A) \[x=3,\,y=15\]
B) \[x=15,\,y=3\]
C) \[x=\frac{1}{15},\,y=\frac{1}{3}\]
D) \[x=\frac{1}{3},\,y=\frac{1}{15}\]
Correct Answer: D
Solution :
Given equations are |
\[x+4y=27xy\]and \[x+2y=21xy\] |
On dividing both sides of the above equations by xy, we get |
\[\frac{1}{y}+\frac{4}{x}=27\] and \[\frac{1}{y}+\frac{2}{x}=21\] |
On putting \[\frac{1}{y}=u\]and \[\frac{1}{x}=v\], we get |
\[u+4v=27\] ...(i) |
and \[u+2v=21\] ...(ii) |
On subtracting Eq. (ii) from Eq. (i), we get |
\[2v=6\Rightarrow v=3\] |
On putting the value of v in Eq. (i), we get |
\[u+12=27\Rightarrow u=15\] |
Now, \[v=3\Rightarrow \frac{1}{x}=3\Rightarrow x=\frac{1}{3}\] |
and \[u=15\Rightarrow \frac{1}{y}=15\Rightarrow y=\frac{1}{15}\] |
Hence, \[x=\frac{1}{3}\]and \[y=\frac{1}{15}\]is the required solution. |
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