# Solving Systems of Equations by Elimination

The method of solving systems of equations by Elimination is also known as Gaussian

Elimination because it is attributed to Carl Friedrich Gauss as the inventor of

the method.

Elimination or involves manipulating the given system of equations such that one

or more of the variables is eliminated leaving a single variable equation which

can then be easily solved. From there, back substitution is used to find the other

variables(s).

## Solving Systems of Equations Using Elimination

When solving systems of equations of two variables, the method of elimination involves

manipulating one equation by multiplying it with a constant such that when this

equation is added to the other equation in the system, one variable gets **eliminated**

and all that is left is a one variable equation which we can easily solve. After

getting the solution to this variable, we then substitute that value into any one

of the equations in order to solve for the other variable.

Given the system of equations below;

The elimination process allows us to multiply either equation (1) or (2) such that

when we add the two equations, one of the variables is eliminated.

For example, if we wanted to eliminate variable **x**, we would achieve this

by multiplying either equation (1) or (2) by some constant which will ensure that

equation (1) + equation (2) is equal to equation(3) which only has one variable

**y**.

If we multiplied equation(2) by the constant ^{-a}⁄_{c} we

would get the following result

which simplifies into

Then if we were to add the new equation(2n) above to equation(1) we would observe

the following:

would result in

The above is a one variable equation in **y** and we can proceed to solve for

**y**. After obtaining the value of **y**, we can substitute it into either

equation(1) or equation(2) from which we can calculate the value of **x**.

Let’s try a few examples to actually see how it all works.

### Example 1

Solve the following system of equations

**Step 1**

In order to keep track of what equation we’re talking about, the first step should

always be labeling the equations in the system.

**Step 2**

The next step is to inspect the equations to see how we can manipulate them to get

rid of one of the variables. This example is fairly simple, in order to get rid

of variable **y**, we don’t need to multiply any of the equations by any constant,

we could simply add the two equations and variable **y** will simply disappear.

This is because in both equations the coefficient of **y** in equation(1) is

the opposite of that in equation(2) i.e. **1** and **-1** so when we add the

two equations y should cancel out.

**Step 3**

The above is a one variable equation in **x**, and we can easily solve for **x**

by dividing through by 5

**Step 4**

Now that we have a solution for **x**, we can substitute for **x** in equation(1)

to get a one variable equation which we can use to solve for **y**.

substituting for x

Hence we have a solution for the system of equations as **x = 5** and **y = -1**.

### Example 2

Solve the following system of equations

**Step 1**

First step is to label the equations so we know which equation we’re referring to.

**Step 2**

Next we inspect the system of equations to see how we can effectively eliminate

one of the variables.

From inspecting the system, we can observe that we can multiply equation (1) by

-3 such that when we add the two equations, we can eliminate **x**.

which results in

**Step 3**

Next we proceed to add the new equation to equation (2)

Which simplifies into

**Step 4**

The value of **y** has been obtained as **-1**. We proceed to use this value

to solve for **x** by substituting **y** into any of the above equations.

Let’s substitute **y** in the original equation (1)

The solution to the system of equation is **{x,y}** = **{-1,-1}**.

### Example 3

Find the values of x and y in the system of equations given below

**Step 1**

When you’re given a system of equations in a disorganized manner as above, it’s

always better to start off by rearranging the equations such that the variables

match up. This will make it easier to work with the equations. Then you can proceed

to label the equations.

**Step 2**

From inspecting the above system, it’s not so obvious which constant to multiply

by either equation (1) or (2) in order to eliminate one of the variables. When such

a situation occurs, you pick one variable and decide that you want to eliminate

that, then you pick one equation and multiply it by a fraction whose numerator is

the negative of the coefficient variable we’re eliminating in the other equation,

and the denominator is the coeffecient of the variable for the equation we’re multiplying.

For this example, let’s eliminate variable **x** and work with equation (2) first:

which simplifies as

**Step 3**

Adding the above to equation (2) results in the following

**Step 4**

Now that we have a value for **y**, we can substitute and solve for **x**

as follows;

Solution to the system of equations is **{x,y}** = **{ ^{148}⁄_{95},^{-43}⁄_{19}}**

## Solving Three Variable Systems of Equations by Elimination Method

Solving three variable systems of equations is not much different from solving those

with two variables. The basic concept remains the same only that in this case we

take it a little further to account for the third variable.

Similar to when we were solving for two variables, when it comes to three variables,

we start off by manipulating two of the three equations such that when they are

added to the other equation, one of the variables disappears completely. We then

focus on those new two variable equations and also manipulate them such that we

remain with a one variable equation.

Given the system of equations below

We can solve for the variables **x**, **y** and **z** as follows:

First we pick equation (1) and use it to eliminate the variable **x** from the

other equations.

We can achieve this in equation (2) by multiplying the equation by a constant such

that when we add equation(2) to equation (1), the **x** variable disappears.

Multiplying the above by a constant ^{a}⁄_{d}

Adding the above to equation (1):

results in:

The above is the new equation (2).

We also eliminate the x variable from equation (3) in a similar manner:

multiplying through by a constant

Adding the above to equation (1)

which results in

The above is the new equation (3)

So now we have two two-variable equations, we can solve these two-variable equations

using as we learned in the section on two-variable systems of equations above.

Lets have an example to see how the above algorithm works

### Example 4

Solve the following system of equations

**Step 1**

We start off by labeling our equations to make it easier to refer to them

**Step 2**

Next, we pick equation (2) variable **y**, we inspect this equation together

with equation (1) to find a way of eliminating this equation.

**Step 3**

From inspection, you can see that simply added equation (1) to equation (2) you

would effectively eliminate the variable **y** from equation (2)

results in

The above is the new equation (2)

**Step 4**

We do the same for equation (3)

But observe that in this case we’re lucky that equation (3) is already a two variable

equation in **z** and **x**.

**Step 5**

So we can proceed to handle equations (2) and (3) as a two variable system of equations

From inspecting the above system, we can choose any variable to eliminate since

they’ll all be relatively simple, but lets go with **x**.

**Step 6**

We can eliminate x by multiplying equation (3) by -1 and then adding the result

to equation 2 as follows:

which gives

which results in

**Step 7**

The above is a relatively simple one variable equation from which we can solve for

**x** by dividing both sides by 3:

Using back substitution, we can then solve for **x** and **y**, starting with

**x**. To find **x**, we substitute the value of **z** into equation (2)

**Step 8**

Now that we have values for **x** and **z**, we can find **y** by substituting

these value into equation (1)

The solution to the system of equations is **{x, y, z} = {1, 2, 3}**

### Example 5

Find x, y and z from the system of equations below

**Step 1**

As always, we begin by labeling the equations so that they’re easier to reference.

**Step 2**

Next, we consider equations (1) and (2) such that we eliminate one of the variables;

in this case **x**

**Step 3**

We can see that if we were to multiply equation (2) by ^{-1}⁄_{3},

and then add the result to equation (1), we can obtain a new equation (2) which

will be a two variable equation in **y** and **z**.

which results in:

**Step 4**

Adding the above result to equation (1)

results in:

The above becomes our new equation (2).

**Step 5**

We then carry out the same procedure for equation (3):

From inspecting the above system, we can see that if we multiply equation (3) by

^{-1}⁄_{6}

results in:

**Step 6**

Then we add the above to equation (1)

which results in:

The above result becomes our new equation (3)

**Step 7**

Now we have obtained a new system of equations which only has two variables:

**Step 8**

We can eliminate **y** from the above system by multiplying equation (3) by ^{
-2}⁄_{3} and then adding the result to equation (2). The

resulting equation will be a one variable equation in **z** and we can solve

for **z** from this.

results in:

**Step 9**

Adding the above to equation (2):

which results in;

We can solve for **z** from the above:

**Step 10**

Using back substitution, we can now solve for **y** then for **x**.

First we find **y** by substituting the value of **z** into equation (2):

**Step 11**

Having obtained **y**, we can find **x** by substituting **y** and **z**

in equation (1):

The solution to the system of equations is: **{x, y, z} = { ^{1}⁄_{14}, ^{4}⁄_{21}, ^{8}⁄_{7}}**