solve if 5x^2-13xy+6y^2=0 then x:y is
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Answered by
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Heyy Buddy
Here's your Answer
Given:-
5x^2 - 13xy + 6y^2 = 0.
To find :-
x : y = ?
Find :-

=> x : y = 2 : 1 or 3 : 5.
✔✔✔
Here's your Answer
Given:-
5x^2 - 13xy + 6y^2 = 0.
To find :-
x : y = ?
Find :-
=> x : y = 2 : 1 or 3 : 5.
✔✔✔
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