Question

find the sol. of pair of equation 3/x+8/y=-1 ;1/x-2/y=2.............​.

Answer 1

Answer:

3/x+8/y=-1............(1)

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

let

1/x=a;1/y=b

3a+8b=-1............[3]

a-2b=2................[4]

(4)*3

3a+8b=-1

3a-6b=+6

(-) + -

--------------

14b=-7

b=-7/14

b=-1/2

sub(b=-1/2)in[4]

a-2*1/2=2

a-1=2

a=2+1

a=3

a=1/x=3

x=1/3

b=1/y

1/y=-1/2

y=-2

Answer 2

SOLUTION:-

Given:

3/x + 8/y= -1

1/x - 2/y= 2

Assume 1/x= R & 1/y= M

Therefore,

3R + 8M= -1.............(1)

R - 2M= 2.................(2)

Using substitution Method:

From equation (1), we get;

=) 3R + 8M= -1

=) 3R= -1 - 8M

=) R= -1-8M/3............(3)

So,

Putting the value of R in equation (2), we get;

$= > \frac{ - 1 - 8</u></strong><strong><u>M</u></strong><strong><u>}{3} - 2</u></strong><strong><u>M</u></strong><strong><u>= 2 \\ \\ = > - 1 - 8</u></strong><strong><u>M</u></strong><strong><u> - 6</u></strong><strong><u>M</u></strong><strong><u> = 6 \\ \\ = > - 1 - 14</u></strong><strong><u>M</u></strong><strong><u> = 6 \\ \\ = > - 14</u></strong><strong><u>M</u></strong><strong><u> = 6 + 1 \\ \\ = > - 14</u></strong><strong><u>M</u></strong><strong><u> = 7 \\ \\ = > </u></strong><strong><u>M</u></strong><strong><u> = - \frac{7}{14} \\ \\ = > </u></strong><strong><u>M</u></strong><strong><u> = - \frac{1}{2}$

Now,

Putting the value of M in equation (3), we get;

$= > R = \frac{ - 1 - 8( - \frac{ 1}{2}) }{3} \\ \\ = > R = \frac{ - 1 + \frac{8}{2} }{3} \\ \\ = > R = \frac{ \frac{ - 2 + 8}{2} }{3} \\ \\ = > R = \frac{ \frac{6}{2} }{3} \\ \\ = > R = \frac{3}{3} \\ \\ = > R = 1$

Therefore,

$= > \frac{1}{x} = R \\ \\ = > \frac{1}{x} = 1 \\ \\ = > x = 1$

&

$= > \frac{1}{y} = M \\ \\ = > \frac{1}{y} = - \frac{1}{2} \\ \\ = > y = - 2$

Hope it helps ☺️