Substitution Method

The substitution method can be defined as a way to solve a linear system algebraically. The substitution method works by substituting one y-value with the other. To put it simply, the method involves finding the value of the x-variable in terms of the y-variable. After this is done, we then end up substituting the value of x-variable in the second equation. This helps us to directly find the value of the y-variable. We can now easily put the value of y in any of the given equations to find x.

The method of substitution involves three steps:

• Step  1: First you need to solve one equation for one of the variables.

• Step 2: Now you need to substitute (plug-in) this expression into the other equation and solve it.

• Step 3: In the last step you need to re-substitute the value into the original equation and you will be able to find the corresponding variable.

Now at first glance, this method might seem complicated, but some helpful tricks will help you to keep things straight. You need to keep in mind that our goal when solving any system is to find the point of intersection. 

What is the Algebraic Method?

An Algebraic method can be defined as a collection of several methods, which are generally used to solve a pair of linear equations that includes two variables supposedly (a,b) or (x,y). Generally, the algebraic method can be further subdivided into three different categories:

• The Substitution Method: We have already seen what the substitution method is all about and the various steps involved in solving questions via this method. 

• The Elimination Method: Combining equations can turn out to be a rather powerful tool for solving a system of linear equations. Students should know that adding or subtracting two equations to eliminate a common variable is what is referred to as the elimination method. Once one variable is eliminated, we can easily find the value of the other one.

• The Cross-Multiplication Method: As the name suggests, the Cross-multiplication method makes the use of multiplication to set up matching terms in equations before they are combined. This ultimately helps us to find a solution to the system of linear equations provided. When using the multiplication method, students ought to multiply all the terms on both sides of the equation. Do not just multiply the term you are trying to eliminate or you will end up with an incorrect answer to your problem. 

Difference between the Substitution and Elimination Method

As we have noted earlier, both the elimination and the substitution methods are the ways to solve linear equations algebraically. The elimination method is a little easier to apply than the substitution method when it comes to dealing with big equations or fractions. That is why the elimination method is often referred to as the substitution method. However, let us take a comprehensive look at the differences between these two methods through the table given below:

• Substitution Method: In this method, we modify the equation to find the value of any one of the variables. After this is done, we substitute its value in the other equation that has been given. Mathematicians have noted that it is better to use the substitution method when equations are given in the form of x = ay + b and y = MX + n.

• Elimination Method: In this method, we multiply or divide either one or both the equations by a number to make the coefficient of one of the two unknown variables the same in both the equations. Then, we move on to add or subtract the equations to eliminate the variable whose coefficient is the same. Thus, our aim is that we try to find the value of one variable which can be substituted in any one of the equations to find the other variable too. In contrast to the substitution method, mathematicians have noted that it is better to use the elimination method when the coefficient of any one of the terms is the same. For example, we can use the elimination method when Ax + By + C = 0 and Px + By + R = 0.

What is the Graphical Method?

The graphical method can also be known as the geometric Method used to solve the system of the linear equation. In this graphical method, the equations are designed based on the constraints and objective function.

To solve the system of linear equations, this method involves different steps to obtain the solutions. In this article, we are going to focus mainly on solving the linear equations using the first algebraic method which is known as the “Substitution Method” in detail.

Let’s go through the example,

Y = 2x + 4,  3x + y = 9

Now we can substitute y in the second equation with the first equation since we can write the second equation in terms of y(y=y) as 

3x + y = 9 

y=2x + 4 

3x + (2x + 4) =9 

5x + 4 =9

5x = 9 - 4

5x=5

x=1

This value of x that is equal to 1 can then be used to find the value of y by substituting 1 with x example in the first equation,

Y = 2x + 4y = 2x + 4 = 2*1 + 4y 

= 2*1+ 4 = 6

Y = 6

Therefore, the value of y is equal to 6.

The solution of the linear system is equal to (1, 6).

We can use the substitution method even if both equations of the linear system are in standard form. We can just begin by solving one of the equations for one of its variables.

Questions to be Solved :

Question 1: Solve for the values of ‘x’ and ‘y’:    x + y = 5.   3x + y = 11.

Ans:  Let’s write down the information given,       

 x + y = 5 …………… Equation (i)       

3x + y = 11 …………Equation (ii)

Since we are given two different equations in terms of two different linear equations, now let us try to solve them using the method of substitution:

From the first equation we find that we can write y = 5 - x.

Substituting the value of y in the Equation (ii),

we get; 3x + (5 – x) = 11⇒2x = 11 - 5⇒ 2x = 6⇒ x = 6/2.

Therefore the value of x = 3.

Now substituting the value x = 3 in the other equation that is y = 5 – x, we get;y = 5- x ⇒y = 5 - 3

Therefore, we get the value of y = 2.

Hence, the value of x = 3 and the value of y = 2

Question 2: Solve for the values of ‘x’ and ‘y’:       2x + 6y = 10,       1x - 2y = 15

 Ans:  Let’s write down the information given,       

2x + 6y = 10 …………… Equation (i)      

1 x - 2y = 15 …………… Equation (ii)

From the given two equations, let us consider first equation and find out value of one of the variables, say ‘x’ from the equation:     2x = 10 - 6y  ⟹ x = 5 - 3y

Substituting the value of x = 5 - 3y in Equation (ii),

we get; (5 - 3y) - 2y = 15 ⟹ -5y = 15 - 5 ⟹ -5y = 10 ⟹ y = -2.

Now substituting the value of  y = -2 in the equation x = 5 - 3y,

we get; x = 5 - 3(-2) ⟹ x = 5 + 6 ⟹ x = 11.

Hence,  the value of x = 11 and y = 2