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Solve the followingsystem of equations:`2/x+3/y=9/(x y), 4/x+9/y=(21)/(x y), `where `x!=0, y!=0` |
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Answer» Multiplying each equation throughout by xy, we get ` 2y + 3x = 9 " " `… (i) `4y + 9x = 21 " " `… ( ii) Multiplying (i) by 3 and subtracting (ii) from the result, we get ` ( 6- 4 ) y = ( 27 - 21 ) rArr 2y = 6 rArr y = 3 `. Putting ` y = 3 ` in (i), we get ` (2 xx 3) + 3x = 9 rArr 6 + 3x = 9 rArr 3x = 3 rArr x= 1 ` Hence, ` x = 1 and y = 3`. |
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