Solve the following system of equations by the method of cross-multiplication:
$(a-b)x+(a+b)y=2a^2-2b^2$
$(a+b)(x+y)=4ab$

Given:

The given system of equations is:

$(a-b)x+(a+b)y=2a^2-2b^2$

$(a+b)(x+y)=4ab$

 To do: 

Here, we have to solve the given system of equations by the method of cross-multiplication.

Solution:  

The given system of equations can be written as,

$(a-b)x+(a+b)y-2(a^2-b^2)=0$....(i)

$(a+b)(x+y)=4ab$

$(a+b)x+(a+b)y-4ab=0$......(ii)

The solution of a linear pair(standard form) of equations $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ is given by,

$\frac{x}{b_1c_2-b_2c_1}=\frac{-y}{a_1c_2-a_2c_1}=\frac{1}{a_1b_2-a_2b_1}$

Comparing the given equations with the standard form of the equations, we get,

$a_1=(a-b), b_1=(a+b), c_1=-2(a^2-b^2)$ and $a_2=(a+b), b_2=(a+b), c_2=-4ab$

Therefore,

$ \begin{array}{l}
\frac{x}{( a+b)( -4ab) -( a+b) \times -2\left( a^{2} -b^{2}\right)} =\frac{-y}{( a-b)( -4ab) -( a+b) \times -2\left( a^{2} -b^{2}\right)} =\frac{1}{( a-b)( a+b) -( a+b)( a+b)}\\
\\
\frac{x}{2( a+b)\left[ -2ab+a^{2} -b^{2}\right]} =\frac{-y}{( a-b)( -4ab) +2( a+b)( a+b)( a-b)} =\frac{1}{( a+b)[ a-b-( a+b)]}\\
\\
\frac{x}{2( a+b)\left[ -2ab+a^{2} -b^{2}\right]} =\frac{-y}{2( a-b)[ -2ab+( a+b)( a+b)]} =\frac{1}{( a+b)[ a-b-a-b]}\\
\\
\frac{x}{2( a+b)\left[ -2ab+a^{2} -b^{2}\right]} =\frac{-y}{2( a-b)\left[ -2ab+a^{2} +2ab+b^{2}\right]} =\frac{1}{( a+b)( -2b)}\\
\\
\frac{x}{2( a+b)\left[ -2ab+a^{2} -b^{2}\right]} =\frac{-y}{2( a-b)\left( a^{2} +b^{2}\right)} =\frac{1}{( a+b)( -2b)}\\
\\
This\ implies,\\
\\
\frac{x}{2( a+b)\left[ -2ab+a^{2} -b^{2}\right]} =\frac{1}{( a+b)( -2b)}\\
\\
x=\frac{2( a+b)\left( -2ab+a^{2} -b^{2}\right)}{( a+b)( -2b)}\\
\\
x=\frac{\left( b^{2} +2ab-a^{2}\right)}{b}\\
\\
\frac{-y}{2( a-b)\left( a^{2} +b^{2}\right)} =\frac{1}{( a+b)( -2b)}\\
\\
y=\frac{-2( a-b)\left( a^{2} +b^{2}\right)}{( a+b)( -2b)}\\
\\
y=\frac{( a-b)\left( a^{2} +b^{2}\right)}{b( a+b)}
\end{array}$

The solution of the given system of equations is $x=\frac{b^2+2ab-a^2}{b}$ and $y=\frac{(a-b))(a^2+b^2)}{b(a+b)}$. 

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Updated on: 10-Oct-2022

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