Question

Solve the following systems of equations:
$
  \dfrac{5}{{x - 1}} + \dfrac{1}{{y - 2}} = 2 \\
  \dfrac{6}{{x - 1}} - \dfrac{3}{{y - 2}} = 1 \\
$

Answer 1

Hint: In this question let $\dfrac{1}{{x - 1}} = a{\text{ and }}\dfrac{1}{{y - 2}} = b$ and simplify this to get two linear equations in variables of a and b, then solve to get the value of a and later substitute it back into one of the equation obtained between a and b to get the value of b. Then since x is earlier computed in terms of a and y is computed in terms of b only, so when a is known and b is known hence x and y can be taken out.

Complete step-by-step answer:
 Let,
$\dfrac{1}{{x - 1}} = a$.................. (1)
And
$\dfrac{1}{{y - 2}} = b$..................... (2)
So the system of equation becomes
$ \Rightarrow 5a + b = 2$................... (3)
$ \Rightarrow 6a - 3b = 1$................... (4)
Now multiply by 3 in equation (3) and add in equation (4) we have,
$ \Rightarrow 3\left( {5a + b} \right) + \left( {6a - 3b} \right) = 3\left( 2 \right) + 1$
Now simplify the above equation we have,
$ \Rightarrow 15a + 3b + 6a - 3b = 6 + 1$
$ \Rightarrow 21a = 7$
$ \Rightarrow a = \dfrac{7}{{21}} = \dfrac{1}{3}$
Now substitute the value of (a) in equation (3) we have,
$ \Rightarrow 5\left( {\dfrac{1}{3}} \right) + b = 2$
$ \Rightarrow b = 2 - \dfrac{5}{3} = \dfrac{{6 - 5}}{3} = \dfrac{1}{3}$
Now substitute the value of (a) and (b) in equation (1) and (2) we have,
$ \Rightarrow \dfrac{1}{{x - 1}} = \dfrac{1}{3}{\text{ and }}\dfrac{1}{{y - 2}} = \dfrac{1}{3}$
$ \Rightarrow x - 1 = 3$
$ \Rightarrow x = 3 + 1 = 4$
And
$ \Rightarrow y - 2 = 3$
$ \Rightarrow y = 3 + 2 = 5$
So the solution of the given system of equations is
$ \Rightarrow \left( {x,y} \right) = \left( {4,5} \right)$
So this is the required answer.

Note – The trick point here was the substitution in the starting phase of the solution the reason behind it was if this would not have done we would be getting a two different equations in terms of some xy terms, which can’t be solved to get the values of x and y, that why the aim was to break the problem into smaller subdivision.