Substitution Method Calculator

We say that we want to solve a system of linear equations if we have several equations and want to find numbers that solve all these equations simultaneously.

Let's recall that an equation is linear if all the variables in that equation are in the first power. This means that they cannot be squared, nor cubed, nor put under any root (in particular, under a square root or cube root). They also cannot be denominators of fractions.

This substitution method calculator works for systems of two linear equations in two variables. These are the systems most commonly encountered in homework! 😉 They take the following form:

a₁x + b₁y = c₁

a₂x + b₂y = c₂

where:

• x and y are the variables;
• a₁, b₁, c₁ are the coefficients of the first equation; and
• a₂, b₂, c₂ are the coefficients of the second equation.

The substitution method is a method of solving systems of linear equations. The main idea behind solving systems with the substitution method is to choose one of the equations, solve it for one of the variables, and plug the result into the other equation. This way, we obtain an equation with one variable, which we can easily solve. Once we have found the value of one variable, we utilize it to find the other variable. This is how we use the substitution method to solve the system of equations. Go to the next section to learn more about the substitution method steps.

Do you know that there are other methods for solving systems of linear equations? You're not forced to solve by substitution method! You can use the elimination method (also known as the linear combination method) or Gaussian elimination, or, once you've mastered matrix determinants, Cramer's rule.

💡 See our elimination method calculator, linear combination calculator, and system of equations calculator to learn more about this!

This system of equation substitution method tools is easy to use:

1. Enter the coefficients of the system of linear equations in their respective fields.

2. The complete solution by substitution method appears below the substitution method calculator!

3. All the substitution method steps are here in case you need them.

4. If you need the system solved with a higher precision than the four significant figures that this substitution method calculator uses by default, simply set your desired precision into the precision input field.

We have already explained what the substitution method is about and what the main idea behind it is. Now, let's discuss in detail how to do the substitution method:

1. Choose one of the equations.

2. In the chosen equation, choose one of the variables.

3. Solve this equation for this variable.

4. Plug the result into the other equation (the one you didn't choose in Step 1). This is the essence of solving systems by the substitution method!

5. You have obtained a one-variable equation – solve it!

6. Substitute the value you got in Step 5 into one of the original equations.

7. Solve this new one-variable equation.

8. That's it! You solved the system of equations by the substitution method. If you want, you may test your solution: substitute the values you obtained into the system and see if everything is OK.

⚠️ It may sometimes happen that you try to solve a system, and suddenly both variables vanish 😱 Keep calm! The variables are no more, but you arrived at some statements about numbers. All you need to do is draw conclusions about the system depending on whether the statement you got is true (like 0 = 0 or 17 = 17) or false (e.g., 0 = 1 or 15 = 17):

• If the statement is false, then the system is inconsistent, which means it has no solution.

• If the statement is true, then the system is dependent, i.e., it has infinitely many solutions.

In this section, we show you step by step how to solve several systems using the substitution method so that you can see how to do the substitution method in practice.

1. Use the substitution method to solve the system of equations:

   3x - 4y = 6

   -x + 4y = 2

   • Solve the second equation for x:

     x = 4y - 2

   • Substitute 4y - 2 for x into the first equation:

     3(4y - 2) - 4y = 6

   • Solve the above equation for y:

     12y - 6 - 4y = 6

     8y = 12

     y = 1.5

   • Substitute y = 1.5 into the second equation:

     -x + 4 × 1.5 = 2

     -x + 6 = 2

   • Solve for x:

     x = 6 - 2

     x = 4

   • Solution: x = 4, y = 1.5

   • Test the solution:

     3 × 4 - 4 × 1.5 = 12 - 6 = 6

     We see that the first equation is OK.

     -4 + 4 × 1.5 = -4 + 6 = 2

     And the second equation is OK as well.

2. Solve using substitution method:

   2x + 3y = 5

   2x + 7y = -3

   • Solve the first equation for x:

     x = -1.5y + 2.5

   • Substitute x = -1.5y + 2.5 into the second equation:

     2×(-1.5y + 2.5) + 7y = -3

   • Solve the above equation for y:

     -3y + 5 + 7y = -3

     4y = -8

     y = -2

   • Substitute y = -2 into the first equation:

     2x + 3×(-2) = 5

   • Solve for x:

     2x = 11

     x = 5.5

   • Solution: x = 5.5, y = -2

3. Now, we will see how to solve the following system of linear equations using the substitution method:

   6x - 3y = 12

   2x - y = 4

   • Solve the second equation for y:

     y = 2x - 4

   • Substitute y = 2x - 4 into the first equation:

     6x - 3×(2x - 4) = 12

   • Solve the above equation for x:

     6x - 6x - 12 = 12

     0 = 0

   • We eliminated both variables and arrived at a true statement. Hence, this system of equations has infinitely many solutions!