Solving Systems of Equations by Using Elimination
By Kathleen Knowles, 23 Sep 2020
In mathematics, an equation is a statement where two mathematical expressions are equal to each other. Because this is algebra, there must be a variable in the equation. A variable is an unknown number, and we end up mostly solving these variables to prove the equation true. In some cases, we'll have to solve an equation that uses more than one variable and one equation. There are plenty of established methods for solving these equations, but one of the more common ways is by using elimination.
Let's first review some key points about equations.
An equal sign separates the two mathematical expressions of an algebraic equation. One expression is on the right-hand side of the equal sign, and the other expression is on the left-hand side of the equal sign.
An example of an equation is:
2x + 3 = 12
The left-hand side, which is 2x + 3, is equal to the right-hand side, 12.
When dealing with equations, you'll often come across these other terms:
- Term: the part(s) of the equation usually separated by a plus or minus sign. In the example given, the terms in the equation are 2x, 3, and 12.
- Coefficient: a number associated with a variable by multiplication. In the example above, the coefficient of x is 2. Only variables have coefficients.
- Constants: terms of an equation that do not have an associated variable.
- Solving the Equation: finding the values of the variables that make the equation true.
Some equations are very simple, and you can solve them without needing elaborate methods, like y = 3 or x + 1 = 3.
However, some equations are complex and require an established method for finding the solution. The elimination method is used for solving equations that have more than one variable and more than one equation. In the elimination method, you eliminate one of the variables to solve for the remaining one. Once you have solved for that variable's value, you can substitute the value into any of the equations to find the other variable.
Generally, if an equation contains two unknown variables, you need at least two equations to solve for the two unknown variables. This is called system equations.
2x + 3y = 7 …. eqn 1
x – y = 2 …. eqn 2
Eqn 1 and Eqn 2 form a system equation. The two unknown variables in the two equations are x and y.
Example 1: Two Equations
Solve the system equation below using the elimination method.
3x – 2y = 9 ….. (eqn 1)
6x – y = 27 ….. (eqn 2)
Let’s call the first equation Eqn 1 and the second equation Eqn 2.
The first step is to choose which variable to eliminate. I am going to eliminate x. Before you can eliminate, the coefficients of the variable in the two equations must be the same. The coefficient of x in eqn 1 must be the same as the coefficient of x in eqn 2. But they are not the same, so we have to make them the same. You can change the coefficients of variables by multiplying the equation with constants. So let's multiply eqn 1 by 2.
2 * (3x – 2y = 9)
6x – 4y = 18 …. (eqn 3)
As you can see, we multiplied all the terms of the equation by 2. You can also choose to divide an equation by a constant if you prefer.
Our eqn 1 is now eqn 3.
Since the coefficients of x are now the same, we can proceed with the elimination.
Let's subtract eqn 3 from eqn 2.
6x – y = 27
– 6x – 4y = 18
0 + 3y = 9 …… (eqn 4)
A few notes on this subtraction:
- The coefficient of x is now 0.
- -y – (- 4y) = -y + 4y = 3y
We now have eqn 4, which is 3y = 9.
Solve the resulting equation to find the remaining variable.
3y = 9
y = 9 / 3
y = 3.
We now know the value of y is 3.
Substitute the value of y = 3 into eqn 2 to find the value of x.
6x – y = 27
Replace y with 3.
6x – 3 = 27
6x = 30
x = 30 / 6
x = 5.
We have solved the system of equations to arrive at x = 5 and y = 3.
Let us look at another example.
Example 2: Three Equations
Solve the systems of equations below.
2x – y + 3z = 14 .... (eqn 1)
x + 5y – 2z = 20 …. (eqn 2)
x – y = 4 …. (eqn 3)
The above system equations contain three variables x, y, and z. The third equation does not have the z variable. This only means that the coefficient of z in eqn 3 is 0. So if you are to subtract, you will simply include 0z in eqn 3.
By looking at the three equations, subtracting any two equations won't leave us with only one variable, because there are three variables. If we eliminate one, we still have two variables left. There is something else we can do, though.
Consider eqn 3. It has only two variables, but we can express y in terms of x.
x – y = 4
x = 4 + y …. (eqn 4)
By moving y to the right side of the equation, we have a new equation to help us solve the problem.
Substitute eqn 4 into eqn 1. This means we will replace the x in eqn 1 with 4 + y
2x – y + 3z = 14
2 (4 + y) – y + 3z = 14
8 + 2y – y + 3z = 14
y + 3z = 14 – 8
y + 3z = 6 …. (eqn 5)
We also have a new equation.
Substitute eqn 4 into eqn 2.
x + 5y – 2z = 20
4 + y + 5y – 2z = 20
6y – 2z = 20 – 4
6y – 2z = 16 ….. (eqn 6)
This makes eqn 6, where there are now two variables.
The next step is to eliminate y. But we first need to make the coefficient of y in eqn 5 the same as in eqn 6. So we multiply eqn 5 by 6.
6 * (y + 3z = 6)
6y + 18z = 36 …. (eqn 7)
We can now subtract eqn 6 from eqn 7.
6y + 18z = 36
– (6y – 2z = 16)
0 + 20z = 20 …. (eqn 8)
Eqn 8 only contains one variable.
Solve for z.
20z = 20
z = 1
Now substitute z = 1 into eqn 5.
y + 3z = 6
y + 3 * 1 = 6
y = 3.
Now substitute y = 3 into eqn 4.
x = 4 + y
x = 4 + 3
x = 7.
The solution to the system equations is x = 7, y = 3 and z = 1.
The elimination method is not difficult to learn, but you must stay organized. Variables and substitutions can get pretty messy and confusing if you don't lay them out on the paper correctly. And, as you can see, some equations take more than a few steps to complete. Just keep your pencil handy and have plenty of scrap paper to show your work.
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