Question

solve the following pair of linear equation?
1/2x -1/y = -1
1/x +1/2Y -8=0
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Answer 1

Question:

Solve the given pair of linear equations;

1/2x - 1/y = -1

1/x + 1/2y - 8 = 0

Answer:

x = 1/6 , y = 1/4

OR (1/6,1/4)

Solution:

Let;

1/2x - 1/y = -1 --------(1)

1/x + 1/2y = 8 --------(2)

Now,

Let us assume that , 1/x = a and 1/y = b .

Now, putting 1/x = a and 1/y = b in eq-(1) and (2),

We have;

=> a/2 - b = -1

=> (a-2b)/2 = -1

=> a - 2b = -2

=> a = 2b - 2 -------(1)

=> a + b/2 = 8

=> (2a+b)/2 = 8

=> 2a + b = 16 ---------(2)

From eq-(1) and (2) , we have;

=> 2a + b = 16

=> 2(2b-2) + b = 16

=> 4b - 4 + b = 16

=> 5b = 16 + 4

=> 5b = 20

=> b = 20/5

=> b = 4

=> 1/y = 4

=> y = 1/4

Now, putting b=4 in eq-(1);

We get;

=> a = 2b - 2

=> a = 2•4 - 2

=> a = 8 - 2

=> a = 6

=> 1/x = 6

=> x = 1/6

Hence,

The solution of the given system of linear equations is :

x = 1/6 and y = 1/4

OR (1/6,1/4)

Answer 2

Step-by-step explanation:

1/2x = - 1 + 1/y

x = (- 1 + 1/y)2

= - 2 + 2/y

Now,

$\frac{1}{ - 2 + \frac{2}{y} } + \frac{1}{2y} - 8 = 0 \\ \\ \frac{2y + ( - 2 + \frac{2}{y} )}{( - 2 + \frac{2}{y} ) \times 2y} = 8 \\ \\ \frac{2y - 2 + \frac{2}{y} }{ - 2y + 4} = 8 \\ \\ 2y - 2 + \frac{2}{y} = 8( - 2y + 4) \\ \\2y + \frac{2}{y} - 2 = - 16y + 32 \\ \\ \frac{ {y}^{2} + 2 }{y} = - 16y + 32 + 2 \\ \\ {y}^{2} + 2 = - 16 {y}^{2} + 34y \\ \\ {y}^{2} + 16 {y}^{2} - 34y + 2 = 0 \\ \\ 17 {y}^{2} - 34y + 2 = 0$

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