Question

Solve the following equations:
${{\left( 3-\dfrac{6y}{x+y} \right)}^{2}}+{{\left( 3+\dfrac{6y}{x-y} \right)}^{2}}=82$ , $3x+7y=26$ .

Answer 1

Hint: For solving this problem first we will simplify the term on the left-hand side in the equation ${{\left( 3-\dfrac{6y}{x+y} \right)}^{2}}+{{\left( 3+\dfrac{6y}{x-y} \right)}^{2}}=82$ and then assume $\dfrac{x-y}{x+y}=t$ . After that, we will find the values of $t$ which will be further used to find the suitable values of $x$ and $y$ from the given equations.

Complete step-by-step solution -
Given:
We have to find the value of $x$ and $y$ from the following two equations:
$\begin{align}
  & 3x+7y=26................................................\left( 1 \right) \\
 & {{\left( 3-\dfrac{6y}{x+y} \right)}^{2}}+{{\left( 3+\dfrac{6y}{x-y} \right)}^{2}}=82..................\left( 2 \right) \\
\end{align}$
Now, we will simplify the term on the left-hand side in the equation (2). Then,
$\begin{align}
  & {{\left( 3-\dfrac{6y}{x+y} \right)}^{2}}+{{\left( 3+\dfrac{6y}{x-y} \right)}^{2}} \\
 & \Rightarrow {{\left( \dfrac{3x+3y-6y}{x+y} \right)}^{2}}+{{\left( \dfrac{3x-3y+6y}{x-y} \right)}^{2}} \\
 & \Rightarrow {{\left( \dfrac{3x-3y}{x+y} \right)}^{2}}+{{\left( \dfrac{3x+3y}{x-y} \right)}^{2}} \\
 & \Rightarrow 9{{\left( \dfrac{x-y}{x+y} \right)}^{2}}+9{{\left( \dfrac{x+y}{x-y} \right)}^{2}} \\
\end{align}$
Now, let $\dfrac{x-y}{x+y}=t$ . Then,
\[\begin{align}
  & {{\left( 3-\dfrac{6y}{x+y} \right)}^{2}}+{{\left( 3+\dfrac{6y}{x-y} \right)}^{2}} \\
 & \Rightarrow 9{{\left( \dfrac{x-y}{x+y} \right)}^{2}}+9{{\left( \dfrac{x+y}{x-y} \right)}^{2}} \\
\end{align}\]
\[\Rightarrow 9{{t}^{2}}+\dfrac{9}{{{t}^{2}}}\]
Now, as it is given that ${{\left( 3-\dfrac{6y}{x+y} \right)}^{2}}+{{\left( 3+\dfrac{6y}{x-y} \right)}^{2}}=82$ . Then,
\[\begin{align}
  & {{\left( 3-\dfrac{6y}{x+y} \right)}^{2}}+{{\left( 3+\dfrac{6y}{x-y} \right)}^{2}}=82 \\
 & \Rightarrow 9{{t}^{2}}+\dfrac{9}{{{t}^{2}}}=82 \\
 & \Rightarrow 9{{t}^{4}}+9=82{{t}^{2}} \\
 & \Rightarrow 9{{t}^{4}}-82{{t}^{2}}+9=0 \\
\end{align}\]
Now, we will solve this equation by splitting the middle term method. Then,
\[\begin{align}
  & 9{{t}^{4}}-82{{t}^{2}}+9=0 \\
 & \Rightarrow 9{{t}^{4}}-81{{t}^{2}}-{{t}^{2}}+9=0 \\
 & \Rightarrow 9{{t}^{2}}\left( {{t}^{2}}-9 \right)-\left( {{t}^{2}}-9 \right)=0 \\
 & \Rightarrow \left( {{t}^{2}}-9 \right)\left( 9{{t}^{2}}-1 \right)=0 \\
\end{align}\]
Now, as we know that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ . Then,
\[\begin{align}
  & \left( {{t}^{2}}-9 \right)\left( 9{{t}^{2}}-1 \right)=0 \\
 & \Rightarrow \left( t+3 \right)\left( t-3 \right)\left( 3t+1 \right)\left( 3t-1 \right)=0 \\
 & \Rightarrow t=\pm 3,\pm \dfrac{1}{3} \\
\end{align}\]
Now, we got four values of $t$ so, we will calculate the value of $x$ and $y$ corresponding to each value of $t$ one by one as written below:
1. When $t=\dfrac{x-y}{x+y}=3$ . Then,
$\begin{align}
  & \dfrac{x-y}{x+y}=3 \\
 & \Rightarrow x-y=3x+3y \\
 & \Rightarrow -2x=4y \\
 & \Rightarrow x=-2y \\
\end{align}$
Now, put $x=-2y$ in the equation (1). Then,
$\begin{align}
  & 3x+7y=26 \\
 & \Rightarrow -6y+7y=26 \\
 & \Rightarrow y=26 \\
 & x=-2y \\
 & \Rightarrow x=-52 \\
\end{align}$
Now, from the above result, we can say that $x=-52$ and $y=26$ will be one of the required values of $x$ and $y$ .
2. When $t=\dfrac{x-y}{x+y}=-3$ . Then,
$\begin{align}
  & \dfrac{x-y}{x+y}=-3 \\
 & \Rightarrow x-y=-3x-3y \\
 & \Rightarrow 2y=-4x \\
 & \Rightarrow y=-2x \\
\end{align}$
Now, put $y=-2x$ in the equation (1). Then,
$\begin{align}
  & 3x+7y=26 \\
 & \Rightarrow 3x-14x=26 \\
 & \Rightarrow -11x=26 \\
 & \Rightarrow x=-\dfrac{26}{11} \\
 & y=-2x \\
 & \Rightarrow y=\dfrac{52}{11} \\
\end{align}$
Now, form the above result we can say that $x=-\dfrac{26}{11}$ and $y=\dfrac{52}{11}$ will be one of the required values of $x$ and $y$ .
3. When $t=\dfrac{x-y}{x+y}=\dfrac{1}{3}$ . Then,
$\begin{align}
  & \dfrac{x-y}{x+y}=\dfrac{1}{3} \\
 & \Rightarrow 3x-3y=x+y \\
 & \Rightarrow 2x=4y \\
 & \Rightarrow x=2y \\
\end{align}$
Now, put $x=2y$ in the equation (1). Then,
$\begin{align}
  & 3x+7y=26 \\
 & \Rightarrow 6y+7y=26 \\
 & \Rightarrow 13y=26 \\
 & \Rightarrow y=2 \\
 & x=2y \\
 & \Rightarrow x=4 \\
\end{align}$
Now, from the above result, we can say that $x=4$ and $y=2$ will be one of the required values of $x$ and $y$ .
4. When $t=\dfrac{x-y}{x+y}=-\dfrac{1}{3}$ . Then,
$\begin{align}
  & \dfrac{x-y}{x+y}=-\dfrac{1}{3} \\
 & \Rightarrow 3x-3y=-x-y \\
 & \Rightarrow -2y=-4x \\
 & \Rightarrow y=2x \\
\end{align}$
Now, put $y=2x$ in the equation (1). Then,
$\begin{align}
  & 3x+7y=26 \\
 & \Rightarrow 3x+14x=26 \\
 & \Rightarrow 17x=26 \\
 & \Rightarrow x=\dfrac{26}{17} \\
 & y=2x \\
 & \Rightarrow y=\dfrac{52}{17} \\
\end{align}$
Now, from the above result we can say that $x=\dfrac{26}{17}$ and $y=\dfrac{52}{17}$ will be one of the required values of $x$ and $y$ .
Now, from all the above results we conclude that if ${{\left( 3-\dfrac{6y}{x+y} \right)}^{2}}+{{\left( 3+\dfrac{6y}{x-y} \right)}^{2}}=82$ , $3x+7y=26$ then $x$ and $y$ will have following four pair of values:
$\begin{align}
  & x=-52;y=26 \\
 & x=-\dfrac{26}{11};y=\dfrac{52}{11} \\
 & x=4;y=2 \\
 & x=\dfrac{26}{17};y=\dfrac{52}{17} \\
\end{align}$

Note: Here, the student should first understand the problem before solving and try to analyse the given equations. After that, we should make assumptions like $\dfrac{x-y}{x+y}=t$ to make calculation smooth. Moreover, as there are many equations so we should take care of the values and signs of each variable while making the substitution and avoid calculation mistakes while solving to get the correct answer.