A linear equation is a first degree equation having one or more variables. A system of linear equations is formed by two or more linear equations with the same variables. Solution of a system of linear equations in two variables is an ordered pair of values of the variables that satisfy each of the equation in the system. The equations of a system are often called simultaneous equations as any solution to the system has to satisfy all the equations simultaneously. The equations cannot be solved independently of one another.

## Solving Systems of Linear Equations Solver

**Consistent system**

A consistent system is subdivided into two types2.

- System with unique solution
- System with many solutions

**Inconsistent system**

- Inconsistent systems have no solution.

## Methods of Solving a System of Linear Equations

**Graphical Method****Substitution Method****Elimination Method****Cramer's Rule****Matrix Row ellimination**

## Graphical Method of Solving Linear Equations

- If the lines
**intersect**at a point, then the system is**consistent**and has a unique solution. - If the
lines drawn
**parallel**, there will not be a point of intersection. The system is**inconsistent**and no solution exists for the system. - If
the lines are
**coincident**, the system is**consistent**and have many solutions.

### Solved Example

**Question:**Solve graphically

2x + y = 5

4x - y = 1

**Solution:**

System of linear equations is given as

2x + y = 5

4x - y = 1

The Graphs of the equations to be straight lines. The solution set to the system is the set of all ordered pairs that satisfy both equations. If we graph each equation on the same set of axes, then we can see the solution set.

Graphical representation of given system:

Point of intersection of two lines is the solution of linear system that is (1, 3).

2x + y = 5

4x - y = 1

The Graphs of the equations to be straight lines. The solution set to the system is the set of all ordered pairs that satisfy both equations. If we graph each equation on the same set of axes, then we can see the solution set.

Graphical representation of given system:

Point of intersection of two lines is the solution of linear system that is (1, 3).

## Solving System of Linear Equations By Substitution

### Solved Example

**Question:**Solve

y = 6x -11

-2x - 3y = -7

**Solution:**

Given linear system

y = 6x -11 .........(1)

-2x - 3y = -7 .........(2)

Here the first equation is already solved for y. The expression for y is substituted in the second equation.

-2x - 3(6x - 11) = -7

=> -2x - 18x + 33 = -7

=> -20x + 33 = -7

=> -20x + 33 - 33 = -7 - 33

=> -20x = -40

=> x = 2

Substituting x =2 in equation (1)

y = 6(2) - 11

= 12 - 11

=1

=> y = 1.

y = 6x -11 .........(1)

-2x - 3y = -7 .........(2)

Here the first equation is already solved for y. The expression for y is substituted in the second equation.

-2x - 3(6x - 11) = -7

=> -2x - 18x + 33 = -7

=> -20x + 33 = -7

=> -20x + 33 - 33 = -7 - 33

=> -20x = -40

=> x = 2

Substituting x =2 in equation (1)

y = 6(2) - 11

= 12 - 11

=1

=> y = 1.

## Solving System of Linear Equations By Elimination

### Solved Example

**Question:**Solve x + y = 13

x - y = 3

**Solution:**

Given

x + y = 13 .........(1)

x - y = 3 .........(2)

Here the variable y can be eliminated by adding the two equations.

(1) + (2)

=> 2x = 16

=> x = 8

substituting x = 8 in equation (1)

x + y = 13

=> 8 + y = 13

=> y = 13 - 8

=> y = 5

x + y = 13 .........(1)

x - y = 3 .........(2)

Here the variable y can be eliminated by adding the two equations.

(1) + (2)

=> 2x = 16

=> x = 8

substituting x = 8 in equation (1)

x + y = 13

=> 8 + y = 13

=> y = 13 - 8

=> y = 5

## Cramer's Rule

Cramer's rule for systems of two linear equations:

$a_1$x + $b_1$y = $c_1$

$a_2$x + $b_2$y = $c_2$

then the solution is given by

x = $\frac{D_1}{D}$ and

y = $\frac{D_2}{D}$

Where

D = $\begin{vmatrix}

a_1 & b_1\\ a_2

& b_2

\end{vmatrix}$ $\neq$ 0

$D_1$ = $\begin{vmatrix}

c_1 & b_1\\ c_2

& b_2

\end{vmatrix}$

$D_2$ = $\begin{vmatrix}

a_1 & c_1\\ a_2

& c_2

\end{vmatrix}$.

### Solved Example

**Question:**Solve linear equations system by using Cramer's rule.

x + y = 3

2x - y = 1

**Solution:**

Given linear equations system

x + y = 3

2x - y = 1

D = $\begin{vmatrix}

1 & 1\\ 2

& -1

\end{vmatrix}$

= -1 - 2 = - 3 $\neq$ 0

$D_1$ = $\begin{vmatrix}

3& 1\\ 1

& -1

\end{vmatrix}$

= - 3 - 1

= - 4

$D_2$ = $\begin{vmatrix}

1& 3\\ 2

& 1

\end{vmatrix}$

= 1 - 6

= - 5

Now by Cramer's Rule:

The solution of given system is

x = $\frac{D_1}{D}$ and

y = $\frac{D_2}{D}$

=> x = $\frac{-4}{-3}$

= $\frac{4}{3}$

and y = $\frac{-5}{-3}$

= $\frac{5}{3}$

x + y = 3

2x - y = 1

D = $\begin{vmatrix}

1 & 1\\ 2

& -1

\end{vmatrix}$

= -1 - 2 = - 3 $\neq$ 0

$D_1$ = $\begin{vmatrix}

3& 1\\ 1

& -1

\end{vmatrix}$

= - 3 - 1

= - 4

$D_2$ = $\begin{vmatrix}

1& 3\\ 2

& 1

\end{vmatrix}$

= 1 - 6

= - 5

Now by Cramer's Rule:

The solution of given system is

x = $\frac{D_1}{D}$ and

y = $\frac{D_2}{D}$

=> x = $\frac{-4}{-3}$

= $\frac{4}{3}$

and y = $\frac{-5}{-3}$

= $\frac{5}{3}$