Question

Solve the following pair of simultaneous equations: $\dfrac{1}{x} + \dfrac{1}{y} = 5;\dfrac{1}{x} - \dfrac{1}{y} = 1$
\[A)x = \dfrac{3}{2},y = \dfrac{3}{4}\]
\[B)x = \dfrac{1}{2},y = \dfrac{2}{3}\]
\[C)x = \dfrac{1}{3},y = \dfrac{1}{2}\]
\[D)x = \dfrac{2}{3},y = \dfrac{1}{2}\]

Answer 1

Hint: Here we are asked to solve the given simultaneous equation. Since the unknown variables are all in the denominator, we will use the substitution method to make the variables easier to solve. We can substitute $\dfrac{1}{x}\& \dfrac{1}{y}$ as any other different temporary variable to make the equation to simpler form then after solving those equation for temporary variables re-substitute them to get the values of the original variables. After that, we will compare our answer with the given options to find the right option.

Complete answer:
Since given that $\dfrac{1}{x} + \dfrac{1}{y} = 5;\dfrac{1}{x} - \dfrac{1}{y} = 1$ and now assume that $\dfrac{1}{x} = u,\dfrac{1}{y} = v$ then substitute the values in the given we have $u + v = 5;u - v = 1$
Now using the addition, we add the both values, then we get $u + v + u - v = 5 + 1$
Using the subtraction, the equal values with opposite signs cancel each other, then we get $u + u = 5 + 1$ and again by the addition we get $2u = 6$
Hence by the division, we have $u = \dfrac{6}{2} = 3$
Now substitute the value in the equation $u - v = 1$ then we get $3 - v = 1 \Rightarrow v = 2$
Again $\dfrac{1}{x} = u,\dfrac{1}{y} = v$ back apply the values we get $\dfrac{1}{x} = 3,\dfrac{1}{y} = 3$
Taking the reciprocal, we get \[x = \dfrac{1}{3},y = \dfrac{1}{2}\]
Now let us compare our answer with the given options, as we can see that no options is matching with our answer other than the option (c).

Therefore, the option \[C)x = \dfrac{1}{3},y = \dfrac{1}{2}\] is correct.

Note:
Mistakes we may commit using substitution method: one can get confused with the original variable and temporary variables, after getting the values of temporary variable we may forgot to re-substitute it to find the original variable. Thus, we must be very careful in while doing the substitution. Solving the equation can be done by any method available here we have used the elimination method to solve the equation.