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## ${2 \over x} + {2 \over 3y} = {1 \over 6}$  ;  ${3 \over x} + {2 \over y} = 0$

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$2({1 \over x}) + {2 \over 3}({1 \over y})={1 \over 6}$  ...(I)

$3({1 \over x}) +2({1 \over y}) = {1 \over 6}$  ...(II)

Replace $1 \over x$ by  a and  $1 \over y$  by b in equations  (I) and (II),

we get

$2a + ({2 \over 3})b = {1 \over 6}$ and 3a + 2b = 0

Consider $2a + ({2 \over 3})b = {1 \over 6}$

Multiplying throughout by 6

$2a(6) + {2 \over 3}b(6)={1 \over 6}(6)$

12a + 4b = 1 ...(III)

Now consider,

3a + 2b =0

Multiplying eq. (iv) by 2

6a + 4b = 0 ...(IV)

Subtracting eq. (III) and (V)

12a +4 b =1

– 6 a +4 b =0

(–)    (–)    (–)

6a          =1

a = $1 \over 6$

Place the value of a in eq. (iv)

3 * $1 \over 6$ + 2b = 0

$1 \over 2$ + 2b = 0

2b = - $1 \over 2$

b = $-1 \over 4$

Resubstituting the values of a and b

$1 \over x$ = $1 \over 6$   and   $1 \over y$ $-1 \over 4$

∴ x =6 and  y = –4

∴ (x, y) = (6, –4) is the required solution.

Linear equation in two variables August 04 , 2018 0 Comments 83 views