In the following questions,two equations numb | vedclass

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(C) Step $1$: Solve equation $I$.$\frac{13}{\sqrt{x}} + \frac{9}{\sqrt{x}} = \sqrt{x}$$\frac{13 + 9}{\sqrt{x}} = \sqrt{x}$$\frac{22}{\sqrt{x}} = \sqrt

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$x < y$

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(C) Step $1$: Solve equation $I$.$\frac{13}{\sqrt{x}} + \frac{9}{\sqrt{x}} = \sqrt{x}$$\frac{13 + 9}{\sqrt{x}} = \sqrt{x}$$\frac{22}{\sqrt{x}} = \sqrt{x}$$22 = \sqrt{x} 	imes \sqrt{x}$$x = 22$Step $2$: Solve equation $II$.$y^4 - \frac{(26)^{9/2}}{\sqrt{y}} = 0$$y^4 = \frac{(26)^{9/2}}{y^{1/2}}$$y^4 	imes y^{1/2} = (26)^{9/2}$$y^{4 + 0.5} = (26)^{9/2}$$y^{9/2} = (26)^{9/2}$Since the exponents are equal and odd,the bases must be equal.$y = 26$Step $3$: Compare $x$ and $y$.$x = 22$ and $y = 26$.Therefore,$x < y$.

In the following questions,two equations numbered $I$ and $II$ are given. You have to solve both the equations and give the answer.
$I: \frac{13}{\sqrt{x}} + \frac{9}{\sqrt{x}} = \sqrt{x}$
$II: y^4 - \frac{(13 \times 2)^{9/2}}{\sqrt{y}} = 0$

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In the following questions,two equations numbered $I$ and $II$ are given. You have to solve both the equations and give the answer.$I: \frac{13}{\sqrt{x}} + \frac{9}{\sqrt{x}} = \sqrt{x}$$II: y^4 - \frac{(13 \times 2)^{9/2}}{\sqrt{y}} = 0$