# How to Solve Systems of Equations by Elimination

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- Solving Systems of Equations by Elimination

### Overview of Solving Systems of Equations by Elimination:

### The Elimination Method

The verb “eliminate” means to remove something. In this lesson, we’ll learn how to solve a system of equations in two variables by elimination, a method that works by eliminating a variable to find the value of the other.

### Solving Systems of Equations by Elimination - Process Explained

Before we explore the process, we need to recall the properties of equality, which forms the basis for eliminating a variable.

The properties of equality state that when we add, subtract, multiply, or divide both sides of an equation with the same number, the statement remains the same.

Let’s get a clear picture of how and where these properties are applied by looking at the steps of solving a system of equations in two variables.

Identify the two pairs of like terms from both the equations.

If the coefficients of one of the pairs are opposites:

Add the equations to eliminate one variable.

Solve for the other variable.

Plug the value so obtained in one of the equations.

Solve for the eliminated variable.

If the coefficients of one of the pairs are common:

Subtract the equations to eliminate a variable.

Solve for the other variable.

Plug the value so obtained in one of the equations.

Solve for the eliminated variable.

If there aren’t any common or opposite coefficients:

Multiply one equation or both the equations by a non-zero constant so you get at least one pair of like terms with the same or opposite coefficients.

Subtract or add the equations to eliminate a variable.

Solve for the other variable.

Substitute this value in one of the equations.

Solve for the eliminated variable.

Let’s look at each of these cases in detail with examples.

### Elimination by Addition

To eliminate a variable from a system that has a pair of like terms with opposite coefficients, we need to add the equations. Let’s take the system of two equations below for instance:

3x – 2y = 14

5x + 2y = 18

The two pairs of like terms of this system are the x-terms: 3x, 5x and the y-terms: –2y, 2y. Now, the coefficients of y-terms are opposites.

Also, both sides of an equation are equal. So when you add one equation to another, it’s perfectly fine to add the expressions on one side and the constants on the other. The same quantity is added to both sides, preserving the addition property of equality.

Adding the two equations, we get:

3x – 2y = 14*5x + 2y = 18*

8x = 32

Thus, we have eliminated the variable y. Now dividing both sides of the above equation by 8, we have:

*8x***8** = *32***8**

⇨ x = 4

Thus, we have obtained the value of x. Now, all we need to do is plug it in one of the equations and solve for y. Substituting x = 4 in 3x – 2y =14, we have:

⇨ 3(4) – 2y = 14

⇨ 12 – 2y = 14

Subtracting 12 from both sides, we get:

⇨ 12 – 2y – 12 = 14 – 12

⇨ –2y = 2

Dividing both sides by –2, we get:

⇨ *–2y***–2** = *2***–2**

⇨ y = –1

∴ The solution is (4, –1).

### Elimination by Subtraction

To solve a system of equations wherein a pair of like terms share a common coefficient, we bring into play the subtraction property of equality. Here’s an example:

6x + 3y = 33

6x – 7y = –17

The x-terms have the same coefficient. So, we should subtract to eliminate them.

6x + 3y = 33*–(6x – 7y = –17)*

10y = 50

Note: Subtracting the equations is the same as multiplying the second equation by –1 and adding it to the first.

Dividing both sides of the equation by 10, we get:

*10y***10** = *50***10**

⇨ y = 5

Substituting y = 5 in one of the equations, we get:

6x + 3(5) = 33

6x + 15 = 33

Subtracting 15 from both sides, we get:

6x + 15 – 15 = 33 – 15

6x = 18

Dividing both sides by 6, we have:

*6x***6** = *18***6**

⇨ x = 3

So, the solution set is (3, 5).

Check Your Solution!

You can cross-check if your solution is right by substituting the x and y values in the equations.

Plugging (3, 5) in equation 1, you get:

⇨ 6(3) + 3(5) = 33

18 + 15 = 33

33 = 33 ✔

Plugging (3, 5) in equation 2, you get:

⇨ 6(3) – 7(5) = –17

18 – 35 = –17

–17 = –17 ✔

### Elimination by Multiplication Followed by Addition or Subtraction

Look at this system of equations:

3x + 2y = 8

2x + 7y = –6

Does it have any coefficients – common or opposite?

The answer is obvious. There are no common or opposite coefficients in the above system of equations.

To find the solution of such a system, we need to apply the multiplication property of equality. We multiply the equation(s) by a non-zero constant, create equivalent equation(s) where the x or y-terms are common or additive inverses, carry out elimination, and solve for the variables.

You can choose to eliminate either x or y. Here’s how the eliminations work:

Eliminating x-terms

To make the coefficients of x equal, we need to multiply the equations as follows:2(3x + 2y = 8) ⇨ 6x + 4y = 16

(–3)(2x + 7y = –6) ⇨ –6x – 21y = 18

Why did we multiply the first equation by 2 and the next by –3?

The coefficients of x are 2 and 3; 6 is their least common multiple. So, we have multiplied the equations by the constants that will give us the opposites 6 and –6.

Now, we have the ground prepared for elimination by addition as follows:

6x + 4y = 16*–6x – 21y = 18*

–17y = 34

Dividing both sides by –17, we get:

*–17y***–17** = *34***–17**

y = –2

Substituting y = 2 in one of the equations, we get:

⇨ 3x + 2(–2) = 8

⇨ 3x – 4 = 8 [simplifying]

⇨ 3x – 4 + 4 = 8 + 4 [adding 4 to both sides]

⇨ 3x = 12 [simplifying]

⇨ *3x***3** = *12***3** [dividing both sides by 3]

⇨ x = 4

∴ The solution set is (4, –2).

Eliminating y-terms

Here we are going to make the coefficients of y equal. The least common multiple of the coefficients 2 and 7 is 14. So we need to multiply the first equation as follows:

7(3x + 2y = 8) ⇨ 21x + 14y = 56

2(2x + 7y = –6) ⇨ 4x + 14y = –12

Now, the above equation has one pair of like terms with a common coefficient - 14. So, let’s eliminate by subtraction as shown below:

21x + 14y = 56*–(4x + 14y = –12)*

17x = 68

Dividing both sides by 17, we get:

*17x***17** = *68***17**

x = 4

Substituting x = 4 in one of the equations, we get:

⇨ 2(4) + 7y = –6

⇨ 8 + 7y = –6 [simplifying]

⇨ 8 + 7y – 8 = –6 – 8

[subtracting 8 from both sides]

⇨ 7y = –14 [simplifying]

⇨ *7y***7** = *–14***7** [dividing both sides by 7]

⇨ y = –2

∴ The solution set is (4, –2).

If you study the steps in the above example, you will notice that the constant multiplied to the equations decides whether the elimination is by addition or subtraction.

Thus no matter which variable you eliminate, the solution remains the same.

### The Lesson in a Nutshell!

Elimination is one of the methods used to solve a system of simultaneous equations.

In the elimination method, we first eliminate one variable and find the value of the other. We then substitute this value and solve for the eliminated variable.

We add up the equations if one of the pairs of like terms has opposite coefficients.

We subtract the equations if one of the pairs of like terms has the same coefficient.

We multiply one or both equations by a non-zero constant and then eliminate one variable if none of the like-term pairs has the same or opposite coefficients.

Elimination by subtraction is the same as multiplying an equation by –1 and performing elimination by addition.

It doesn’t matter which method you choose or which variable you eliminate - the solution of a system of equations is unique.

Shape up your prowess with our free printable Solving Systems of Linear Equations worksheets!