| The value of x and y of the following pair of equation is |
| \[\frac{2}{x}+\frac{3}{y}=13;\,\frac{5}{x}-\frac{4}{y}=-2\] |
A) \[x=2,\,y=3\]
B) \[x=\frac{1}{3},\,y=\frac{1}{2}\]
C) \[x=\frac{1}{2},\,y=\frac{1}{3}\]
D) \[x=3,\,y=2\]
Correct Answer: C
Solution :
| Given, pair of equations is not linear. |
| \[\therefore \] Put \[\frac{1}{x}=p\]and \[\frac{1}{y}=q\], we get |
| \[2p+3q=13\] ...(i) |
| and \[5p-4q=-2\] ...(ii) |
| which is a pair of linear equations. |
| Now, on multiplying Eq. (i) by 4 and Eq. (ii) by 3 and then adding both of them, we get |
| \[\left( 8p+12q \right)+\left( 15p-12q \right)=52-6\] |
| \[\Rightarrow \,\,\,8p+15p=52-6\Rightarrow 23p=46\] |
| \[\Rightarrow \,\,\,p=\frac{46}{23}=2\] |
| On putting p = 2 in Eq. (i), we get |
| \[2\left( 2 \right)+3q=13\] |
| \[\Rightarrow \,\,3q=13-4=9\,\,\Rightarrow q=\frac{9}{3}=3\] |
| Since, \[p=\frac{1}{x}\]and \[q=\frac{1}{y}\] |
| \[\therefore \,\,\,\,\,\,\,2=\frac{1}{x}\] and \[3=\frac{1}{y}\] |
| \[\Rightarrow x=\frac{1}{2}\] and \[y=\frac{1}{3}\], which is the required solution. |
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