The set of values of $x$ which satisfy $5x + 2 < 3x + 8$ and $\frac{x + 2}{x - 1} < 4$ is:

If $|{x^2} - x - 6| = x + 2$,then the values of $x$ are

If $\alpha$ and $\beta$ are the roots of the equation $x^{2}+px+2=0$ and $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ are the roots of the equation $2x^{2}+2qx+1=0$,then $\left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right)\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)$ is equal to:

Solve the given two equations and select the correct answer from the given options.
$I.$ $\sqrt{x} - \frac{(18)^{15/2}}{x^2} = 0$
$II.$ $\sqrt{y} = \frac{(19)^{9/2}}{y}$

If $ax^2 + bx + c = 0$ and $bx^2 + cx + a = 0$ have a common root and $a, b, c$ are non-zero real numbers,then $\frac{a^3 + b^3 + c^3}{abc} = $

If $x^{2}+y^{2}+z^{2}=xy+yz+zx$,then the value of $\frac{3x^{4}+7y^{4}+5z^{4}}{5x^{2}y^{2}+7y^{2}z^{2}+3z^{2}x^{2}}$ is