Answer:
(x, y) = (3/7, 1/7) or (3, 4)
Step-by-step explanation:
The second equation can be rearranged to ...
4x^2 +9y^2 = 15xy . . . . . . multiply by xy
4x^2 -15xy +9y^2 = 0 . . . subtract 15xy
(4x -3y)(x -3y) = 0 . . . . . . factor
This resolves to two linear equations:
4x -3y = 0
x -3y = 0
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Combined with the first equation, we get two systems of equations. We can solve both of them at once by substituting for y and using a stand-in for the coefficient of x.
y = (3x -1)/2 . . . . first equation solved for y
Cx -3(3x -1)/2) = 0 . . . . substitute for y in the 2nd/3rd equations (C = 1 or 4)
2Cx -9x +3 = 0 . . . . . multiply by 2
x(2C -9) = -3 . . . . . . .subtract 3
x = -3/(2C -9) = 3/(9 -2C) . . . . . remembering that C = 1 or 4
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For C=1, x = 3/(9 -2) = 3/7
y = (3(3/7) -1)/2 = 1/7
For C=4, x = 3/(9 -8) = 3
y = (3(3) -1)/2 = 4
The solutions are (x, y) = (3/7, 1/7) and (3, 4).
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Additional comment
We could have substituted for y in the quadratic form of the second equation. This would give a quadratic in x that could be solved in any of the usual ways.
