Question

Solve by cross-multiplication method:
$\begin{align}
  & \left( a-b \right)x+\left( a+b \right)y=2{{a}^{2}}-2{{b}^{2}} \\
 & \left( a+b \right)\left( x+y \right)=4ab \\
\end{align}$

Answer 1

Hint: First take all the terms in the given equations to the L.H.S and then assume the two equations as equation (i) and equation (ii). Now, assume that coefficients of x, y and constant term of equation (i) are ${{a}_{1}},{{b}_{1}}$ and ${{c}_{1}}$ respectively and coefficients of x, y and constant term of equation (ii) are ${{a}_{2}},{{b}_{2}}$ and ${{c}_{2}}$ respectively. Now, apply the formula for method of cross-multiplication:
\[\dfrac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\dfrac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\dfrac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\] and equate, \[\dfrac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\dfrac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\] to find the value of x and \[\dfrac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\dfrac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\] to find the value of y.

Complete step-by-step answer:
We have been provided with two equations:
$\begin{align}
  & \left( a-b \right)x+\left( a+b \right)y=2{{a}^{2}}-2{{b}^{2}} \\
 & \left( a+b \right)\left( x+y \right)=4ab \\
\end{align}$
Taking all the terms to the L.H.S, we get,
$\begin{align}
  & \left( a-b \right)x+\left( a+b \right)y+2{{b}^{2}}-2{{a}^{2}}=0 \\
 & \left( a+b \right)\left( x+y \right)-4ab=0 \\
\end{align}$
Now, $\left( a+b \right)\left( x+y \right)=4ab$, by multiplying the L.H.S terms it can be written as,
$\left( a+b \right)x+\left( a+b \right)y=4ab$
Now, we have two equations:
$\begin{align}
  & \left( a-b \right)x+\left( a+b \right)y+2{{b}^{2}}-2{{a}^{2}}=0.............(i) \\
 & \left( a+b \right)x+\left( a+b \right)y-4ab=0..............(ii) \\
\end{align}$
Assuming the coefficients of x, y and constant term of equation (i) as ${{a}_{1}},{{b}_{1}}$ and ${{c}_{1}}$ respectively and coefficients of x, y and constant term of equation (ii) as ${{a}_{2}},{{b}_{2}}$ and ${{c}_{2}}$ respectively, we have,
\[\begin{align}
  & {{a}_{1}}=\left( a-b \right),{{b}_{1}}=\left( a+b \right)\text{ and }{{c}_{1}}=2{{b}^{2}}-2{{a}^{2}} \\
 & {{a}_{2}}=\left( a+b \right),{{b}_{2}}=\left( a+b \right)\text{ and }{{c}_{2}}=-4ab \\
\end{align}\]
Applying the formula of cross-multiplication method, we have,
\[\dfrac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\dfrac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\dfrac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\]
Therefore, substituting all the values of coefficients, we get,
\[\dfrac{x}{\left( a+b \right)\left( -4ab \right)-\left( a+b \right)\left( 2{{b}^{2}}-2{{a}^{2}} \right)}=\dfrac{y}{\left( 2{{b}^{2}}-2{{a}^{2}} \right)\left( a+b \right)-\left( -4ab \right)\left( a-b \right)}=\dfrac{1}{\left( a-b \right)\left( a+b \right)-\left( a+b \right)\left( a+b \right)}\]
Equating, \[\dfrac{x}{\left( a+b \right)\left( -4ab \right)-\left( a+b \right)\left( 2{{b}^{2}}-2{{a}^{2}} \right)}=\dfrac{1}{\left( a-b \right)\left( a+b \right)-\left( a+b \right)\left( a+b \right)}\], we get,
\[x=\dfrac{\left( a+b \right)\left( -4ab \right)-\left( a+b \right)\left( 2{{b}^{2}}-2{{a}^{2}} \right)}{\left( a-b \right)\left( a+b \right)-\left( a+b \right)\left( a+b \right)}\]
Cancelling the common terms, we get,
\[\begin{align}
  & x=\dfrac{\left( a+b \right)\left( -4ab \right)-\left( a+b \right)\left( 2{{b}^{2}}-2{{a}^{2}} \right)}{\left( a-b \right)\left( a+b \right)-\left( a+b \right)\left( a+b \right)} \\
 & \Rightarrow x=\dfrac{\left( -4ab \right)-\left( 2{{b}^{2}}-2{{a}^{2}} \right)}{\left( a-b \right)-\left( a+b \right)} \\
 & \Rightarrow x=\dfrac{-4ab-2{{b}^{2}}+2{{a}^{2}}}{a-b-a-b} \\
 & \Rightarrow x=\dfrac{-4ab-2{{b}^{2}}+2{{a}^{2}}}{-2b} \\
 & \Rightarrow x=\dfrac{-2ab-{{b}^{2}}+{{a}^{2}}}{-b} \\
 & \Rightarrow x=\dfrac{2ab+{{b}^{2}}-{{a}^{2}}}{b} \\
\end{align}\]
Now equating, \[\dfrac{y}{\left( 2{{b}^{2}}-2{{a}^{2}} \right)\left( a+b \right)-\left( -4ab \right)\left( a-b \right)}=\dfrac{1}{\left( a-b \right)\left( a+b \right)-\left( a+b \right)\left( a+b \right)}\], we get,
\[\Rightarrow y=\dfrac{\left( 2{{b}^{2}}-2{{a}^{2}} \right)\left( a+b \right)-\left( -4ab \right)\left( a-b \right)}{\left( a-b \right)\left( a+b \right)-\left( a+b \right)\left( a+b \right)}\]
Using the algebraic identity: \[\left( {{x}^{2}}-{{y}^{2}} \right)=\left( x+y \right)\left( x-y \right)\], we get,
\[\begin{align}
  & y=\dfrac{2\left( b-a \right)\left( b+a \right)\left( a+b \right)-\left( -4ab \right)\left( a-b \right)}{\left( a-b \right)\left( a+b \right)-\left( a+b \right)\left( a+b \right)} \\
 & \Rightarrow y=\left( a-b \right)\times \dfrac{2\left( -1 \right)\left( b+a \right)\left( a+b \right)-\left( -4ab \right)}{\left( a-b \right)\left( a+b \right)-\left( a+b \right)\left( a+b \right)} \\
 & \Rightarrow y=\left( a-b \right)\times \dfrac{-2{{\left( a+b \right)}^{2}}+4ab}{\left( {{a}^{2}}-{{b}^{2}} \right)-{{\left( a+b \right)}^{2}}} \\
 & \Rightarrow y=\left( a-b \right)\times \dfrac{-2{{a}^{2}}-2{{b}^{2}}-4ab+4ab}{\left( {{a}^{2}}-{{b}^{2}} \right)-\left( {{a}^{2}}+{{b}^{2}}+2ab \right)} \\
 & \Rightarrow y=\left( a-b \right)\times \dfrac{-2{{a}^{2}}-2{{b}^{2}}}{-2{{b}^{2}}-2ab} \\
 & \Rightarrow y=\left( a-b \right)\times \dfrac{-2\left( {{a}^{2}}+{{b}^{2}} \right)}{-2b\left( b+a \right)} \\
 & \Rightarrow y=\dfrac{\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}} \right)}{b\left( b+a \right)} \\
\end{align}\]

Note: One may note that we can also solve this question easily by the help of substitution or elimination method but it is asked in the question to solve it with the help of cross-multiplication method. While solving this question, we have to be careful about the formula and calculation. Any one sign mistake can lead us to a wrong answer. We can check our answer by substituting the obtained values of x and y, in the provided equations.