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Re: Lines 3x + 4ky + 6 = 0, and kx -3y +9 = 0 intersect in the second quad [#permalink]
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Expert Reply
The given equations are
3x + 4ky = -6 ------1
kx - 3y = -9 ------2
Try solving the two equations:
1. Multiply eq 1 by k, 3kx + 4k^2y = -6k ---------3
2. Multiply eq 2 by 3, 3kx - 9y = -27 ---------4
Solving eq, 3 and 4, we can get the value of y.
where y = (27-6k)/(4k^2 + 9)..
Since y should be > 0, 27-6k > 0, solving k<4.5
Similarly, solving for x gives or sub y in eq 2 , x = -6(6k+3) / 4k^2 + 9
since x < 0, 6k+3 should be > 0 => 6k+3 > 0, solving k < -0.5
Hence k has 5 Integral values 0, 1, 2, 3 and 4.
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Re: Lines 3x + 4ky + 6 = 0, and kx -3y +9 = 0 intersect in the second quad [#permalink]
Expert Reply
Not a GRE question.

Well could be but is quite hard to see such one on the actual exam
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Re: Lines 3x + 4ky + 6 = 0, and kx -3y +9 = 0 intersect in the second quad [#permalink]
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