solve this by eliminated by substitute 2xy+3y=9; 3x+4y=5
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from eqn (1)
y(2x+3)=9
y=9/(2x+3)
put in. (2)
3x +4(1/(2x+3))=5
3x(2x+3)+4=5(2x+3)
solve this quadratic u will get x... and then put values of X in y=1/(2x+3)..u will get y
y(2x+3)=9
y=9/(2x+3)
put in. (2)
3x +4(1/(2x+3))=5
3x(2x+3)+4=5(2x+3)
solve this quadratic u will get x... and then put values of X in y=1/(2x+3)..u will get y
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