- Written By Sushma_P
- Last Modified 19-07-2022

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:

**Substitution method****Elimination method****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.

**Practice 10th CBSE Exam Questions**

**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** \(5x+y=11\)

**\(3x+2y=15\)**

*and***Given, \(5x+y=11\) and \(3x+2y=15\)**

*Ans:*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:** \(3x-2y=3\)

**\(2x+y=9\)**

*and***\(x\)**

*substitute***\(y.\)**

*in terms of***\(3x – 2y = 3 \ldots \ldots .{\rm{(i)}}\)**

*Ans:*\(2x + y = 9 \ldots \ldots ..{\rm{(ii)}}\)

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** \(x-y=2\)

**\(3x – 8y = 21 \ldots \ldots .\left( {\rm{i}} \right)\)**

*Ans:*\(x – y = 2 \ldots \ldots \ldots {\rm{(ii)}}\)

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** \(x\)

**\(y,\)**

*and*

*using the substitution method.*\(\frac{x}{2} + \frac{y}{3} = \frac{1}{6}\)

**\(\frac{{2x}}{3} + \frac{y}{4} = \frac{5}{{12}}\)**

*and***Given, \(\frac{x}{2} + \frac{y}{3} = \frac{1}{6}\) and \(\frac{{2x}}{3} + \frac{y}{4} = \frac{5}{{12}}\)**

*Ans:*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** \(2x-3y=4.\)

**Given,**

*Ans:*\(4x+5y=-3…….(1)\)

\(2x-3y=4…….(2)\)

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.

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**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.

Learn About Linear Equations Here

*Now that you have a detailed article on the Method of substitution to solve simultaneous linear equations, we hope you study well. If you get stuck somewhere do let us know in the comments sections. We will get back to you at the earliest.*

## FAQs

### How do you solve simultaneous linear equations using substitution? ›

2 we have 3 y plus x equals to 5 but in this case we know that x equals to 3 minus y from Equation 1

### What are the 5 steps in solving equations by substitution? ›

**Solving Systems of Equations By Substitution:**

- Step 1: Rearrange one of the equations to get 'y' by itself. ...
- Step 2: Substitute the rearranged equation into its partner. ...
- Step 3: Solve for x. ...
- Step 4: Substitute the solution for x into either of the initially given equations to find y. ...
- Step 5: Write final answer out as a point.

### What are the steps of substitution method? ›

The method of substitution involves three steps: Solve one equation for one of the variables. Substitute (plug-in) this expression into the other equation and solve. Resubstitute the value into the original equation to find the corresponding variable.

### What is substitution method with example? ›

The first step in the substitution method is to **find the value of any one of the variables from one equation in terms of the other variable**. For example, if there are two equations x+y=7 and x-y=8, then from the first equation we can find that x=7-y. This is the first step of applying the substitution method.

### What are the 3 ways to solve simultaneous equations? ›

If you have two different equations with the same two unknowns in each, you can solve for both unknowns. There are three common methods for solving: **addition/subtraction, substitution, and graphing**. This method is also known as the elimination method.

### How do you solve two simultaneous equations? ›

Simultaneous Equations Math Lesson - YouTube

### What is the first step to solve the following system of equations using substitution? ›

Solving Systems of Equations... Substitution Method (NancyPi)

### How do you solve a system of linear equations by elimination and substitution? ›

**To Solve a System of Equations by Elimination**

- Write both equations in standard form. ...
- Make the coefficients of one variable opposites. ...
- Add the equations resulting from Step 2 to eliminate one variable.
- Solve for the remaining variable.
- Substitute the solution from Step 4 into one of the original equations.

### What is solving systems of equations by substitution? ›

The method of solving "by substitution" works by solving one of the equations (you choose which one) for one of the variables (you choose which one), and then plugging this back into the other equation, "substituting" for the chosen variable and solving for the other.

### What is the definition of the substitution method? ›

The substitution method is **the algebraic method to solve simultaneous linear equations**. As the word says, in this method, the value of one variable from one equation is substituted in the other equation.

### What are the correct order of steps to solve the following system by substitution? ›

**Steps for Using the Substitution Method in order to Solve Systems of Equations**

- Solve 1 equation for 1 variable. ( Put in y = or x = form)
- Substitute this expression into the other equation and solve for the missing variable.
- Substitute your answer into the first equation and solve.
- Check the solution.

### How do you solve a substitution equation in standard form? ›

Solving Systems of Equations with Substitution (both standard form)

### How many types of simultaneous equations are there? ›

There are **three** different approaches to solve the simultaneous equations such as substitution, elimination, and augmented matrix method.

### What do you mean by simultaneous equation? ›

**a set of two or more equations, each containing two or more variables whose values can simultaneously satisfy both or all the equations in the set**, the number of variables being equal to or less than the number of equations in the set.

### What are simultaneous linear equation in two variables? ›

**Two or more linear equations that all contain the same unknown variables** are called a system of simultaneous linear equations. Solving such a system means finding values for the unknown variables which satisfy all the equations at the same time.

### What is the rule for simultaneous equations? ›

**If the signs are different, add the equations together.** **If the signs are the same, subtract them**.

### How Do You Solve 3 linear equations with 2 variables? ›

Solving Systems of 3 Equations Elimination - YouTube

### Why do we use simultaneous equations? ›

Simultaneous equations can be used **when considering the relationship between the price of a commodity and the quantities of the commodity people want to buy at a certain price**. An equation can be written that describes the relationship between quantity, price and other variables, such as income.

### How do you solve systems of equations with three variables using substitution? ›

How to Solve Systems by Substitution With 3 Variables : Algebra - YouTube

### How do you solve word problems using substitution and elimination? ›

Solving a word problem using substitution and elimination

### What is elimination method with example? ›

**Both the variables get eliminated**. For example, let us solve two equations 2x - y = 4 → (1) and 4x - 2y = 7 → (2) by the elimination method. In order to make the x coefficients equal in both the equations, we multiply equation (1) by 2 and equation (2) by 1. By doing so we get, 4x - 2y = 8 → (3) and 4x - 2y = 7 → (4).

### Which of the following is an example of a linear system of equations? ›

Some of the examples of linear equations are **2x – 3 = 0**, 2y = 8, m + 1 = 0, x/2 = 3, x + y = 2, 3x – y + z = 3.

### What is the definition of elimination method? ›

The elimination method is the process of eliminating one of the variables in the system of linear equations using the addition or subtraction methods in conjunction with multiplication or division of coefficients of the variables.

### How do you solve systems of equations with three variables using substitution? ›

How to Solve Systems by Substitution With 3 Variables : Algebra - YouTube

### How do you do the substitution method of integration? ›

According to the substitution method, **a given integral ∫ f(x) dx can be transformed into another form by changing the independent variable x to t**. This is done by substituting x = g (t). Now, substitute x = g(t) so that, dx/dt = g'(t) or dx = g'(t)dt.

### How do you solve linear equations in two variables? ›

Linear Equations in 2 Variables - Review - YouTube

### How do you solve a system of equations with 3 variables and 2 equations? ›

Solving a System of 2 Equations with 3 Unknowns - YouTube

### How Do You Solve 3 linear equations with 2 variables? ›

Solving Systems of 3 Equations Elimination - YouTube