Theory:

Let us consider two linear equations.
 
\(a_1x+b_1y+c_1 = 0\) ----- (\(1\))
 
\(a_2x+b_2y+c_2 = 0\) ----- (\(2\))
 
Where \(\frac{a_1}{a_2}\neq \frac{b_1}{b_2}\).
 
Now, let us solve these two equations.
 
Multiply equation (\(1\)) by \(b_2\) and equation (\(2\)) by \(b_1\).
 
Thus, we have:
 
a1b2x+b1b2y+b2c1=0()a2b1x+b1b2y+b1c2=0¯x(a1b2a2b1)+b2c1b1c2=0
 
\(x(a_1b_2-a_2b_1) =\)\(b_1c_2-b_2c_1\)
 
\(x=\frac{(b_1c_2-b_2c_1)}{(a_1b_2-a_2b_1)}\)
 
Similarly, let us multiply equation (\(1\)) by \(a_2\) and equation (\(2\)) by \(a_1\).
 
Thus, we have:
 
a1a2x+b1a2y+c1a2=0()a1a2x+b2a1y+c2a1=0¯0+y(b1a2b2a1)+c1a2c2a1=0
 
\(y(b_1a_2-b_2a_1) =\)\(c_2a_1-c_1a_2\)
 
\(y=\frac{(c_2a_1-c_1a_2)}{(b_1a_2-b_2a_1)}\)
 
Therefore, the solution to the given system of linear equations is
\(x = \frac{(b_1c_2-b_2c_1)}{(a_1b_2-a_2b_1)}\) and \(y = \frac{(c_1a_2-c_2a_1)}{(b_1a_2-b_2a_1)}\).
 
The other way to represent the solution is given by:
 
\(\frac{x}{b_1c_2-b_2c_1} = \frac{y}{c_1a_2-c_2a_1}\)\( = \frac{1}{a_1b_2-a_2b_1}\)
 
It can easily remembered as:
 
4.png
 
Let us consider an example to understand the concept clearly.
Example:
Solve the simultaneous linear equations \(2x+y = 8\) and \(x+2y = 10\) by cross multiplication method.
 
Solution:
 
Let us write the given equations in the form of \(ax+by+c = 0\).
 
\(2x+y-8 = 0\) ---- (\(1\))
 
\(x+2y-10 = 0\) ---- (\(2\))
 
Now, let us write the coefficients in the form of
 
4.png
 
Thus, we have:
 
5.svg
 
\(\frac{x}{(1)(-10)-(2)(-8)}=\)\(\frac{y}{(-8)(1)-(-10)(2)} = \)\(\frac{1}{(2)(2)-(1)(1)}\)
 
\(\frac{x}{-10-(-16)} = \frac{y}{-8-(-20)} = \)\(\frac{1}{4-1}\)
 
\(\frac{x}{-10+16} = \frac{y}{-8+20}=\frac{1}{3}\)
 
\(\frac{x}{6} = \frac{y}{12}=\frac{1}{3}\)
 
\(\frac{x}{6} = \frac{1}{3}\) and \(\frac{y}{12} = \frac{1}{3}\)
 
\(x = \frac{6}{3}\) and \(y = \frac{12}{3}\)
 
\(x = 2\) and \(y = 4\)
 
Therefore, the solution is \(x = 2\) and \(y = 4\).