Solve the following equations:
$x^{2} + 4 y^{2} - 15 x = 10 (3 y - 8), x y = 6$ .

(4, 5)

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(2, 3)

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(3, 2)

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(4, 3)

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Solution

The correct options are
B $(3, 2)$
C $(2, 3)$
Given equations are $x y = 6$
$\Rightarrow y = \frac{6}{x}$ ......(i)
and $x^{2} + 4 y^{2} - 15 x = 10 (3 y - 8)$
$\Rightarrow x^{2} + 4 y^{2} - 15 x - 30 y + 80 = 0$

Substituting value of $y$ in above equation

$x^{2} + 4 {(\frac{6}{x})}^{2} - 15 x - 30. \frac{6}{x} + 80 = 0 \Rightarrow x^{2} + \frac{144}{x^{2}} - 15 x - \frac{180}{x} + 80 = 0 \Rightarrow \frac{x^{4} + 144 - 15 x^{3} - 180 x + 80 x^{2}}{x^{2}} = 0 \Rightarrow x^{4} - 15 x^{3} + 80 x^{2} - 180 x + 144 = 0$

Solving by hit and trial method.

Put $x = 2$ , we have

$16 - 120 + 320 - 360 + 144 = 0$

Hence, $(x - 2)$ is the root of the equation.

$\Rightarrow (x - 2) (x^{3} - {12 x}^{2} + 54 x - 72) = 0 \Rightarrow x - 2 = 0$ .........(ii)
$\Rightarrow x^{3} - {13 x}^{2} + 54 x - 72 = 0$

Put $x = 3$

$27 - 117 + 162 - 72 = 0$

Hence, $(x - 3)$ is the root of the equation.

$(x - 3) (x^{2} - 10 x + 24) = 0 \Rightarrow x - 3 = 0$ ...........(iii)
$x^{2} - 10 x + 24 = 0 \Rightarrow x^{2} - 6 x - 4 x + 24 = 0 \Rightarrow x (x - 6) - 4 (x - 6) = 0 \Rightarrow (x - 4) (x - 6) = 0 \Rightarrow x = 4, 6$ ............(iv)

From (ii),(iii) and $i v$

$x = 2, 3, 4, 6$

Now from (i), we have

$y = \frac{6}{x} \Rightarrow y = 3, 2, \frac{3}{2}, 1$

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