Solve the following equations :
$2 x y - 4 x + y = 17, 3 y z + y - 6 z = 52, 6 x z + 3 z + 2 x = 29.$

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Solution

The equations can be rewritten as
$2 x (y - 2) + (y - 2) = 17 - 2$
or $(2 x + 1) (y - 2) = 15$
Similarly $(y - 2) (3 z + 1) = 50$ and $(3 z + 1) (2 x + 1) = 30$
Multiplying $(1), (2)$ and $(3)$ , we get
${[(2 x + 1) (y - 2) (3 z + 1)]}^{2} = 15 \times 50 \times 30$
$∴ (2 x + 1) (y - 2) (3 z + 1) = \pm 150$
Now from $(1)$ and $(4)$ ,
$3 z + 1 = \pm \frac{150}{15} = \pm 10$ so that $z = 3, - \frac{11}{3}$ .
From $(2)$ and $(4)$ ,
$2 x + 1 = \pm \frac{150}{50} = \pm 3$ or $z = 1, - 2$
From $(3)$ and $(4)$ ,
$y - 2 = \pm \frac{150}{30} = \pm 5$ or $z = 7, - 3$
Hence the solution sets are
$x = 1, y = 7, z = 3$ or $x = - 2, y = - 3, z = - \frac{11}{3}$ .

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