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Solve the following system of equations by the method of cross-multiplication:
$bx+cy=a+b$
$ax(\frac{1}{a-b}-\frac{1}{a+b})+cy(\frac{1}{b-a}-\frac{1}{b+a})=\frac{2a}{a+b}$
Given:
The given system of equations is:
$bx+cy=a+b$
$ax(\frac{1}{a-b}-\frac{1}{a+b})+cy(\frac{1}{b-a}-\frac{1}{b+a})=\frac{2a}{a+b}$
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,
$bx+cy-(a+b)=0$....(i)
$ax(\frac{1}{a-b}-\frac{1}{a+b})+cy(\frac{1}{b-a}-\frac{1}{b+a})=\frac{2a}{a+b}$
$ax(\frac{a+b-(a-b)}{(a-b)(a+b)})+cy(\frac{b+a-(b-a)}{(b-a)(b+a)})-\frac{2a}{a+b}=0$
$\frac{1}{a+b}[ax(\frac{a+b-a+b}{a-b})+cy(\frac{b+a-b+a}{b-a})-2a]=0$
$ax(\frac{2b}{a-b})-cy(\frac{2a}{a-b})-\frac{2a(a-b)}{a-b}=0$
$\frac{ax(2b)-cy(2a)-2a(a-b)}{a-b}=0$
$2abx-2acy-2a(a-b)=0$...(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,
$a_1=b, b_1=c, c_1=-(a+b)$ and $a_2=2ab, b_2=-2ac, c_2=-2a(a-b)$
Therefore,
$\frac{x}{-2 a c(a-b)-[-(a+b)][-2 a c]}=\frac{-y}{-2 a b(a-b)-[-(a+b)][2 a b]}=\frac{1}{-2 a b c-2 a b c}$
$\frac{x}{-2 a^{2} c+2 a b c-\left[2 a^{2} c+2 a b c\right]}=\frac{-y}{-2 a^{2} b+2 a b^{2}+\left[2 a^{2} b+2 a b^{2}\right]}=\frac{1}{-4 a b c}$
$\frac{x}{-2 a^{2} c+2 a b c-2 a^{2} c-2 a b c}=\frac{-y}{-2 a^{2} b+2 a b^{2}+2 a^{2} b+2 a b^{2}}=\frac{-1}{4 a b c}$
$\frac{x}{-4 a^{2} c}=\frac{-y}{4 a b^{2}}=\frac{-1}{4 a b c}$
This implies,
$\frac{x}{-4 a^{2} c}=\frac{-1}{4 a b c}$
$x=\frac{4 a^{2} c}{4 a b c}=\frac{a}{b}$
$\frac{-y}{4 a b^{2}}=\frac{-1}{4 a b c}$
$y=\frac{4 a b^{2}}{4 a b c}=\frac{b}{c}$
The solution of the given system of equations is $x=\frac{a}{b}$ and $y=\frac{b}{c}$.
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