Linear Combination Calculator

Our linear combination calculator is here whenever you need to solve a system of equations using the linear combination method (also known as the elimination method). If you want to learn what the linear combination method is or how to use the linear combination method, check the article below.

We explain what combining linear equations means and how to use the linear combination method to solve systems of linear equations. Most importantly, we show you several very detailed step-by-step examples of systems solved with the linear combination method.

For an equation to be linear, all its variables must be in the first power: they cannot be squared/cubed, nor under a root, nor placed in the denominator. If some numbers satisfy several linear equations at once, we say that these numbers are a solution to the system of those linear equations.

In school, we most often encounter systems of two linear equations in two variables. In general, such a system takes the 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 linear combination of equations is a method for solving systems of linear equations. The key idea is to combine the equations into a system of fewer and simpler equations. If we deal with two linear equations in two variables, we want to combine these equations into one equation with a single variable.

By combining linear equations we mean multiplying one or both equations by suitably chosen numbers and then adding the equations together. Multipliers must allow the operation of addition to cause one of the variables to vanish. We then end up with a single equation in one variable, which we can easily solve. Once we have solved this equation, we substitute the value we've found into one of the original equations. This leads to another equation in one variable, which we quickly solve. This way, we've solved the system using linear combination!

Are there other methods for solving systems of linear equations? Sure! You can discover them in Omni's substitution method calculator and in the Cramer's rule calculator.

If you want to quickly solve a system of equations using linear combination, our tool is the best choice! To use it, follow the steps below:

1. Tell us the coefficients of your system: type them in the appropriate fields.
2. Our linear combination calculator solves the system using the linear combination method...
3. ...and displays and explains all the intermediate steps!
4. Go to the advanced mode of the linear combination calculator if you want to compute the solution with higher precision (with a higher number of sig figs). By default, our linear combination calculator uses six sig figs.

Did you know you can use this method to solve a linear programming problem algebraically? Read more about it in our corner point calculator.

For the system of equations:

a₁x + b₁y = c₁

a₂x + b₂y = c₂

we can always use the least common multiple of a₁ and a₂. Namely, put:

m₁ := LCM(a₁, a₂) / a₁

m₂ := LCM(a₁, a₂) / a₂

and multiply the first equation by m₁ and the second equation by -m₂. This leads to the following system:

LCM(a₁, a₂)x + [LCM(a₁, a₂)b₁/a₁]y = LCM(a₁,a₂)c₁/a₁

-LCM(a₁, a₂)x - [LCM(a₁, a₂)b₂/a₂]y = -LCM(a₁, a₂)c₂/a₂

We have created opposite coefficients for the variable x! All we need to do is to add these equations together: we will obtain an equation with the y variable only.

In some particular situations, this LCM approach boils down to elementary operations:

• When the coefficients of one variable are opposite numbers, both multipliers are equal to 1.

• When the coefficients of one variable are equal, one multiplier is equal to 1 and the other to -1.

It may sometimes happen that you eliminate both variables at once. First of all, do not panic. You arrived at a statement about numbers. Depending on whether the statement you got is true, like:

0 = 0 or 7 = 7,

or false, e.g.:

0 = 1 or 5 = 7,

draw conclusions about the system:

• If the statement is false, then the system has no solution.

• If the final statement is true, then the system has infinitely many solutions.

Wow, we've learned a lot about what the linear combination method is! It's time to solve a few systems of linear equations using linear combinations.

1. Consider the system of equations:

   x - 4y = 1

   -2x + 4y = 2

   and solve it using linear combination.

   • Add the two equations together:

     -x = 3

   • Solve for x:

     x = -3

   • Substitute x = -3 into the first equation:

     -3 - 4y = 1

   • Solve for y:

     -4y = 4

     y = -1

   • Solution: x = -3, y = -1

2. Solve using linear combination:

   2x + 3y = 3

   2x - y = -3

   • First, multiply the first equation by -1:

     -2x - 3y = -3

     2x - y = -3

   • Add the equations, which results in eliminating x:

     -4y = -6

   • Solve for y:

     y = 1.5

   • Substitute y = 1.5 into the second equation:

     2x - 1.5 = -3

   • Solve for x:

     2x = -1.5

     x = -0.75

   • Solution: x = -1.5, y = -0.75

3. Solve the system using linear combination:

   3x - 7y = 1

   4x + 4y = -2

   • Use the LCM approach: find the calculate the least common multiplicity of 3 and 4:

     LCM(3, 4) = 12.

     The multipliers are:

     m₁ = 4 and m₂ = -4 and the system is:

     12x - 28y = 4

     -12x - 12y = 6

   • Add the equations:

     -40y = 10

   • We solve for y:

     y = -0.25

   • We substitute y = -0.25 into the second equation:

     4x - 4 ⋅ (-0.25) = 0

   • We solve for x:

     4x = -1

     x = -0.25

   • Solution: x = -0.25, y = -0.25

In vector algebra, a linear combination of vectors refers to adding two or more vectors multiplied by a scalar value. For example, if a, b, and c are vectors, then a vector d is a linear combination of a, b, and c, if:

d = x⋅a + y⋅b + z⋅c

where x, y, and z are scalars. We can use a linear combination of vectors to represent a system of linear equations.

Consider two linear equations in two variables x and y: a₁⋅x + b₁⋅y = c₁ and a₂⋅x + b₂⋅y = c₂.

1. Find the LCM of a₁ and a₂:

   LCM(a₁, a₂) = L

2. Multiply the first equation by L/a₁ to get:

   L⋅x + L⋅b₁/a₁⋅y = L⋅c₁/a₁

3. Multiply the second equation by -L/a₂ to get:

   -L⋅x - L⋅b₂/a₂⋅y = -L⋅c₂/a₂

4. Add these two equations together to obtain an equation in y-term only:

   (L⋅b₁/a₁ - L⋅b₂/a₂)⋅y = L⋅c₁/a₁ -L⋅c₂/a₂

5. Solve this equation for y:

   y = (L⋅c₁/a₁ - L⋅c₂/a₂)/(L⋅b₁/a₁ - L⋅b₂/a₂)

6. Substitute this solution for y in any one of the original equations to get the x value.

The solution for the equations 5x + 2y = 12 and 8x + 12y = 28 is x = 2, y = 1. To solve this, follow these steps:

1. Find the LCM of 2 and 12:

   LCM(2, 12) = 12

2. Multiply the first equation by -12/2 = -6 to get:

   -30x - 12y = -72

3. Add these two equations to obtain:

   -22x = -44
   ⇒x = -44/-22 = 2

4. Substitute x = 2 in the first equation to get the y value:

   5(2) + 2y = 12
   ⇒2y = 12 - 10
   ⇒y = 2/2 = 1

5. Verify this result with our linear combination calculator.