The value of x and y of the following pairs of equations by reducing them to a pair of linear equations is |
\[\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2;\,\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1\] |
A) \[x=4,\text{ }y=9\]
B) \[x=2,\,\,y=3\]
C) \[x=8,\,\,y=18\]
D) None of the above
Correct Answer: A
Solution :
Given pair of equations is |
\[\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2\] and \[\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1\] |
On putting \[\frac{1}{\sqrt{x}}=u\]and \[\frac{1}{\sqrt{y}}=v\]in the given equations, we get |
\[2u+3v=2\] ...(i) |
and \[4u-9v=-1\] ...(ii) |
On multiplying Eq. (i) by 3 and then adding both of them, we get |
\[3\left( 2u+3v \right)+4u-9v=2\times 3-1\] |
\[\Rightarrow \,\,6u+4u=5\Rightarrow 10u=5\Rightarrow u=\frac{1}{2}\] |
On putting \[u=\frac{1}{2}\]in Eq. (i), we get |
\[2\times \frac{1}{2}+3v=2\,\,\,\Rightarrow \,\,\,\,3v=2-1\,\,\Rightarrow v=\frac{1}{3}\] |
If \[u=\frac{1}{2},\]then by \[u=\frac{1}{\sqrt{x}}\], we get |
\[\sqrt{x}=\frac{1}{u}\] |
\[\Rightarrow \,\,\,\,\,\sqrt{x}=2\] |
\[\Rightarrow \,\,\,\,x=4\] [squaring both sides] |
If \[v=\frac{1}{3}\], then by \[v=\frac{1}{\sqrt{y}}\], we get |
\[\sqrt{y}=\frac{1}{v}\Rightarrow \sqrt{y}=3\Rightarrow y=9\] |
[squaring both sides] |
\[\therefore \] \[x=4\]and y = 9 is the required solution. |
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