PUMPA - THE SMART LEARNING APP

AI system creates personalised training plan based on your mistakes

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:

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

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

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

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

Thus, we have:

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

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

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

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 $$y = 4$$.