In mathematics, the system of linear equations contains two or more equations in which the power of each variable is one and there will be two or more variables involved. There are various methods for solving the system of linear equations.
These methods are the substitution method, elimination method, graphical method, cross-multiplication method, and so on. In this article, we’ll discuss two methods of solving linear equations by elimination method and the substitution method along with solved examples.
Table of Contents
What is the system of linear equations?
In mathematics, a collection of one or more linear equations in which the same variables are involved is said to be a system of linear equations. The equations of the system of linear equations are simultaneously satisfied.
In linear algebra, the system of linear equations is the basis and fundamental part that is helpful in modern mathematics. There is an infinite number of solutions for the system of the linear equation.
For assistance, 3x + 2y = 12 and 4x – 3y = 20 are two linear equations and simultaneously both equations are the form of the system of linear equations. Algebraically, there are two well-known methods of the system of linear equations.
Method for solving the system of linear equations
There are two well-known and most-used methods for solving the system of linear equations.
- Elimination method
- Substitution method
Let us discuss both methods for solving linear equations.
Elimination Method
The elimination method is one of the best and most widely used techniques for solving the system of linear equations. According to this method, either add the equation or subtract equations to eliminate one variable and get the equation in one variable.
There are two possibilities for adding or subtracting the system of linear equations.
For same coefficients
When there are two same coefficients in a system of linear equations, then you have to add or subtract the equations on the basis of the signs of the equations. If both the equations have the same sign such as plus or minus then you have to subtract the equations.
If the signs of the same variables in the system of linear equations are different then you have to add the equations.
For different coefficients
When there are two different coefficients in a system of linear equations, then you have to multiply one or both the equations to make the same coefficients of any variable of the system of linear equations. If both the equations have the same sign such as plus or minus then you have to subtract the equations.
If the signs of the same variables in the system of linear equations are different then you have to add the equations.
Examples of elimination method
Let’s take a few examples of solving the system of linear equations by the elimination method.
Example 1
Solve the system of linear equations to evaluate the undefined variables of the equations.
11x + 6y = 22
11x + 2y = 16
Solution
Step 1: The system of linear equations has the same coefficients for one variable with the same sign, so subtract both equations.
11x + 6y = 22
-(11x + 2y = 16)
0 + 4y = 6
4y = 6
y = 6/4 = 3/2 = 1.5
Step 2: Now put the above value of the y to any linear equation to evaluate the other variable of the equation.
11x + 6 (3/2) = 22
11x + 3 (3) = 22
11x + 9 = 22
11x = 22 – 9
11x = 13
x = 13/11 = 1.1818
The problems of the elimination method can also be done with the help of the elimination calculator by AllMath. Let us solve the above system of linear equations using online calculator.
Example 2
Solve the system of linear equations to evaluate the undefined variables of the equations.
3x + 3y = 4
4x + 2y = 6
Solution
Step 1: As there are no similar coefficients of any variable in the system of linear equations by multiplying equation (1) by 2 and equation (2) by 3.
2 (3x + 3y = 4) → 6x + 6y = 8
3 (4x + 2y = 6) → 12x + 6y = 12
Step 2: Now the system of linear equations has the same coefficients for one variable with the same sign, so subtract both equations.
6x + 6y = 8
-(12x + 6y = 12)
-6x + 0 = -4
-6x = -4
6x = 4
x = 4/6 = 2/3
Step 3: Now put the above value of the y to any linear equation to evaluate the other variable of the equation.
6 (2/3) + 6y = 8
2 (2) + 6y = 8
4 + 6y = 8
6y = 8 – 4
6y = 4
y = 4/6
y = 2/3
Substitution Method
The substitution method is another method for solving the system of a linear equation that is a frequently used technique. It is an algebraic method in which the value of one variable is taken from one equation and substituted into the other linear equation.
In this method, you have to transform one equation and put it into the other equation for solving the system of linear equations. Below are a few steps to follow for solving the system of linear equations.
- Solve any equation of the system of linear equations for finding the value of x or y.
- Substitute the evaluated value of x or y for the other equation
- Evaluate the value of the variable and put it in the first equation.
- Both variables are evaluated by the substitution method
Examples of Substitution Method
Let’s take a few examples for solving the system of linear equations by substitution method.
Example 1
Solve the system of linear equations to evaluate the undefined variables of the equations.
5x + 3y = 20
2x + 5y = 12
Solution
Step 1: Take the given linear equations
5x + 3y = 20
2x + 5y = 12
Step 2: Now find the value x from the first equation.
5x + 3y = 20
5x = 20 – 3y
x = [20 – 3y] / 5
x = 20/5 – 3y/5
x = 4 – 3y/5
Step 3: Now substitute the value of x in the second linear equation.
2 (4 – 3y/5) + 5y = 12
8 – 6y/5 + 5y = 12
8 – 1.2y + 5y = 12
8 + [-1.2 + 5]y = 12
8 + 3.8y = 12
3.8y = 12 – 8
3.8y = 4
y = 4/3.8
y = 1.05
Step 4: Now substitute the value of y in the first equation.
x = 4 – 3y/5
x = 4 – 3(1.05)/5
x = 4 – 3.15/5
x = 4 – 0.63
x = 3.37
Example 2
Solve the system of linear equations to evaluate the undefined variables of the equations.
x + 4y = 28
5x + 3y = 15
Solution
Step 1: Take the given linear equations
x + 4y = 28
5x + 3y = 15
Step 2: Now find the value x from the first equation.
x + 4y = 28
x = 28 – 4y
Step 3: Now substitute the value of x in the second linear equation.
5 (28 – 4y) + 3y = 15
140 – 20y + 3y = 15
140 – 17y = 15
– 17y = 15 – 140
– 17y = -125
17y = 125
y = 125/17
y = 7.35
Step 4: Now substitute the value of y in the first equation.
x = 28 – 4y
x = 28 – 4(7.35)
x = 28 – 4(7.35)
x = 28 – 29.4
x = -1.4
Summary
Now you can learn how to evaluate the system of linear equations by using the elimination method and substitution method. We have discussed the system of linear equations and its methods for solving linear equations with solved examples.