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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