A system of linear equations with two variables can be solved both algebraically and graphically. The solution of the system is an ordered pair of values corresponding to the two variables in the system. Graphically the solution is given by the coordinates of the point of intersection of the two lines representing the two equations. Graphical method does not always guarantee an accurate solution, where as the algebraic methods land on exact solutions of the systems. The two commonly used algebraic methods applied for solving a system of equations in two variables are
  1. Substitution method
  2. Elimination method

Steps done in solving a system of linear equation using substitution method

  1. Solve one of the equations literal for one of the variables. The solution will consist of an expression containing the other variable.
  2. Substitute the expression for the variable solved in the other equation.This after simplification leads you to an equation with one variable.
  3. Solve the equation to find the value of the variable.
  4. Substitute this value in the expression for the other variable got in step 1 and simplify to get its value.
  5. Verify the solutions got by substituting in the original equations.

When do we use substitution method?

A system of equation can be solved using many methods. We need to choose the method that will work efficient and easy for a given system.
Substitution method is used in the following situations:
  1. One of the equations given is already solved for a variable.
  2. It is easy to solve for a variable in any one of the equations. Normally the equation will contain a variable with coefficient 1 or -1.
  3. Try to avoid substitution method, if the solved expression for the variable is to contain fractions.
  4. Use substitution method if the problem direction says so.

Examples for substitution method

1. Solve the system of equations:
    $x+2y=30$      .............(1)
    $y=x+3$         ..............(2)
This system is a candidate for applying substitution method as the second equation is given solved for y.
$x+2(x+3)=30$                                           Substitution of y =x+3 in equation (1)
$x+2x+6=30$                                              Distributive Property.  Now we have an equation in one variable to solve.
     $-6    -6$
Now substitute x =8 in the solved equation (2) for y.
Hence the solution to the system is x=8, y=11 or (8,11).

2. Solve the system of equations:
    $2x-3y =7$        ..................(1)
   $-2x+y=-1$        ..................(2)
Even though no equation is given solved for a variable, the second equation can be easily solved for y.
 $+2x          +2x$                                               Addition property to undo subtraction.
      $ y = 2x-1$
Now substitution y =2x-1 in equation (1) we get an equation in one variable x.
$2x-3(2x-1) = 7$
$2x-6x+3 = 7$                                            Distributive Property
$-4x +3 = 7$
       $-3    -3$
$4x  = 4$
$x=-1$                                                     Solution for x
Now substituting x=-1 in equation solved for y,
Therefore x =-1 and y=-3 is a solution of the system.

Substitution method with no solution and many solutions

We will not always get a unique solution for a Linear system. If the given system is not solvable for a unique system, how to determine whether the system is consistent or consistent? The following rules are followed to determine the situation of many solutions or no solution.
  1. If the process of solving lead to a true statement without variables, then the system is consistent and has infinitely many solutions.
  2. If the process of solving lead to a false statement without variables, then the system is inconsistent and has no solution.