Question

1\x +1\y=2 and 1\y-1\x=6
solve this equation by elimination and substitution method
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Answer 1

Answer:

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Answer 2

Given Equations :

$\sf \dfrac{1}{x} + \dfrac{1}{y} = 2 \: \: \: ...(1)$

$\sf\dfrac{1}{y} - \dfrac{1}{x} = 6 \: \: \: ...(2)$

let 1/x be u and 1/y be v

equation 1 will become,

$\sf u + v = 2 \: \: \: ...(3)$

equation 2 will become,

$\sf v - u = 6 \: \: \: ...(4)$

by substituting both equation,

put v = 6 + u in equation (3)

$\dashrightarrow \sf u + v = 2$

$\dashrightarrow \sf u + (6 + u) = 2$

$\dashrightarrow \sf 2u = 2 - 6$

$\dashrightarrow \sf 2u = - 4$

$\dashrightarrow \sf u = - 2$

by putting u = -2 in equation (4)

$\dashrightarrow \sf v - u = 6$

$\dashrightarrow \sf v - ( - 2) = 6$

$\dashrightarrow \sf v + 2 = 6$

$\dashrightarrow \sf v = 6 - 2$

$\dashrightarrow \sf v = 4$

we know,

$\sf \dfrac{1}{x} = u$

$\sf \dfrac{1}{x} = - 2$

$\sf x = - \dfrac{1}{ 2}$

Similarly,

$\sf \dfrac{1}{y} = v$

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

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

Hence,

$\boxed{ \sf x = - \dfrac{1}{ 2} } \sf \: \: \: and \: \: \: \boxed{ \sf y = \dfrac{1}{4} }$