Question

eliminate x from the following equation
x + 1 /x= 2p, x- 1/x = 2q + 1​

Answer 1

GIVEN

eq1 and eq2 respectively :

$x + \frac{1}{x} = 2p \: .......(1) \\ and \\ x - \frac{1}{x} = 2q + 1.........(2)$

On adding (1) + (2) we get ;

$2x = 2p + 2q + 1 \\ \\ x = \frac{2p + 2q + 1}{2}$

Substitute value of x in (1) :

$x + \frac{1}{x} = 2p \\ \\ \implies( \frac{2p + 2q + 1}{2} ) + ( \frac{1}{ \frac{2p + 2q + 1}{2} } ) = 2p \\ \\ \implies \: \frac{2p + 2q + 1}{2} + \frac{2}{2p + 2q + 1} = 2p \\ \\ \implies \frac{{(2p + 2q + 1)}^{2} + {2}^{2} }{(2p + 2q + 1)2} = 2p \\ \\ \implies \frac{4 {p}^{2} + 4 {q}^{2} + 1 + 8pq + 4p + 4q}{4p \: + 4q + 2} = 2p$

Then

$\implies \: 8 {p}^{2} + 8pq + 4p = 4 {p}^{2} + 4 {q}^{2} + 1 + 8pq \: + 4p + 4q \\ \\ \implies \: 4 {p}^{2} - 4 {q}^{2} - 1 - 4q = 0$

Thus after eliminating x we get the above expression

#answerwithquality

#BAL

Answer 2

Answer:

2q + 1 = 2p or 2q + 1 = - 2p

Step-by-step explanation:

Given---> x + 1 / x = 2 p and x - 1 / x = 2q + 1

To find ---> Eliminate x from given equation.

Solution--->We know that

1) ( a + b )² = a² + b² + 2ab

2) ( a - b )² = a² + b² - 2ab

ATQ,

x + 1 / x = 2p

Squaring both sides we get,

=> ( x + 1 / x )² = ( 2 p )²

Applying first formula, we get,

=> x² + 1 / x² + 2 ( x ) ( 1 / x ) = 4p²

=> x² + 1 / x² + 2 = 4p²

=> x² + 1 / x² = 4p² - 2 ...................( 1 )

ATQ, x - 1 / x = 2q + 1

Squaring both sides we get,

=> ( x - 1 / x )² = ( 2q + 1 )²

Applying second formula , we get,

=> x² + 1 / x² - 2 x ( 1 / x ) = 4q² + 1 + 4q

=> x² + 1 / x² = 4q² + 4q + 1 ..................( 2 )

By equation ( 1 ) and ( 2 ) , we get,

4q² + 4q + 1 = 4p²

=> ( 2 q + 1 )² = ( 2p )²

Taking square root of both sides we get,

=> ( 2q + 1 ) = ± 2p

Taking + sign we get,

2q + 1 = 2p

Taking ( - ) sign we get

2q + 1 = -2p

#Answerwithquality

#BAL