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Download now on Google PlayLet 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:

$\begin{array}{l}\phantom{\rule{1.176em}{0ex}}{a}_{1}{b}_{2}x+{b}_{1}{b}_{2}y+{b}_{2}{c}_{1}=0\\ \underset{\xaf}{(-){\phantom{\rule{0.294em}{0ex}}a}_{2}{b}_{1}x+{b}_{1}{b}_{2}y+{b}_{1}{c}_{2}=0}\\ x({a}_{1}{b}_{2}-{a}_{2}{b}_{1})+{b}_{2}{c}_{1}-{b}_{1}{c}_{2}=0\end{array}$

\(x(a_1b_2-a_2b_1) =\)\(b_1c_2-b_2c_1\)

\(x=\frac{(b_1c_2-b_2c_1)}{(a_1b_2-a_2b_1)}\)

Similarly, let us multiply equation (\(1\)) by \(a_2\) and equation (\(2\)) by \(a_1\).

Thus, we have:

$\begin{array}{l}\phantom{\rule{1.029em}{0ex}}{a}_{1}{a}_{2}x+{b}_{1}{a}_{2}y+{c}_{1}{a}_{2}=0\\ \underset{\xaf}{(-){a}_{1}{a}_{2}x+{b}_{2}{a}_{1}y+{c}_{2}{a}_{1}=0}\\ 0+y({b}_{1}{a}_{2}-{b}_{2}{a}_{1})+{c}_{1}{a}_{2}-{c}_{2}{a}_{1}=0\end{array}$

\(y(b_1a_2-b_2a_1) =\)\(c_2a_1-c_1a_2\)

\(y=\frac{(c_2a_1-c_1a_2)}{(b_1a_2-b_2a_1)}\)

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:

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

Thus, we have:

\(\frac{x}{(1)(-10)-(2)(-8)}=\)\(\frac{y}{(-8)(1)-(-10)(2)} = \)\(\frac{1}{(2)(2)-(1)(1)}\)

\(\frac{x}{-10-(-16)} = \frac{y}{-8-(-20)} = \)\(\frac{1}{4-1}\)

\(\frac{x}{-10+16} = \frac{y}{-8+20}=\frac{1}{3}\)

\(\frac{x}{6} = \frac{y}{12}=\frac{1}{3}\)

\(\frac{x}{6} = \frac{1}{3}\) 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 \(y = 4\).