Linear algebra is a branch of mathematics, but the truth of it is that linear algebra is the mathematics of data. Linear algebra is the study of vector spaces, lines and planes, mappings that are required for linear transforms and many more. Dealing with linear equations is also a very prominent part of linear algebra. In this section students will learn about how to solve linear equations using different methods. There are three possible outcomes for a system of linear equations: one unique solution, infinitely many solutions, and no solution.

Any system that can be written in the form.

ax + by = m

cx + dy = n

Where a,b,c,d,m and n are real numbers and a,b,c,d are non zero, is a system of linear equations in two variables.

cx + dy = n

Where a,b,c,d,m and n are real numbers and a,b,c,d are non zero, is a system of linear equations in two variables.

Solving linear equations means to find the values of unknown variables, here x and y, which satisfies both equations. With the help of algebraic methods such as substitution method, elimination method, graphing method and matrix method find the solution set for the pair of two linear equations. These methods also get the answers for such questions, How many solution has a linear equation in two variable? Will get to know all possible conditions if system of equations has, unique solution, infinitely many solutions or no solution.

Example 1:

10x - 20y = 110 ...(1)

5x - 14y = 71 ...(2)

Use elimination method to solve above system of equations:

Multiply equation (1) by number 5 and equation 2 by number 10, we get

Use elimination method to solve above system of equations:

Multiply equation (1) by number 5 and equation 2 by number 10, we get

5{10x - 20y = 110 }

50x - 100y = 550 ....(3)

50x - 100y = 550 ....(3)

10{5x - 14y = 71 }

50x - 140y = 710 ...(4)

Coefficient of x in both the equations is same, so we can eliminate x from the system of equation by subtracting them.

(Equation 4 ) - (Equation 3)

50x - 140y = 710

50x - 100y = 550

- + -

______________

0 - 40y = 160

______________

or -40y = 160

or y = -4 (dividing both the sides by -4)

To get the value of x, put y = -4 in equation (2)

5x - 14y = 71

5x - 14(-4) = 71

5x + 56 = 71

5x = 71 - 56

5x = 15

x = 15/5

This implies x = 3

**Answer:**

50x - 140y = 710 ...(4)

Coefficient of x in both the equations is same, so we can eliminate x from the system of equation by subtracting them.

(Equation 4 ) - (Equation 3)

50x - 140y = 710

50x - 100y = 550

- + -

______________

0 - 40y = 160

______________

or -40y = 160

or y = -4 (dividing both the sides by -4)

To get the value of x, put y = -4 in equation (2)

5x - 14y = 71

5x - 14(-4) = 71

5x + 56 = 71

5x = 71 - 56

5x = 15

x = 15/5

This implies x = 3

3x - 2y = 17

6x + 3y = 27

3x-2y = 17

6x+3y = 27

Solution: 3x - 2y = 17

Solution: 3x - 2y = 17

6x + 3y = 27

$\begin{bmatrix} 3 & -2\\ 6 & 3\end{bmatrix} \begin{bmatrix} x\\ y\end{bmatrix} = \begin{bmatrix} 17\\ 27 \end{bmatrix}$

i.e $A\ X$ = $B$

A = $\begin{bmatrix} 3 & -2\\ 6 & 3 \end{bmatrix}$

Now, Write matrix of cofactors:

$cof(A)$ = $\begin{bmatrix} 3 & -6\\ 2 & 3\end{bmatrix}$

$Adj(A)$ = Transpose of $cof(A)$

= $\begin{bmatrix} 3 & 2\\ -6 & 3\end{bmatrix}$

Determinant of A = |A| = 9 -(-12) = 9 + 12= 21

$A^{-1}$ = $\frac{adj(A)}{|A|}$

= $\begin{bmatrix} \frac{1}{7} & \frac{2}{21}\\ \frac{-2}{7} & \frac{1}{7} \end{bmatrix}$

X = $A^{-1}\ B$

$\begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} \frac{1}{7} & \frac{2}{21}\\ -\frac{2}{7} & \frac{1}{7} \end{bmatrix} \times \begin{bmatrix} 17\\ 27 \end{bmatrix}$ = $\begin{bmatrix} \frac{17}{7} + \frac{18}{7}\\ \frac{-34}{7}+\frac{27}{7} \end{bmatrix}$

$\begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} 5\\ -1 \end{bmatrix}$

Therefore, the value of x and y is:

**x = 5**