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

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

This lesson describes the Substitution method to solve the linear system of two equations with two unknowns.
The method is to express one variable via another using one equation and then to substitute this expression into another equation. In this way you reduce the original system of two equations with two unknowns to the single equation with one unknown. When this reduction is done, simplify the obtained equation and solve it.
After completing this, back-substitute the found value to either of the original equations and find the value of the remaining variable.

Examples below show how this method works.

Example 1

Solve the system of equations



Express from the first equation:
.

Substitute this to the second equation:
.

Simplify the last equation by collecting common terms and solve it for :
,
,
,
,
.

Back-substitute this value of to the first original equation:
.

Simplify and solve this equation for :
.

Thus, the solution of the original system of equations is , .

Now, check the calculated solution. Simply substitute the found values of and into 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.

The check shows that the solution is correct.

Example 2

Solve the system of equations



Express from the second equation:
.

Substitute this to the first equation:
.

Simplify the last equation by collecting common terms and solve it for :
.
.
.

The solution of the last equation is .

Back-substitute this value of to the expression :
.

The solution for 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.

The check shows that the solution is correct.

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

Example 3

Solve the system of equations



Express from the first equation:

.

Substitute this expression for to the second equation:
, or, step by step simplifying,
.
.

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

Since the last equation does not produce any restriction to , the original system of equations 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



Express from the first equation. You will get:

.

Substitute this expression for to the second equation. You will get:
, or, step by step simplifying,
.
.

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 Substitution method for solving the system of two linear equations with two unknowns is to express one variable via another using one equation and then to substitute this expression into another equation. In this way you reduce the original system of two equations with two unknowns to the single equation with one unknown. When this reduction is done, simplify the obtained equation and solve it.
After completing this, back-substitute the found value to either one of the original equations and find the value of the remaining variable.

When solving the linear system equations by the Substitution method, you may have one of the three following typical cases:
Case 1. The system of equations is consistent and equations are independent.
            The system of equations has the solution in this case and it is unique.
            The substitution procedure works smoothly in this case both at the direct-substitution and back-substitution steps.
Case 2. The system of equations is consistent, but equations are dependent.
            The system of equations has infinitely many solutions in this case.
            Typically the substitution procedure leads to the equation (identity) of the form or that have infinitely many solutions.
            This is the case when the coefficients and the right side of one equation are proportional to that of the second equation.
Case 3. The system of equations is inconsistent, equations are dependent.
            The system of equations has no solutions in this case.
            Typically the substitution procedure leads to the unsolvable equation of the form or .
            This is the case when the coefficients of one equation are proportional to that of the second equation, but the right sides are out of this proportionality.