Question

If we are given two equations $\dfrac{x+y}{xy}=2$ and $\dfrac{x-y}{xy}=6$ then what is $y$ ?
A. $\dfrac{1}{2}$
B. $\dfrac{1}{4}$
C. $\dfrac{1}{6}$
D. $\dfrac{1}{8}$

Answer 1

Hint: We are given two equations $\dfrac{x+y}{xy}=2$ and $\dfrac{x-y}{xy}=6$, split $xy$ and simplify the equation. After that assume the value of $\dfrac{1}{x}$ and $\dfrac{1}{y}$ as $p$ and $q$ and simplify the equation. Try it, you will definitely get the value of $x$ and $y$.

Complete step-by-step answer:
We are given two equations $\dfrac{x+y}{xy}=2$ and $\dfrac{x-y}{xy}=6$.
Now let us split $xy$ so that we can simplify the equation in simple manner, we get,
$\dfrac{x}{xy}+\dfrac{y}{xy}=2$
Simplifying we get,
$\dfrac{1}{y}+\dfrac{1}{x}=2$ ……………..(1)
Now let us simplify another equation i.e. $\dfrac{x-y}{xy}=6$.
So simplifying in simple form we get,
$\dfrac{x}{xy}-\dfrac{y}{xy}=6$
Simplifying we get,
$\dfrac{1}{y}-\dfrac{1}{x}=6$ ……………..(2)
Now substituting $\dfrac{1}{x}=p$ and $\dfrac{1}{y}=q$ in both the equation (1) and (2).
In equation (1), we get,
$q+p=2$ ………….. (3)
and that of equation (2),
$q-p=6$ …………. (4)
Now adding equation (3) and (4) we get,
$2q=8$
so simplifying above, that is dividing the above equation by $2$.
$q=4$
Now substituting $q=4$ in equation (3) we get,
$4+p=2$
so we get the value of $p=-2$ .
Now substituting $p$ and $q$.
$\dfrac{1}{y}=q=4$ and $\dfrac{1}{x}=p=-2$
Now taking reciprocal we get,
$x=-\dfrac{1}{2}$ and $y=\dfrac{1}{4}$
Therefore, we get the values of $x$ and $y$ are $-\dfrac{1}{2}$ and $\dfrac{1}{4}$ respectively.
If we are given two equations $\dfrac{x+y}{xy}=2$ and $\dfrac{x-y}{xy}=6$ then $y=\dfrac{1}{4}$.
We get the correct answer as option (B).

Additional information:
We have used a substitution method to get a solution of the equation. Substitution method is as follows: First, solve one linear equation for $y$ in terms of $x$ . Then substitute that expression for $y$ in the other linear equation. You'll get an equation in $x$ . Solve this, and you have the $x$ -coordinate of the intersection. Then plug in $x$ to either equation to find the corresponding $y$-coordinate. (If it's easier, you can start by solving an equation for $x$ in terms of $y$, also – same difference!)

Note: While going through the problem you should understand that the prober assumption and substitution must be taken. As we took the substitution as $\dfrac{1}{x}=p$ and $\dfrac{1}{y}=q$ you can assume any value not that $p$ and $q$ is necessary. Also, while substituting solves the problem in step by step way so no confusion occurs.