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Question

Solve the following systems of equations.
2xy3xy=15,xy+xy=15

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Solution

Solve the following systems of equations,
2xy3xy=15...(i)
xy+xy=15....(ii)
To solve these equations, put xy=a and xy=b
Then, eqn (i) 2a3b=15...(iii)
eqn (ii) a+b=15...(iv)
Solve for a & b.
Huetiplying eqn (iv) with 2.
2a+2b=30
we have, 2a3b=15(), subtracting these.
_________________
0+5b=15
b=15/5=3
Substituting is eqn (iv), we get a=153=12
Thus, a=xy=12 and b=xy=3
AS xy=3,x=3y
Also, xy=12(3y)×y=12
3y2=12
y2=4
y=±2
As, x=3y,x=±6

1172922_885968_ans_b0c39a8181a44d1589c0ae187d05d115.jpg

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