Question

solve the pair of linear eqations 1/2x - 1/y=-1 and 1/x + 1/2y=8​

Answer 1

Given

Two linear equations are

$(1) \dfrac{1}{2x} - \dfrac{1}{y} = - 1 \\ (2) \dfrac{1}{x} + \frac{1}{2y} = 8$

To find:

Values of x and y

EXPLANATION:

$\sf let \: a=\dfrac{1}{x} \\ \sf b=\dfrac{1}{y}$

then the equations become

$(1) \dfrac{a}{2} - b = - 1 \\ = > a - 2b = - 2 \\ \\ (2)a + \dfrac{b}{2} = 8 \\ = > 2a + b = 16$

Multiplying eq (2) with 2

it becomes

$(2)4a + 2b = 32$

(1)+(2)

$= > a - 2b + 4a + 2b = - 2 + 32 \\ \\ = > 5a = 30\\ \\ = > a = \dfrac{30}{5} = 6$

$but \: a = \dfrac{1}{x} \\ \boxed{ \sf \: x = \dfrac{1}{6} }$

Substituting the value of a in (1)

$a - 2b = - 2 \\ \\ = > 6- 2b = - 2 \\ \\ = > b = \dfrac{8}{2} =4\\ \\ but \: b = \dfrac{1}{y}$

$\boxed {\sf \: y = \dfrac{1}{4} }$

Therefore the values of x and y are obtained

Answer 2

SOLUTION:-

Given:

Linear equation:

1/2x - 1/y = -1

1/x +1/2y = 8

Assume 1/x be R & 1/y be M.

Now,

$= > \frac{R}{2} - M = - 1...........(1) \\ \\ = > R + \frac{M}{2} = 8..............(2)$

From equation (1), we get;

$= > \frac{R}{2} - M = - 1 \\ \\ = > R - 2M = - 2 \\ \\ = > R = - 2 + 2M..................(3)$

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

$= > - 2 + 2M + \frac{M}{2} = 8 \\ \\ = > - 4 + 4M + M = 16 \\ \\ = > - 4 + 5M= 16 \\ \\ = > 5M = 16 + 4 \\ \\ = > 5M = 20 \\ \\ = > M = \frac{20}{5} \\ \\ = > M = 4$

Also,

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

=) R = -2 + 2(4)

=) R= -2 + 8

=) R= 6

Now,

$\frac{1}{x} = R \\ \\ = > \frac{1}{x} = 6 \\ \\ = > x = \frac{1}{6}$

&

$\frac{1}{y} = M \\ \\ = > \frac{1}{y} = 4 \\ \\ = > y = \frac{1}{4}$

Hope it helps ☺️