In the following question two equations are given in variables x and y. You have to solve these equations and determine relation between x and y. I. \(\frac{9}{{\sqrt x }} + \frac{{19}}{{\sqrt x }} = \sqrt x\) II. \({y^5} - \frac{{{{\left( {2 \times 14} \right)}^{\frac{{11}}{2}}}}}{{\sqrt y }} = 0\)
In the following question two equations are given in variables x and y. You have to solve these equations and determine relation between x and y.
I. \(\frac{9}{{\sqrt x }} + \frac{{19}}{{\sqrt x }} = \sqrt x\)
II. \({y^5} - \frac{{{{\left( {2 \times 14} \right)}^{\frac{{11}}{2}}}}}{{\sqrt y }} = 0\)
1). x > y
2). x < y
3). x ≥ y
4). x ≤ y
1 answers
I. $(\frac{9}{{\sqrt x }} + \frac{{19}}{{\sqrt x }} = \sqrt x)$
$(\begin{array}{l} \Rightarrow \;\frac{{9\; + \;19}}{{\sqrt x }} = \sqrt x \\ \Rightarrow \;\frac{{28}}{{\sqrt x }} = \sqrt x \\ \Rightarrow \;28\; = \;x \end{array})$
Then, x = 28
II. $({y^5} - \frac{{{{\left( {2 \times 14} \right)}^{\frac{{11}}{2}}}}}{{\sqrt y }} = 0)$
We know, $(\sqrt y = {\left( y \right)^{\frac{1}{2}}})$
$(\begin{array}{l} \Rightarrow \;{y^5} = \frac{{{{\left( {2 \times 14} \right)}^{\frac{{11}}{2}}}}}{{\sqrt y }}\\ \Rightarrow \;{y^5} + {\left( y \right)^{\frac{1}{2}}} = {\left( {2 \times 14} \right)^{\frac{{11}}{2}}} \end{array})$
As per the rule of exponents, $(\;{a^x} \times {a^y} = {a^{x + y}})$
$(\begin{array}{l} \Rightarrow \;{\left( y \right)^{5 + \frac{1}{2}}} = {\left( {2 \times 14} \right)^{\frac{{11}}{2}}}\\ \Rightarrow \;{\left( y \right)^{\frac{{11}}{2}}} = {\left( {28} \right)^{\frac{{11}}{2}}} \end{array})$
Then, y = 28
So, when x = 28, x = y for y = 28
So, we can observe that x = y.