148x + 231y = 527xy and 231x + 149y = 610xy.

1.	148x + 231y = 527xy and 231x + 149y = 610xy.
Answer» 148x + 231y = 527xy ......(1) 231x + 148y = 610xy ......(2) On adding both equations, we get 379x + 379y = 1137xy ⇒ x + y = 3xy .......(3) Subtract equation (2) from (1), we get 83x - 83y = 83xy ⇒ x - y = xy .......(4) From (3), \(\frac1y + \frac1x = 3\) .....(5) From (4), \(\frac1y - \frac1x = 1\) ......(6) On adding equations (5) & (6), we get \(\frac2y=4\) ⇒ \(y = \frac24 = \frac12\) \(\therefore 1x = 3 -\frac1y\) = 3 - 2 = 1 ⇒ x = 1 Hence, solution of given system of equations is x = 1, y = \(\frac12\)

Answer»

148x + 231y = 527xy ......(1)

231x + 148y = 610xy ......(2)

On adding both equations, we get

379x + 379y = 1137xy

⇒ x + y = 3xy .......(3)

Subtract equation (2) from (1), we get

83x - 83y = 83xy

⇒ x - y = xy .......(4)

From (3),

\(\frac1y + \frac1x = 3\) .....(5)

From (4),

\(\frac1y - \frac1x = 1\) ......(6)

On adding equations (5) & (6), we get

\(\frac2y=4\)

⇒ \(y = \frac24 = \frac12\)

\(\therefore 1x = 3 -\frac1y\)

= 3 - 2 = 1

⇒ x = 1

Hence, solution of given system of equations is x = 1, y = \(\frac12\)

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