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Solve The Following System by The Substitution Method Calculator

Reviewed by Calculator Editorial Team

This calculator solves systems of linear equations using the substitution method. It provides step-by-step solutions and visual representations of the results.

How to Use This Calculator

To solve a system of equations using the substitution method:

  1. Enter the coefficients and constants for each equation in the calculator form.
  2. Click the "Calculate" button to solve the system.
  3. Review the solution steps and the graphical representation of the equations.
  4. If needed, adjust the values and recalculate.

Note: The calculator currently supports systems of two equations with two variables. For larger systems, consider using matrix methods.

Substitution Method Explained

The substitution method is a technique for solving systems of linear equations. It involves solving one equation for one variable and substituting that expression into the other equation.

Steps in the Substitution Method

  1. Choose one equation and solve for one variable in terms of the other.
  2. Substitute this expression into the other equation.
  3. Solve the resulting equation for the remaining variable.
  4. Substitute this value back into one of the original equations to find the other variable.

For a system of equations:

a₁x + b₁y = c₁

a₂x + b₂y = c₂

The substitution method involves solving one equation for x or y and substituting into the other.

Worked Example

Let's solve the following system using the substitution method:

2x + 3y = 8

4x - y = 10

Solution Steps

  1. Solve the second equation for y: y = 4x - 10
  2. Substitute into the first equation: 2x + 3(4x - 10) = 8
  3. Simplify: 2x + 12x - 30 = 8 → 14x = 38 → x = 2.714
  4. Substitute x back into y = 4x - 10: y = 4(2.714) - 10 = 0.536

The solution to the system is x ≈ 2.714 and y ≈ 0.536.

Frequently Asked Questions

What is the substitution method?
The substitution method is a technique for solving systems of linear equations by solving one equation for one variable and substituting that expression into the other equation.
When should I use the substitution method?
The substitution method is particularly useful when one of the equations can be easily solved for one variable, making substitution straightforward.
What if the system has no solution?
If the equations are parallel (same slope but different y-intercepts), the system has no solution. The calculator will indicate this case.