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Solve The Following System of Linear Equations by Substitution Calculator

Reviewed by Calculator Editorial Team

Solving systems of linear equations is a fundamental skill in algebra. This calculator helps you solve systems using the substitution method, which involves expressing one variable in terms of another and substituting it into the second equation.

How to Use This Calculator

To solve a system of linear equations using substitution:

  1. Enter the coefficients and constants for both equations in the calculator form.
  2. Click the "Calculate" button to solve the system.
  3. Review the solution and the step-by-step process.

The calculator will display the solution if it exists, or indicate if the system is inconsistent or dependent.

The Substitution Method Explained

The substitution method involves three main steps:

  1. Solve one equation for one variable in terms of the other.
  2. Substitute this expression into the second equation.
  3. Solve the resulting equation for the remaining variable.
  4. Substitute this value back into the expression from step 1 to find the other variable.
Example: Equation 1: 2x + 3y = 8 Equation 2: x - y = 1 Step 1: Solve Equation 2 for x: x = y + 1 Step 2: Substitute into Equation 1: 2(y + 1) + 3y = 8 2y + 2 + 3y = 8 5y + 2 = 8 5y = 6 y = 6/5 Step 3: Find x using x = y + 1: x = (6/5) + 1 = 11/5

Worked Example

Let's solve the following system using substitution:

3x + 2y = 12 x - y = 2
  1. Solve the second equation for x:
    x = y + 2
  2. Substitute into the first equation:
    3(y + 2) + 2y = 12 3y + 6 + 2y = 12 5y + 6 = 12 5y = 6 y = 6/5
  3. Find x using x = y + 2:
    x = (6/5) + 2 = 16/5

The solution is x = 16/5 and y = 6/5.

Frequently Asked Questions

What is the substitution method?

The substitution method is a technique for solving systems of linear equations where one variable is expressed in terms of the other and then substituted into the second equation.

When should I use substitution?

Use substitution when one of the equations can be easily solved for one variable, making it simple to substitute into the other equation.

What if the system has no solution?

If the equations are inconsistent (they represent parallel lines), the system has no solution. The calculator will indicate this case.