Question

Solve each of the following pairs of equations by reducing them to a pair of linear equations:
(i) \[\dfrac{5}{x-1}+\dfrac{1}{y-2}=2\]
     \[\dfrac{6}{x-1}-\dfrac{3}{y-2}=1\]
(ii) \[\dfrac{x+y}{xy}=2\]
     \[\dfrac{x-y}{xy}=6\]
(iii) \[\dfrac{2}{\sqrt{x}}+\dfrac{3}{\sqrt{y}}=2\]
      \[\dfrac{4}{\sqrt{x}}-\dfrac{9}{\sqrt{y}}=-1\]
(iv) \[6x+3y=6xy\]
       \[2x+4y=5xy\]

Answer 1

Hint: There are four parts in question so we will solve each part separately. To reduce the equations into a pair of linear equations, we will put a term equal to a variable such that after doing this, we will get a pair of linear equations. Then we will solve these pairs of linear equations by method of substitution.

Complete step-by-step answer:
In the question, there are four parts, so we will solve each part one by one. To reduce the equations into a pair of linear equations, we will first take terms that are common to both the equation and then we will equate these terms to some other variables. After doing this, we will solve these equations with the help of a substitution method. After finding the values of assumed variables, we will put back these variables equal to the term and find the values of x and y.

(i) \[\dfrac{5}{x-1}+\dfrac{1}{y-2}=2\]
     \[\dfrac{6}{x-1}-\dfrac{3}{y-2}=1\]
Here we can see that the terms \[\left( \dfrac{1}{x-1} \right)\] and \[\left( \dfrac{1}{y-2} \right)\] are common to both the equations. So, we will assume that, \[\dfrac{1}{x-1}=a\] and \[\dfrac{1}{y-2}=b\]. Thus, the new equations we will get are:
\[5a+b=2\] - (1)
\[6a-3b=1\]- (2)
From (1), we can say that:
\[\begin{align}
  & 5a+b=2 \\
 & \Rightarrow b=2-5a \\
\end{align}\]
Now, we will put the value of b in equation (2). Thus, we will get:
\[\begin{align}
  & 6a-3\left( 2-5a \right)=1 \\
 & \Rightarrow 6a-6+15a=1 \\
 & \Rightarrow 6a+15a=1+6 \\
 & \Rightarrow 21a=7 \\
 & \Rightarrow a=\dfrac{7}{21} \\
\end{align}\]
\[\Rightarrow a=\dfrac{1}{3}\] - (3)
Now, we will put this value of a in equation (2).
Thus, we will get:
\[\begin{align}
  & 5\left( \dfrac{1}{3} \right)+b=2 \\
 & \Rightarrow b=2-5\left( \dfrac{1}{3} \right) \\
 & \Rightarrow b=2-\dfrac{5}{3} \\
 & \Rightarrow b=\dfrac{6-5}{3} \\
\end{align}\]
\[\Rightarrow b=\dfrac{1}{3}\] - (4)
Now, the value of a is \[\dfrac{1}{x-1}\]. Therefore:
\[\begin{align}
  & a=\dfrac{1}{x-1} \\
 & x-1=\dfrac{1}{a} \\
 & \Rightarrow x=1+\dfrac{1}{a} \\
\end{align}\]
Now, we will put the value of the a in above equation:
\[\begin{align}
  & \Rightarrow x=1+\dfrac{1}{\left( \dfrac{1}{3} \right)} \\
 & \Rightarrow x=1+3 \\
 & \Rightarrow x=4 \\
\end{align}\]
Now, we had assumed that:
\[\begin{align}
  & \dfrac{1}{y-2}=b \\
 & \Rightarrow y-2=\dfrac{1}{b} \\
 & \Rightarrow y=2+\dfrac{1}{b} \\
\end{align}\]
Now, we will put the value of b from (4) to the above equation.
\[\begin{align}
  & \Rightarrow y=2+\dfrac{1}{\left( \dfrac{1}{3} \right)} \\
 & \Rightarrow y=2+3 \\
 & \Rightarrow y=5 \\
\end{align}\]
Thus, the solutions of the equations are x = 4 and y = 5.

(ii) \[\dfrac{x+y}{xy}=2\]
    \[\begin{align}
  & \Rightarrow \dfrac{x}{xy}+\dfrac{y}{xy}=2 \\
 & \Rightarrow \dfrac{1}{y}+\dfrac{1}{x}=2 \\
 & \dfrac{x-y}{xy}=6 \\
 & \Rightarrow \dfrac{x}{xy}-\dfrac{y}{xy}=6 \\
 & \Rightarrow \dfrac{1}{y}-\dfrac{1}{x}=6 \\
\end{align}\]
Here, we can see that the terms \[\left( \dfrac{1}{y} \right)\] and \[\left( \dfrac{1}{x} \right)\] are common to both. So, I will assume \[\dfrac{1}{x}=a\] and \[\dfrac{1}{y}=b\]. Thus, the new equations formed are:
\[b+a=2\] - (1)
\[b-a=6\] - (2)
From (2), we can say that:
\[\begin{align}
  & b-a=6 \\
 & \Rightarrow b=a+6 \\
\end{align}\]
Now, we will put this value of b in equations (1). Thus, we will get:
\[\begin{align}
  & a+b+a=2 \\
 & 2a=2-b \\
 & 2a=-4 \\
\end{align}\]
\[\Rightarrow a=-2\] - (3)
Now, we had assumed earlier that:
\[\begin{align}
  & \dfrac{1}{x}=a \\
 & \Rightarrow x=\dfrac{1}{a} \\
\end{align}\]
On putting the values of a from (3) to above equation, we will get:
\[\begin{align}
  & \Rightarrow x=\dfrac{1}{-2} \\
 & \Rightarrow x=-\dfrac{1}{2} \\
\end{align}\]
Now, we will put the value of a from (3) to (1):
\[\Rightarrow b-2=2\]
\[\Rightarrow b=4\] - (4)
We had assumed earlier that:
\[\begin{align}
  & \Rightarrow \dfrac{1}{y}=b \\
 & \Rightarrow y=\dfrac{1}{b} \\
\end{align}\]
On putting the value of b (4) to above equation, we will get:
\[\Rightarrow y=\dfrac{1}{4}\]
Thus, the solution of the equations is \[x=\dfrac{-1}{2}\] and \[y=\dfrac{1}{4}\].

(iii) \[\dfrac{2}{\sqrt{x}}+\dfrac{3}{\sqrt{y}}=2\]
      \[\dfrac{4}{\sqrt{x}}-\dfrac{9}{\sqrt{y}}=-1\]
Here, we will assume \[\dfrac{1}{\sqrt{x}}=a\] and \[\dfrac{1}{\sqrt{y}}=b\]. Thus, the new equations formed will be:
\[2a+3b=2\] - (1)
\[4a-9b=-1\] - (2)
Now, from equation (1), we have:
\[\begin{align}
  & 2a=2-3b \\
 & \Rightarrow a=\dfrac{2-3b}{2} \\
\end{align}\]
We will put this value of a in equation (2). Thus, we will get:
\[\begin{align}
  & \Rightarrow 4\left( \dfrac{2-3b}{2} \right)-9b=-1 \\
 & \Rightarrow 2\left( 2-3b \right)-9b=-1 \\
 & \Rightarrow 4-6b-9b=-1 \\
 & \Rightarrow 6b+9b=4+1 \\
 & \Rightarrow 15b=5 \\
 & \Rightarrow b=\dfrac{5}{15} \\
\end{align}\]
\[\Rightarrow b=\dfrac{1}{3}\] - (3)
We had assumed that \[\dfrac{1}{\sqrt{y}}=b\].
So, \[\sqrt{y}=\dfrac{1}{b}\]
On squaring both sides, we will get:
\[\begin{align}
  & {{\left( \sqrt{y} \right)}^{2}}={{\left( \dfrac{1}{b} \right)}^{2}} \\
 & \Rightarrow y=\dfrac{1}{{{b}^{2}}} \\
\end{align}\]
We will now put the value of b from (3) to above equation:
\[\begin{align}
  & \Rightarrow y=\dfrac{1}{{{\left( \dfrac{1}{3} \right)}^{2}}} \\
 & \Rightarrow y=\dfrac{1}{\left( \dfrac{1}{9} \right)} \\
 & \Rightarrow y=9 \\
\end{align}\]
Now, we will put the value of b from (3) to (1):
\[\begin{align}
  & \Rightarrow 2a+3\left( \dfrac{1}{3} \right)=2 \\
 & \Rightarrow 2a+1=2 \\
 & \Rightarrow 2a=1 \\
 & \Rightarrow a=\dfrac{1}{2} \\
\end{align}\]
We had assumed that, \[\dfrac{1}{\sqrt{x}}=a\]
\[\sqrt{x}=\dfrac{1}{a}\]
On squaring both sides, we will get:
\[\begin{align}
  & \Rightarrow x=\dfrac{1}{{{a}^{2}}} \\
 & \Rightarrow x=\dfrac{1}{{{\left( \dfrac{1}{2} \right)}^{2}}} \\
 & \Rightarrow x=\dfrac{1}{\left( \dfrac{1}{4} \right)} \\
 & \Rightarrow x=4 \\
\end{align}\]
Thus, the solution of the equations is \[x=4\] and \[y=9\].

(iv) \[6x+3y=6xy\]
\[\begin{align}
  & \Rightarrow \dfrac{6x+3y}{xy}=6 \\
 & \Rightarrow \dfrac{6}{y}+\dfrac{3}{x}=6 \\
 & \Rightarrow \dfrac{2}{y}+\dfrac{1}{x}=2 \\
 & 2x+4y=5xy \\
 & \Rightarrow \dfrac{2x+4y}{xy}=5 \\
 & \dfrac{2}{y}+\dfrac{4}{x}=5 \\
\end{align}\]
From (1), we have:
\[\begin{align}
  & \Rightarrow 2b+a=2 \\
 & \Rightarrow a=2-2b \\
\end{align}\]
We will put this value of a in (2). Thus, we have:
\[\begin{align}
  & 2b+4\left( 2-2b \right)=5 \\
 & 2b+8-8b=5 \\
 & -6b=5-8 \\
 & -6b=-3 \\
 & \Rightarrow b=\dfrac{-3}{-6} \\
\end{align}\]
\[\Rightarrow b=\dfrac{1}{2}\] - (3)
We had assumed that, \[\dfrac{1}{y}=b\]
\[\Rightarrow y=\dfrac{1}{b}\]
\[\Rightarrow y=\dfrac{1}{\left( \dfrac{1}{2} \right)}\] (from (3))
\[\Rightarrow y=2\]
Now, we will put the value of b from (3) to (1):
\[\begin{align}
  & 2\left( \dfrac{1}{2} \right)+a=2 \\
 & 1+a=2 \\
\end{align}\]
\[\Rightarrow a=1\] - (4)
We had assumed that, \[\dfrac{1}{x}=a\]
\[\Rightarrow x=\dfrac{1}{a}\]
From (4), we will put the value of a in above equation:
\[\Rightarrow x=1\]
Thus, the solution of the equations is \[x=1\] and \[y=2\].

Note: We have solved the pair of linear equations by the method of substitution. Instead, we can also use other methods like elimination method, graphical method and cross multiplication. No matter what method we use to calculate the values of x and y, these values will always be the same for any method.