Solve the following system of equations by the method of cross-multiplication:
$ax+by=\frac{a+b}{2}$
$3x+5y=4$

Given:

The given system of equations is:

$ax+by=\frac{a+b}{2}$

$3x+5y=4$

 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,

$ax+by=\frac{a+b}{2}$

$2(ax+by)=a+b$     (On cross multiplication)

$2ax+2by-(a+b)=0$.......(i)

$3x+5y-4=0$......(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,

$a_1=2a, b_1=2b, c_1=-(a+b)$ and $a_2=3, b_2=5, c_2=-4$

Therefore,

$\frac{x}{2b\times(-4)-5\times-(a+b)}=\frac{-y}{2a\times(-4)-3\times-(a+b)}=\frac{1}{2a\times(5)-3\times (2b)}$

$\frac{x}{-8b+5a+5b}=\frac{-y}{-8a+3a+3b}=\frac{1}{10a-6b}$

$\frac{x}{5a-3b}=\frac{-y}{-5a+3b}=\frac{1}{10a-6b}$

$x=\frac{5a-3b}{2(5a-3b)}$ and $-y=\frac{-(5a-3b)}{2(5a-3b)}$

$x=\frac{1}{2}$ and $-y=\frac{-1}{2}$

$x=\frac{1}{2}$ and $y=\frac{1}{2}$

The solution of the given system of equations is $x=\frac{1}{2}$ and $y=\frac{1}{2}$. 

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Updated on: 10-Oct-2022

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