Solving Systems of Linear Equations by Elimination

 

Learning Objective(s)

·         Define the elimination method of solving systems of equations.

·         Combine equations in a system to eliminate variables.

 

 

Equivalence is a powerful force in algebra. For example, the substitution method of replacing one quantity with an equivalent expression can help solve a system of equations. The elimination method provides another way to apply the concept of equivalence to solve a system of equations. In this method, a variable is eliminated by combining the individual equations within a system.

 

Confused? Okay—let’s get started.

 

 

An imaginary trip to a local pizzeria can help lay the groundwork for understanding the elimination method. Picture that it’s lunchtime, and we walk into a restaurant that advertises the following specials:

 

Special 1: 2 slices and 1 drink for $3.50

Special 2: 5 slices and 2 drinks for $8.25

 

We are hungry, but we also want to figure out how much each slice of pizza and each drink costs. But how? Well, both specials involve the same two items—drinks and slices—in different amounts. We can subtract the smaller special from the bigger one until we get rid of one of those items. Then we’ll know the cost of the other one.

 

If we start with Special 1, 5 slices and 2 drinks for $8.25, and subtract Special 2, 2 slices and 1 drink for $3.50, we get 3 slices and 1 drink for $4.75.

 

This new combination still has two items, so we’ll subtract one more time:  3 slices and 1 drink is $4.75, minus 2 slices and 1 drink for $3.50. That leaves just 1 slice for $1.25.

 

Now we know that a single slice of pizza costs $1.25, so we can substitute that value back into either special to find out the cost of a drink. Putting it into Special 1, we find that two slices cost $2.50, so a drink must be $1.

 

How about that? We just applied the thinking behind the elimination method. We eliminated one quantity (in this case, the cost of the drinks) from an equation to isolate one of the variables (the cost of the slices) in the problem. Let’s revisit what we did, only this time we’ll do it algebraically:

 

 

Let’s eliminate Special 1 from Special 2 (notice how we have flipped the order of the specials so that Special 2 is above Special 1):

 

 

Using subtraction, we are able to find our new total: 3s + d = $4.75. Notice that we are subtracting like terms from each other, and that we are subtracting all of Special 1 from Special 2. Now let’s repeat the elimination process, taking another Special 1 away from our new total:

 

 

Again, we subtracted like terms from each other, resulting in a value for the variable s: $1.25. We’ve done it again—we’ve used the elimination method to find the price of one slice.

 

The elimination method is a useful way of using whole equations to eliminate one variable in a system in order to find the value of another variable in that same system. Once you know the value of one variable, you can substitute that value back into the system and solve for the other one.

 

Sometimes it will make sense to add equations together, and sometimes it will make sense to subtract one from the other. Your decision about whether to add or subtract will depend on the variable that you want to eliminate, as well as the numbers in the equation itself.

 

 

This system is a good example of when to use the elimination method. Looking at the two equations, you can see that a term of 3h appears in one, and a term of -3h appears in the other. If we add these two equations together we will be able to eliminate the h term, enabling us to easily solve for g.

 

 

 

Now that we know g = 1.5, we can substitute this value back into one of the equations to find a value for h.

 

 

After solving a system this way, it is good practice to substitute both values into the original system to make sure that they are correct. If both equations resolve to true statements, then you can be sure you have identified the correct solution. If the resulting statements are false, however, that is a good indication that computation errors have been made along the way.

 

 

Adding and subtracting equations from each other is often an effective way of eliminating a variable from a system. There are times, however, where adding or subtracting does not yield a helpful result.

 

 

The equations in this system have no terms that will easily be eliminated if the equations are added or subtracted. So if we add the equations, we will be left with a third equation that still has two variables.

 

This is where multiplication comes in handy. Notice that the first equation contains the term 4y, and the second equation contains the term -2y. If we double -2y, when we add both equations the y terms will cancel out.

 

 

By multiplying the equation by two, we now have a term of -4y. Multiplying (or dividing) the entire equation by the same number maintains the equation and the relationship between the variables. We just have to be very careful to multiply both sides of the equation when we use this method.

 

Because both the original Equation B and New Equation B have the same variables in the same relationship, they are equivalent. We can substitute New Equation B into the system in place of Equation B. Then we can add it to Equation A.

 

 

Knowing that x = -7, we would just have to substitute -7 in for x in either Equation A or Equation B to find the corresponding value for y.

 

There are other ways to solve this system. Instead of multiplying one equation in order to eliminate a variable when the equations were added, we could have multiplied both equations by different numbers. Watch and see.

 

Let’s get rid of the variable x this time. We’ll multiply Equation A by 5 and equation B by 3:

 

 

We chose to multiply these equations by 5 and 3, respectively, because that gave us numbers that would cancel out, -15x and 15x. Multiplying both equations is a good strategy to use, but again, we must be diligent about multiplying all of the terms within an equation.

 

We could also have left the first equation alone, and multiplied the second equation by  to yield 3x.

 

By now you may be getting the idea that the elimination method is very flexible. We can multiply or divide the equations in a system by any number that is convenient (except for 0). The goal of using this strategy is to manipulate the equations so that adding or subtracting them eliminates one of the variables.

 

 

 

 

Combining equations is a powerful tool for solving a system of equations. Adding or subtracting two equations in order to eliminate a common variable is called the elimination method. Once one variable is eliminated, it becomes much easier to solve for the other one. Multiplication and division can be used to set up matching terms in equations before they are combined. When using the multiplication or division method, it is important to multiply or divide all the terms on both sides of the equation—not just the one you are trying to eliminate.