Simultaneous linear equations are pairs of linear equations with two variables. To find the value of those unknown variables, we solve them. The solution of simultaneous linear equations can be carried out by two methods: The Graphical method and the Algebraic method. The method of substitution to solve simultaneous linear equations falls under the category of the algebraic method. In this article, we will look at the meaning of the substitution method, the steps for solving simultaneous equations by substitution, and some examples to help you understand the concept. Let us quickly get an idea of the algebraic method and graphical method.

Methods to Solve Simultaneous Linear Equations

Graphical Method

Another name for the graphical method is the geometric method used to solve the system of linear equations. In this method, the equations are designed based on the objective function and constraints. Different steps are involved in obtaining the solution of simultaneous equations by graphical method.

Algebraic Method

This method mainly involves algebraic operations to solve the pair of equations with two variables. The algebraic method is classified into three categories:

1. Substitution method
2. Elimination method
3. Cross-multiplication method

This article will discuss one of the algebraic methods called the “Substitution Method” in detail.

Definition of Substitution Method

The substitution method is one of the categories of the algebraic method to solve the pair of linear equations. From the word substitution, we understand that the method majorly includes substituting certain values. In this method, we write the value of one variable in terms of another variable. Then we have to substitute the value of the variable in another equation. In this way, the equation becomes a linear equation in one variable that can easily be solved.

Steps involved in Substitution Method

Suppose we are given a pair of linear equations in two variables, say \(x\) and \(y.\) To solve these equations by the method of substitution, we follow the below-given steps:

Step 1: Consider any one equation out of the two and express \(y\) in terms of \(x\) or vice versa.

Step 2: Substitute this value of \(y\) in terms of \(x\) in another equation. This will give us a linear equation in one variable, i.e. \(x.\)

Step 3: Solve the linear equation in \(x\) in step \(2.\)

Step 4: Substitute back the value of \(x\) in the equation taken in Step \(1\) to obtain the linear equation in \(y.\)

Step 5: Solve the linear equation in \(y\) to get the value of \(y.\)

Note: We may swap the role of \(x\) and \(y\) in the above steps.

Verification

To verify whether the solution obtained is correct or not, substitute the values of \(x\) and \(y\) in any of the given systems of equations.

If the pair of linear equations has no solution, then after the substitution, you won’t get the exact value of LHS and RHS. 

Both sides of the equation will be equal to the same constant in the case of infinite solutions.

You will get a unique solution only when you get a proper value of the unknown variable after substitution.

Example

Let us assume the system of linear equations:

\(4x+5y=23\) and \(x-2y=-4\)

Given:

\(4x+5y=23 … (1)\) 

\(x-2y=-4…(2)\)

The equation \((2)\) can be written as,

\(⇒x=2y-4…….(3)\)

Now, in equation \((1)\) eliminate \(x\) by substituting the equation \((3).\)

Hence, equation \((1)\) becomes,

\(4(2y-4)+5y=23\) 

Applying the distributive property for the above equation,

\(8y-16+5y=23\)

Now, solve the above equation for \(y\),

\(13y –16=23\)

\(⇒13y=23+16\) 

\(⇒13y=39\) 

\( \Rightarrow y = \frac{{39}}{{13}}\) 

\(⇒y=3\) 

Hence, the value of \(y\) is \(3.\)

Now, substituting \(y=3\) in the equation \((2),\) we get,

\(⇒x-2(3)=-4\)

\(⇒x-6=-4\) 

\(⇒x=-4+6\) 

\(⇒x=2\)

Therefore, the value of \(x\) is \(2.\)

Hence, the solution for the system of linear equations is \(x=2\) and \(y=3.\)

Verification:

Use equation \((2)\) to verify the solution:

\(x-2y=-4\)  

Now, substitute \(x=2\) and \(y=3\),

\(⇒2-2(3)=-4\) 

\(⇒2-6=-4\) 

\(⇒-4=-4\) 

Here, L.H.S = R.H.S

Hence, we have obtained the correct solution.

Solved Examples

Q.1. Solve \(5x+y=11\) and \(3x+2y=15\)
Ans: Given, \(5x+y=11\) and \(3x+2y=15\)
For solving simultaneous equations,
Let, \(5x+y=11……..(1)\)
and \(3x+2y=15……..(2)\)
From Equation \((2)\) we get,
\(y = \left( {\frac{{15 – 3x}}{2}} \right) \ldots \ldots \ldots ..(3)\)
In the substitution method, we find the value of one variable in terms of other variables and then substitute it back.
Substitute the value of \(y\) in terms of \(x\) in equation \((1),\) we get
\(5x+y=11\)
\( \Rightarrow 5x + \left( {\frac{{15 – 3x}}{2}} \right) = 11\)
\(⇒10x +15-3x=22\)
\(⇒7x=7\)
\(⇒x=1\)

Now, the value of \(y\) can be found out using \(x=1\) in equation \((3).\)
So, \(y = \left( {\frac{{15 – 3x}}{2}} \right)\)
\( \Rightarrow y = \left( {\frac{{15 – 3 \times 1}}{2}} \right)\)
\( \Rightarrow y = \frac{{12}}{2} = 6\)
Hence the solution of the simultaneous equation will be \(x=1\) and \(y=6.\)
In this way, we can determine the value of the unknown variables \(x\) and \(y\) using the substitution method.

Q.2. Solve the pair of linear equations by substitution method: \(3x-2y=3\) and \(2x+y=9\) substitute \(x\) in terms of \(y.\)
Ans: \(3x – 2y = 3 \ldots \ldots .{\rm{(i)}}\)
\(2x + y = 9 \ldots \ldots ..{\rm{(ii)}}\)
Finding the value of \(x\) in terms of \(y\) from equation \({\rm{(i)}},\) we get-
\(3x-2y=3\)
\(⇒3x=3+2y\)
\( \Rightarrow x = \left( {\frac{{3 + 2y}}{3}} \right)……………{\rm{(iii)}}\)

Using this method, substituting the value of \(x\) in equation \({\rm{(ii)}},\) we get-
\(2x+y=9\)
\( \Rightarrow 2\left( {\frac{{3 + 2y}}{3}} \right) + y = 9\)
\(⇒6+4y+3y=27\)
\(⇒7y=21\)
\(⇒y=3\)

Finding the value of \(x,\) substitute the value of \(y\) in equation \({\rm{(iii)}},\) we get-
\( \Rightarrow x = \left( {\frac{{3 + 2y}}{3}} \right)\)
\( \Rightarrow x = \left( {\frac{{3 + 2 \times 3}}{2}} \right)\)
\( \Rightarrow x = \frac{9}{3} = 3\)
Hence the value of both \(x\) and \(y\) is \(3.\)

Q.3. Solve the following systems of simultaneous linear equations by the method of substitution.
\(3x-8y=21\) and \(x-y=2\)
Ans: \(3x – 8y = 21 \ldots \ldots .\left( {\rm{i}} \right)\)
\(x – y = 2 \ldots \ldots \ldots {\rm{(ii)}}\)
Finding the value of \(y\) in terms of \(x\) from equation \(\left( {\rm{i}} \right),\) we get-
\(3x-8y=21\)
\(⇒-8y=21-3x\)
Multiplying the whole equation by \(-1\) to remove the negative coefficient
\( \Rightarrow – 1[ – 8y] = – 1[21 – 3x]\)
\(⇒8y=\,-21+3x\)
\( \Rightarrow y = \left( {\frac{{3x – 21}}{8}} \right) \ldots \ldots \ldots {\rm{(iii)}}\)

Using this method, substituting the value of \(y\) in equation \(\left( {\rm{ii}} \right),\) we get-
\(x-y=2\)
\( \Rightarrow x – \left( {\frac{{3x – 21}}{8}} \right) = 2\)
\( \Rightarrow 8x – 3x + 21 = 16\)
\(⇒5x=-5\)
\(⇒x=-1\)

Finding the value of \(y,\) substitute the value of \(x\) in equation \(\left( {\rm{iii}} \right),\) we get-
\( \Rightarrow y = \left( {\frac{{3x – 21}}{8}} \right)\)
\( \Rightarrow y = \left( {\frac{{3x – 1 – 21}}{8}} \right)\)
\( \Rightarrow y = \frac{{ – 24}}{8} = \, – 3\)
Hence the value of \(x=-1\) and \(y=-3.\)

Q.4. Solve for \(x\) and \(y,\) using the substitution method.
\(\frac{x}{2} + \frac{y}{3} = \frac{1}{6}\) and \(\frac{{2x}}{3} + \frac{y}{4} = \frac{5}{{12}}\)
Ans: Given, \(\frac{x}{2} + \frac{y}{3} = \frac{1}{6}\) and \(\frac{{2x}}{3} + \frac{y}{4} = \frac{5}{{12}}\)
The given pair of equations may be written as
\(3x+2y=1……(1)\)
\(8x+3y=5……(2)\)
Finding the value of \(y\) in terms of \(x\) from equation \(\left( {\rm{i}} \right),\) we get-
\(3x+2y=1\)
\(⇒2y=1-3x\)
\( \Rightarrow y = \left( {\frac{{1 – 3x}}{2}} \right)………..{\rm{(iii)}}\)

Using this method, substituting the value of \(y\) in equation \(\left( {\rm{ii}} \right),\) we get-
\(8x+3y=5\)
\( \Rightarrow 8x + 3\left( {\frac{{1 – 3x}}{2}} \right) = 5\)
\(⇒16x+3-9x=10\)
\(⇒7x=7\)
\(⇒x=1\)
Finding the value of \(y,\) substitute the value of \(x\) in equation \(\left( {\rm{iii}} \right),\) we get-
\( \Rightarrow y = \left( {\frac{{1 – 3x}}{2}} \right)\)
\( \Rightarrow y = \left( {\frac{{1 – 3 \times 1}}{2}} \right)\)
\( \Rightarrow y = \frac{{ – 2}}{2} = \, – 1\)
Hence the value of \(x=1\) and \(y=-1.\)

Q.5. Solve the system of linear equations. Also, verify your answer.
\(4x+5y=-3\) and \(2x-3y=4.\)
Ans: Given,
\(4x+5y=-3…….(1)\)
\(2x-3y=4…….(2)\)
The equation \((2)\) can be written as
\( \Rightarrow x = \left( {\frac{{4 + 3y}}{2}} \right)………(3)\)
Now, in equation \((1)\) eliminate \(x\) by substituting in equation \((3).\)
Hence, equation \((1)\) becomes
\(4\left( {\frac{{4 + 3y}}{2}} \right) + 5y = \, – 3\)
Applying the distributive property for the above equation,
\(8+6y+5y=-3\)

Now, solve the above equation for \(y\)
\(⇒11y+8=-3\)
\(⇒11y=-11\)
\( \Rightarrow y = \frac{{ – 11}}{{11}}\)
\(⇒y=-1\)
Hence, the value of \(y\) is \(-1.\)
Now, substituting \(y=-1\) in the equation \((2),\) we get
\(2x-3y=4\)
\(⇒2x-3(-1)=4\)
\(⇒2x+3=4\)
\(⇒2x=1\)
\( \Rightarrow x = \frac{1}{2}\)
Therefore, the value of \(x\) is \(\frac{1}{2}.\)
Hence, the solution for the system of linear equations is \(x = \frac{1}{2}\) and \(y=-1.\)

Verification:
Use equation \((2)\) to verify the solution
\(2x-3y=4\)
Now, substitute \(x = \frac{1}{2}\) and \(y=-1\)
\( \Rightarrow 2 \times \frac{1}{2} – 3( – 1) = 4\)
\(⇒1+3=4\)
\(⇒4=4\)
Here, L.H.S = R.H.S
We have obtained the correct answer.
Hence verified.

Summary

In this article, we learned that there are two methods of solving simultaneous linear equations. Then, we studied one of the algebraic methods of solving the pair of equations: the substitution method. We understood the meaning of the substitution method and then learnt the steps involved in solving pair of linear equations by the substitution method. Lastly, we have solved examples with positive, negative and fractional coefficients of variables in the pair of linear equations. This article will help students to learn the method of substitution quickly.

Frequently Asked Questions (FAQs)

Q.1. What is systems of equations by substitution?
Ans: In mathematics, the system of equations by substitution are pair of linear equations that have to be solved simultaneously using the substitution method. Here, we solve the equation by converting it into the equation in one variable by substitution. Then, back substitute the value of the variable in another equation.

Q.2. What are the 3 methods for solving systems of equations?
Ans: There are majorly two methods of solving the system of the equation: Graphical method and Algebraic method
There are three algebraic methods of solving the pair of linear equations in two variables:
1. Substitution method
2. Elimination method
3. Cross-multiplication method

Q.3. When would you use the substitution method?
Ans: The substitution method can be applied when we have smaller coefficients in terms or when the equations are given in form \(x=ay+c,\) and \(y=bx+p\) it is advisable to use the substitution method
This method may be applied to any simultaneous linear equations with two variables.

Q.4. What is the benefit of using the substitution method?
Ans: The benefit of using the substitution method is that this method gives us the exact values of the variables (preferably \(x\) and \(y,\) which coincide at the point of intersection.

Q.5. Can the substitution method be used to solve the system of equations in three variables?
Ans: Yes, the substitution method can be used to solve the pair of linear equations in three variables. In general, while solving the system of equations with three variables, either we can use the substitution method or the elimination method to convert the system into the system of two equations with two variables first, and from there, we use the elimination or substitution method again and solve them.

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