Lesson Solution of the linear system of two equations with two unknowns by the Elimination method

Solution of the linear system of two equations with two unknowns by the Elimination method

This lesson describes the Elimination method to solve the linear system of two equations with two unknowns.
The method is to multiply both sides of the first equation to some number, to multiply both sides of the second equation to another number and then to add both these equations (or to subtract one equation from another). The multipliers are selected to cancel co-named variable terms in the final equation for some variable.
When this is done, you can easily solve the obtained equation for the remaining variable, back-substitute the found value to either of the original equations and to find the value of the variable that was eliminated at the first step.

Examples below show how this method works.

Example 1

Solve the system of equations



Multiply the first equation by 2 (which means "Multiply both sides of the first equation by 2"). Leave the second equation as is. You will get:

.

Add the first and the second equations (which means "Add left sides and add right sides of the first and the second equations"). You will get:
.

The solution of this equation is .

Back-substitute this value of to the first (original) equation. You will get:
, and then
.

The solution of this equation is .

Now, check the calculated solution. Simply substitute the found values of and to the original equations. You will get
                  for the left side of the first equation, and this is identical to its right side;
for the left side of the second equation, and this is identical to its right side.

So, the check shows that the solution is correct.

Example 2

Solve the system of equations



Multiply the second equation by 2 (which means "Multiply both sides of the second equation by 2"). Leave the first equation as is. You will get:

.

Subtract the second equation from the first one (which means "Subtract the left side and the right side of the first equation from those of the second equation").
You will get:
,
.

The solution of the last equation is .

Back-substitute this value of to the first (original) equation. You will get:
, and then
.

The solution of the last equation is .

Now, check the calculated solution. Simply substitute the found values of and to the original equations. You will get
      for the left side of the first equation, and this is identical to its right side;
for the left side of the second equation, and this is identical to its right side.

So, the check shows that the solution is correct.

Examples 1 and 2 are cases, when the linear equation system has the unique solution. These are cases of the consistent equation systems with independent equations.

Example 3

Solve the system of equations



Multiply the first equation by 4 (which means "Multiply both sides of the first equation by 4"). Leave the second equation as is. You will get:

.

Subtract the second equation from the first one (which means "Subtract the left side and the right side of the second equation from those of the first one").
You will get:
, or, after canceling common terms,
.

The last equation has infinitely many solutions: each and any value of satisfies to this equation.

So, the original system of equations is equivalent to



Since the last equation does not produce any restriction to , our system is equivalent to the single equation
,
which has infinitely many solutions: for every value of the value of

is the solution. For example, the following pairs of variables values are solutions:
, ,
, ,
, ,
, ,
, , ... .

In this example, the linear equation system has infinitely many solutions, because the coefficients and the right side of one equation are proportional to that of the second equation. This is the case of the consistent equation system with dependent equations.

Example 4

Solve the system of equations



Multiply the first equation by 4 (which means "Multiply both sides of the first equation by 4"). Leave the second equation as is. You will get:

.

Subtract the second equation from the first one (which means "Subtract the left side and the right side of the second equation from those of the first one").
You will get:
, or, after canceling common terms,
.

The last equation has no solutions because its left side is always equal to zero, while the right side is not equal to zero.

So, the original system of equations has no solutions.

In this example, the linear equation system has no solutions, because the coefficients of one equation are proportional to that of the second equation, while the right sides are out of this proportionality. This is the case of the inconsistent equation system with dependent equations.

Summary

The Elimination method of solving the system of two linear equations with two unknowns is to multiply both sides of equations to the appropriate numbers (multipliers) and then to add or to subtract the equations to cancel co-named variable terms in the final equation for some variable.

When it is done, you may have one of the three following typical cases:
Case 1. You got a single equation with the non-zero coefficient at the remaining variable.
            In this case you can easily solve the obtained equation for that variable, back-substitute the found value to the original equation and to find
            the value of the variable that was eliminated at the first step.
            The solution always exists and is unique in this case.
            The system of equations is consistent in this case, and equations are independent.
Case 2. You got an equation with the zero coefficient at the remaining variable and the zero right side.
            In this case the original system has infinitely many solutions.
            The system of equations is consistent in this case, and equations are dependent.
Case 3. You got an equation with the zero coefficient at the remaining variable and the non-zero right side.
            In this case the original system has no solutions.
            The system of equations is inconsistent in this case, and equations are dependent.