System of linear equations is a set or collection of variables involving the same set of variables. A linear equation in variables $x_{1},x_{2},........,x_{n}$ is an equation of the form $a_{1}x_{1},a_{2}x_{2},........,a_{n}x_{n}$ = b. It will have two or more equations in one or more variables. System of linear equations are used to make accurate predictions in real world problems.

## System of Linear Equations Definition

A system of linear equations is one which may be written in the form:

$a_{11}x_{1}$ + $a_{12}x_{2}$ + .........+ $a_{1n}x_{n}$ = $b_{1}$
$a_{21}x_{1}$ + $a_{22}x_{2}$ + .........+ $a_{2n}x_{n}$ = $b_{2}$
...... ........
....... .......
$a_{m1}x_{1}$ + $a_{m2}x_{2}$ + .........+ $a_{mn}x_{n}$ = $b_{m}$

All of the coefficients $a_{ij}$ and $b_{i}$ are assumed to be known constants. All the $x_{i}$'s are assumed to be unknowns that we are to solve for.

A system of linear equations means two or more linear equations. If two linear equations intersect then the point of intersection is called the solution to the system of linear equations.

## Consistent and Inconsistent System of Linear Equations

Solution of the system of equations is an ordered pair that satisfies each equation in the system. If there are three equations with three variables then the ordered pair is (a, b, c). System of linear equations are of three types with
• One Solution
• Infinite Solution and
• No Solution
Consistent System

If the system has at least one solution then it is known as consistent system of linear equations and they Intersect at one point and have one solution. Also known as independent system of equations.

Inconsistent System

If the system has no solution then it is said to be inconsistent system and has no solutions. In this case the lines will be parallel and never intersect.

## System of Linear Equations in Two Variables and Three Variables

A linear system of two equations with two variables in any system is written as

ax + by = p
cx + dy = q

where any of the constants can be zero with the exception that each equation must have at least one variable in it. The variables should be to the first power only.

Similarly an equation of three variables with three equations can be written as

ax + by = p
cx + dy = q
ex + fy = r
Here also the variables should be in first power only.

## Solving Systems of Linear Equations by Substitution

Solve one of the equations with one of the variables and substitute in the remaining equations which gives one equation with one variable which can easily be solved. Once solved substitute the value back in one of the given equations and find the value of the remaining variables.

Number of equations should be equal to number of variables.

### Solved Example

Question: Solve the given system of equations using substitution method

l  + m  = 2
5l + 2m = 5
Solution:

From l + m = 2  isolate l.
$\Rightarrow$  l = 2 - m

Substitute l = 2 - m  in   5l + 2m = 5  and solve for m
5( 2 - m) + 2m = 5

10 - 5m + 2m = 5

10 - 3m = 5

$\Rightarrow$  -3m = - 5

$\Rightarrow$   m = $\frac{5}{3}$

Substituting m = $\frac{5}{3}$ in one of the given equations

l + $\frac{5}{3}$ = 2

$\Rightarrow$ 3l + 5 = 6

$\Rightarrow$  l = $\frac{1}{3}$

Thus l = $\frac{1}{3}$   and m = $\frac{5}{3}$

## Solving Systems of Linear Equations by Graphing

In graphical method there can be one, none or infinitely many solutions. If for the given system of equations, we can graph a straight line then it possible to solve them graphically.
Graph the two lines and look for the point where they intersect(cross). The intersection point is termed as the solution.

### Solved Example

Question: By using the method of graphing solve:

x - y = 6
-2x + 2y = 1
Solution:

Consider the system of given equations
x - y = 6
-2x + 2y = 1

Now put y = 0, 1, 2  in  x - y = 6 (isolate x) we get

 y 0 1 2 x 6 7 8

Consider the second equation and isolate y again

-2x + 2y = 1

$\Rightarrow$    y = $\frac{1}{2}$ + x

Now put x = 0, 1, 2  in  -2x + 2y = 1

 x 0 1 2 y 0.5 1.5 2.5

Graph the above points

From the graph it appears that the two lines are parallel. As the lines are parallel they have different intercepts and will not intersect. So there will be no solution to the given system of equations.

## Systems of Linear Equations by Elimination

Elimination method is also known as Addition or Subtraction method. Transform the variables in such a way that one variable cancels out when solved simultaneously.

### Solved Example

Question: Solve by elimination method:
x  - 7y  = 9
-2x + 3y  = 3
Solution:

The given system of equations are:
x  - 7y  = 9                      --------->   1
-2x + 3y  = 3                      --------->   2

Multiply the first equation by 2.
2x - 14y = 18                        --------->  3

Solve third and second equation simultaneously.
2x - 14y = 18
-2x + 3y  = 3
_______________
-11y  = 21
_______________

$\Rightarrow$      y  = $\frac{-21}{11}$

Substitute y = $\frac{-21}{11}$ in the first equation and solve for y

x  -  7 ($\frac{-21}{11}$) = 9
$\Rightarrow$            x + $\frac{147}{11}$  = 9
$\Rightarrow$             11x = 99 - 147

$\Rightarrow$   x = $\frac{-48}{11}$

Therefore x = $\frac{-48}{11}$  and y = $\frac{-21}{11}$

Verification can be done by substituting  x and y values in one of the given equations.

## System of Linear Equations Word Problems

### Solved Example

Question: A total of 12,000 is invested in two funds paying 9% and 11% simple interest. If the yearly interest is 1,180. How much of the 12,000 is invested at each rate?
Solution:

Let x: Investment in fund for 9 years.
Let y: Investment in fund for 11 years.
Then the above statement can be written in the mathematical form as:
x  +      y    =     12000
0.09x + 0.11y     =      1180

Multiply the first equation by -0.09, we get
-0.09x   -   0.09y  = -1080
0.09x    +  0.11y  =  1180
_________________________
0.02y =  100
______________________
$\Rightarrow$  y = 5000

To find x, substitute y =  5000 in   x + y = 12000
x + 5000 =    12000
$\Rightarrow$  x = 12000 - 5000
$\Rightarrow$  x = 7000

Therefore, x = 7000 is the investment for 9 years and y = 5000 is the  investment for 11 years.

## System of Linear Equations Examples

### Solved Examples

Question 1: By using the method of addition solve:

3x  + 5y  =  8
8x  -  5y  =  12
Solution:

Consider the given equations
3x  + 5y  =  8
8x  -  5y  =  12

3x  + 5y  =  8
8x  - 5y  =  12
__________________
11x = 20
_______________

$\Rightarrow$  x = $\frac{20}{11}$

Substitute x = $\frac{20}{11}$ in   3x  + 5y  =  8

We get,        3($\frac{20}{11}$)  + 5y  =  8

Solve for y

$\Rightarrow$  5y = 8 - ($\frac{60}{11}$)

55y = 88 - 60

y = ($\frac{28}{55}$)

Therefore, x= $\frac{20}{11}$ and y = ($\frac{28}{55}$)

Question 2: Using substitution method solve :  x + y = 9
x - y  = 3
Solution:

Isolate x from x + y = 9
$\Rightarrow$  x = 9 -  y
Substitute x = 9 -  y in x - y  = 3
We get,
9 - y - y = 3
9 - 2y = 3
-2y = -6
$\Rightarrow$ y = 3

Now you can substitute y = 3 in any one of the given equations to find the value of x

In x + y = 9  substitute y = 3 and solve for x
x + 3 = 9
$\Rightarrow$ x = 6

Therefore (x, y) = (6, 3).