In algebra, the system of equations can be defined as the set of systems of equations that seek common answers. The use of the equations can be seen in our everyday life where the values which are unknown are expressed with the help of variables. The process of solving an equation to get an appropriate answer using different methods can be regarded as solving a system of equations. As mentioned, there are various methods such as substitution method, elimination method, method of the graph. In this article, we will try to cover some basic concepts regarding solving systems of equations such as types of methods to solve, types of solutions, and do a detailed analysis about them.
Different Methods of Solving System of Equation
In the paragraph mentioned above, we dealt with the system of equations. There are various methods to solve an equation. In the next few points, we may cover the types of methods for solving any equation. Some of them are as follows:
- The substitution method of solving the system of equations is a type where one variable is substituted with the second one in order to get an appropriate answer. Let us solve some examples so that you can understand the concept clearly.
Example 1: Find the values of A and B, if the two equations are, 3a – b = 23 and 4a + 3b = 48
According to the question,
Equation 1 = 3a – b = 23
Equation 2= 4a + 3b = 48
From the equation 1, we get the values of,
b = 3a − 23, equation 3
Now use the equation of b in equation 2
4a + 3 (3b − 23) = 48
13a − 69 = 48
13a = 117
a = 9
Now, use the value of a = 9 in equation 2, thus the value of b = 4.
Therefore, the values of a and b are 9 and 4 respectively.
- The elimination method is a type of method where one unknown number is eliminated by multiplying the given equations by some suitable numbers. This will make the coefficients similar to that of variables. Let us solve some examples of it so that you understand the topic clearly.
Example 2:
Find the values of a and b if the equations given are 2a+ 3b = 4 and 3a+ 2b= 11?
Given that,
Equation 1 = 2a + 3b = 4
Equation 2 = 3a + 2b = 11
Now,
Multiplying Equation 1 by 2 and equation 2 by 3 as the LCM of coefficients are 3 and 2 respectively.
Equation 3 = 4a + 6b = 8
Equation 4 = 9a + 6b = 33
On subtracting equation 3 from 4, the result is,
5a = 25
a = 5
Now, substituting the value in equation 2 to get,
y = -2
Therefore, the values of a and b for the given equation are 5 and -2 respectively.
- In this method of system of equations we plot the given equations in a bar or coordinate graph to get an appropriate answer.
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