Solve the following system of equations by the method of cross-multiplication:
$5ax+6by=28$
$3ax+4by=18$

Given:

The given system of equations is:

$5ax+6by=28$

$3ax+4by=18$

 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,

$5ax+6by-28=0$

$3ax+4by-18=0$

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,

$a_1=5a, b_1=6b, c_1=-28$ and $a_2=3a, b_2=4b, c_2=-18$

Therefore,

$\frac{x}{6b\times(-18)-(4b)\times(-28)}=\frac{-y}{5a\times(-18)-3a\times(-28)}=\frac{1}{5a\times(4b)-3a\times (6b)}$

$\frac{x}{-108b+112b}=\frac{-y}{-90a+84a}=\frac{1}{20ab-18ab}$

$\frac{x}{4b}=\frac{-y}{-6a}=\frac{1}{2ab}$

$x=\frac{4b\times1}{2ab}$ and $-y=\frac{(-6a)\times1}{2ab}$

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

$x=\frac{2}{a}$ and $y=\frac{3}{b}$

The solution of the given system of equations is $x=\frac{2}{a}$ and $y=\frac{3}{b}$.

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Updated on: 10-Oct-2022

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