 \\ \frac{4}{\sqrt{x}} - \frac{9}{\sqrt{y}} = - 1 \end{array}$

Accepted Answer

Given $\begin{array}{l} \frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2 \\ \frac{4}{\sqrt{x}} - \frac{9}{\sqrt{y}} = - 1 \end{array}$ -----(i)

and $\frac{4}{\sqrt{x}} - \frac{9}{\sqrt{y}} = - 1$ -----(ii)

Writing $\frac{1}{\sqrt{x}} = p$ and $\frac{1}{\sqrt{y}} = q$ in both the equations (i) and (ii), we get,

$2 p + 3 q = 2$ -----(iii)

and $4 p - 9 q = - 1$ -----(iv)

From (iii), we get, $2 p + 3 q = 2 \Rightarrow 3 q = 2 - 2 p \Rightarrow q = \frac{2 - 2 p}{3}$ ----- (v)

Putting this value of $q$ from (v) in (iv), we get, $4 p - 9 q = - 1$

$\Rightarrow 4 p - 9 \times \frac{2 - 2 p}{3} = - 1$

$\Rightarrow 4 p - 3 (2 - 2 p) = - 1$

$\Rightarrow 4 p - 6 + 6 p = - 1$

$\Rightarrow 10 p = - 1 + 6$

$\Rightarrow 10 p = 5$

$\Rightarrow p = \frac{5}{10} = \frac{1}{2}$ .

From (v), we get, $q = \frac{2 - 2 p}{3} = \frac{2}{3} - \frac{2}{3} \times p = \frac{2}{3} - \frac{2}{3} \times \frac{1}{2} = \frac{2}{3} - \frac{1}{3} = \frac{2 - 1}{3} = \frac{1}{3}$ .

Now, $\frac{1}{\sqrt{x}} = p \Rightarrow \frac{1}{\sqrt{x}} = \frac{1}{2} \Rightarrow \sqrt{x} = 2 \Rightarrow x = 4$

and $\frac{1}{\sqrt{y}} = q \Rightarrow \frac{1}{\sqrt{y}} = \frac{1}{3} \Rightarrow \sqrt{y} = 3 \Rightarrow y = 9$

Therefore, the required solution is $x = 4$ and $y = 9$ .

Manipur Board Solutions for Chapter: Pair of Linear Equations in Two Variables, Exercise 2: EXERCISE 4.2

Manipur Board Mathematics Solutions for Exercise - Manipur Board Solutions for Chapter: Pair of Linear Equations in Two Variables, Exercise 2: EXERCISE 4.2

Questions from Manipur Board Solutions for Chapter: Pair of Linear Equations in Two Variables, Exercise 2: EXERCISE 4.2 with Hints & Solutions

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