Solve the following system of equations by the method of cross-multiplication:
$ax\ +\ by\ =\ a^2$
$bx\ +\ ay\ =\ b^2$

Given:

The given system of equations is:

$ax\ +\ by\ =\ a^2$

$bx\ +\ ay\ =\ b^2$

 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,

$ax+by-a^2=0$

$bx+ay-b^2=0$

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,

$a_1=a, b_1=b, c_1=-a^2$ and $a_2=b, b_2=a, c_2=-b^2$

Therefore,

$\frac{x}{b\times(-b^2)-a\times(-a^2)}=\frac{-y}{a\times(-b^2)-b\times(-a^2)}=\frac{1}{a\times a-b\times b}$

$\frac{x}{-b^3+a^3}=\frac{-y}{-ab^2+a^2b}=\frac{1}{a^2-b^2}$

$\frac{x}{a^3-b^3}=\frac{1}{a^2-b^2}$ and $\frac{-y}{ab(a-b)}=\frac{1}{a^2-b^2}$

$x=\frac{(a-b)(a^2+ab+b^2)\times1}{(a+b)(a-b)}$ and $-y=\frac{ab(a-b)\times1}{(a+b)(a-b)}$

$x=\frac{a^2+ab+b^2}{a+b}$ and $-y=(\frac{ab}{a+b})$

$x=\frac{a^2+ab+b^2}{a+b}$ and $y=-\frac{ab}{a+b}$

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

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Updated on: 10-Oct-2022

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