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# Solve the following equation:${x^2} + xy + xz = 18, \\ {y^2} + yz + yz + 12 = 0, \\ {z^2} + zx + zy = 30. \\ \\ {\mathbf{A}}.\,x = \pm 2,y = \mp 2,z = \pm 4 \\ {\mathbf{B}}.\,x = \pm 3,y = \mp 2,z = \pm 5 \\ {\mathbf{C}}.\,\,x = \pm 4,y = \pm 5,z = \pm 5 \\ {\mathbf{D}}.\,{\text{None of the above}} \\$

Last updated date: 29th May 2024
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Hint: In order to solve this question, we have to solve the three equations given in the question. We can take something common then solve it by eliminating or dividing.

The given equations are,
${x^2} + xy + xz = 18$,
Taking $x$ as common from LHS we get,
$x(x + y + z) = 18\,\,\,\,\,\,...({\text{i}}) \\ \\ {y^2} + yz + yz + 12 = 0 \\$
Taking $y$ as common from LHS we get,
$y(x + y + z) = - 12\,\,\,\,\,\,\,\,\,...({\text{ii}})$
${z^2} + zx + zy = 30$
Taking $z$ as common from LHS we get,
$z(x + y + z) = 30\,\,\,\,\,\,\,\,\,...({\text{iii}})$
Dividing (i) by (ii) we get,
$\dfrac{x}{y} = \dfrac{{ - 18}}{{12}} = \dfrac{{ - 3}}{2} \\ y = - \dfrac{2}{3}x\;\;{\text{ }}\;.......\left( a \right) \\$
Dividing (i) by (iii) we get,
$\dfrac{x}{z} = \dfrac{{18}}{{30}} = \dfrac{3}{5}$
Then,
$z = \dfrac{5}{3}x\,\,\,\,\,\,......(b)$
Substituting $(a)$ and $(b)$ in (i) we get,
$x\left( {x - \dfrac{2}{3}x + \dfrac{5}{3}x} \right) = 18$
On solving above we get,
$2{x^2} = 18 \\ {x^2} = 9 \\$
Therefore either $x = 3{\text{ }}$ or $x = - 3$
That is $x = \pm 3$
On putting the value of $x$ in $(a)$ and $(b)$ we get,
$y = - \dfrac{2}{3}( \pm 3) = \mp 2 \\ \\ z = \dfrac{5}{3}( \pm 3) = \pm 5 \\$
Therefore,
$x = \pm 3,y = \mp 2,z = \pm 5$
Hence the correct option is B.

Note: In this question we have taken $x,y,z$ common from the equation ${\text{(i),(ii),(iii)}}$ then we got various equations which can be used to find the values of $x,y,z$ 5as done above . Therefore we can find the values of various variables . We can multiply, divide, add or subtract between these equations to get the desired results.