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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 52 views