Question & Answer

What is the value of $x$ and $y$ if $\dfrac{7x-2y}{xy}=5$ and $x=2y$.
(a) $x=\dfrac{12}{5}$, $y=\dfrac{6}{5}$
(b) $x=\dfrac{2}{5}$, $y=\dfrac{1}{5}$
(c) $x=9$, $y=\dfrac{9}{2}$
(d) $x=2$, $y=1$

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Hint: For solving this question we will use two given equations to find the value of $x$ and $y$ which satisfies both of the given equations. Then we will solve using the substitution method.

Complete step-by-step answer:
We have two equations:
  & \dfrac{7x-2y}{xy}=5............\left( 1 \right) \\
 & x=2y...............\left( 2 \right) \\
Now, directly substitute the value of $x$ in terms of $y$ from equation (2) into equation (1). Then,
  & \dfrac{7x-2y}{xy}=5 \\
 & \Rightarrow \dfrac{7\times 2y-2y}{2y\times y}=5 \\
 & \Rightarrow 14y-2y=10{{y}^{2}} \\
 & \Rightarrow 12y-10{{y}^{2}}=0 \\
 & \Rightarrow 2y\left( 6-5y \right)=0 \\
 & \Rightarrow y=0\text{ and }y=\dfrac{6}{5} \\
Now, from the above results first, we substitute the value of $y=0$ in equation (2) to get the value of $x$. Then,
  & x=2y \\
 & \Rightarrow x=0 \\
Now, we have $x=0$ and $y=0$ as one solution of the given equation. But there are no such options. So, as we have got two values of $y$ above. Now, we substitute $y=\dfrac{6}{5}$ in equation (2) to get the value of $x$ . Then,
  & x=2y \\
 & \Rightarrow x=\dfrac{12}{5} \\
Now, we have one more solution for the given set of equations and that is $x=\dfrac{12}{5}$ and $y=\dfrac{6}{5}$ .
Now, after going through options we can say that option (a) is matching with our answer.
Hence, option (a) is the correct option.

Note: Although the question is very easy to solve, students should do the correct substitution and avoid making the calculation mistakes while solving the question. There can be one other method also by which we can solve such problems, even that question in which equations are tough to solve. But we need options to be there, then only we can solve it using that method and that method is to put the value of variables directly from the options and check whether it is correct or not.