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Solving Systems of Equations with Linear Combinations

Instructor: Maria Blojay

Maria has taught College Algebra and has a master's degree in Education Administration.

This lesson will show how to solve Systems of Equations with Linear Combinations. We will use some sample problems and properties to prove the solution is the point of intersection.

Have Something in Common

Where we work, we meet different people. Those who we meet may have a common bond. When linear equations work together, they can have a point of intersection that is in common between the lines.

What are Systems of Linear Equations Using Linear Combinations?

These are two equations with two variables that have a point or an ordered pair as a solution that is in common between the two linear equations. Yes, there could be more equations and variables, but we'll stick with the basics..

There are three key methods (graphing, substitution, and elimination) to solve systems of equations. Simply, the linear combinations method, which is this lesson's focus, includes adding or subtracting variable terms to eliminate one of the variables and then combining the two equations to solve for the common point.

Also, with the help of two properties of equality, the examples below will prove that a solution is the same regardless of how we combine our linear equations.

What Are Some Ways to Prove a Solution of System Of Equations?

Suppose we want to prove that (-1,3) is the solution to the linear system of equations:

-2x + y = 5 (Equation A)

x + y = 2 (Equation B)

Here are a few ways to prove the solution is (-1,3).

Substitution

If we substitute -1 for x and 3 for y, each equation would provide true statements.

-2(-1) + 3 = 5 (Equation A) which is 2 +3 = 5 or 5 = 5 is a true statement

-1 + 3 = 2 (Equation B) which is 2 = 2 is a true statement

Graphing

To easily graph the equations above we would rewrite the equations like this:

Equations A and B Graphs

y = 2x + 5 (Equation A rewritten) RED

y = -x + 2 (Equation B rewritten) GREEN

Our graph would show that our solution or point of intersection (x,y) would be (-1,3).

Algebraically - Elimination

If we lay these equations out like this:

2x + 5 = y = -x + 2

Then this is the same as : 2x + 5 = -x + 2 applying the Addition Property of Equality, by adding x on each side of the equation.

We then get 3x + 5 = 2. Since -x and +x cancel out, add -5 on each side of the equation. They cancel out, we are left with 3x = -3. Applying the Multiplication Property of Equality, by multiplying each side of the equation by 1/3. Leaving us with:

x = -1.

If we substitute the x value into either equation, we find that our y value is 3. So, we find our point of intersection or solution is (-1,3)

What if we did not lay these equations out this way? What would happen if we multiplied Equation A by 3 and added to Equation B, would we still get the same solution (-1,3)? Let's try it and see if we do.

If we multiply Equation A, -2x + y = 5 by 3 we would get 3(-2x + y) = 3(5) or

-6x + 3y = 15 then when we add or combine it to Equation B, we would have -5x + 4y = 17 then when we substitute -1 for x and 3 for y we would get -5(-1) + 4(3) = 17, a true statement.

This demonstrates that it does not matter how we apply these properties to our combined equations. We still have the same solution or point of intersection. The appearance of our equations look different but the point of intersection is the same.

But you still may be wondering, why would we need to do this?

Let's take a look at our original system of equations:

-2x + y = 5 (Equation A)

x + y = 2 (Equation B)

These equality properties would apply. We just need to be more strategic in using them.

Notice, if we strategically choose what we multiply each side by, when we combine our equations, one of our variables will drop out …

If we MULTIPLIED Equation B by 2, we would have 2(x + y) = 2(2) or 2x + 2y = 4 and COMBINED it with Equation A we would get:

-2x + y = 5 (Equation A)

2x + 2y = 4 (Equation B) after multiplying the equation on each side by 2

Then combine or add it with Equation A

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