CBSE Notes Class 10 Maths Chapter 3 - Pair of Linear Equations in Two Variables

In Class 10 mathematics, the topic Pair of Linear Equations in Two Variables is a fundamental concept that forms the foundation for algebra and higher-level mathematics. This chapter delves into how two linear equations can be solved simultaneously using different methods, ensuring students master the techniques and concepts crucial for their exams. Let’s dive into the details of this chapter by using CBSE Notes, breaking down the concepts, methods, and important questions step by step.

1.0Download CBSE Notes for Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables - Free PDF !!

Students can download a complimentary PDF containing CBSE Notes for Class 10 Maths chapter 3 "Pair of Linear Equations in Two Variables." This readily accessible resource is designed to facilitate a clear understanding of the chapter's concepts.

2.0What is a Pair of Linear Equations in Two Variables?

A pair of linear equations in two variables refers to two equations of the form:

$a_{1}x+b_{1}y+c_1=0$

$a_{2}x+b_{2}y+c_2=0$

Here, x and y are the two variables, and a₁, b₁, c₁, a₂, b₂, c₂ are constants. The objective is to find values of x and y that satisfy both equations simultaneously.

3.0Class 10 Math’s Pair of Linear Equations in Two Variables: Key Concepts

In Class 10, you will learn to solve the pair of linear equations in two variables using some main methods:

1. Graphical Method: Plotting the two equations on a graph to find the point of intersection, which gives the solution to the system.
2. Substitution Method: Solve one equation for a variable and substitute its value into the second equation to find the solution.
3. Elimination Method: By addition or subtraction, the equations, one variable is eliminated, allowing for easier solving of the remaining variable.
4. Cross-Multiplication Method: A direct method to solve equations without modifying or simplifying them, based on a formula.

4.0Graphical Representation of Linear Equations

In the graphical method, the two equations are plotted on the coordinate plane. There are three possible outcomes: 

1. Intersecting Lines: The lines intersect at one point, which represents the unique solution of the system. This is called a consistent system.
2. Parallel Lines: The lines never meet, implying there is no solution. This is known as an inconsistent system.
3. Coincident Lines: The two lines are identical, meaning they have infinitely many solutions. This is also a consistent system.

If a pair of linear equations is given by $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$ , then

1. If $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$ then it is Intersecting Lines.
2. If $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$ then it is Parallel Lines.
3. If $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$, then it is Coincident Lines.

5.0Solutions to a Pair of Linear Equations in Two Variables: Class 10 Methods

1. Substitution Method

This method involves solving one equation for either x or y and substituting this value into the second equation. Here's how it works:

Step-by-step:

• Solve for one variable: Change one of the equations to find one variable (like ( x )) in terms of the other (like ( y )).
• Substitute: Put this expression into the other equation.
• Solve for the second variable: Solve the new equation to find the value of the second variable.
• Find the first variable: Substitute this value in the first equation to find the value of the first variable.

2. Elimination Method

In this method, the goal is to eliminate one variable by addition or subtraction the two equations after manipulating their coefficients.

Step-by-step:

• Make coefficients equal: If needed, multiply the equations so that the coefficients (numbers in front) of one variable are the same.
• Add or subtract: Add or subtract the equations to cancel out that variable.
• Solve for the remaining variable: Solve the new equation to find the value of the remaining variable.
• Find the other variable: Substitute this value back into one of the original equations to find the value of the other variable.

3. Cross-Multiplication Method

The cross-multiplication formula for solving the pair of linear equations in two variables is derived from the coefficients of the variables in the two equations:

For equations of the form:

$a_{1}x+b_{1}y+c_1=0$

$a_{2}x+b_{2}y+c_2=0$

The solution is given by:

$x=\frac{b_1{c}_2-b_2{c}_1}{a_1{b}_2-a_2{b}_1}$

$y=\frac{c_1{a}_2-c_2{a}_1}{a_1{b}_2-a_2{b}_1}$

This method is particularly useful when the equations are complex, and it's easier to use the formula directly.

6.0MCQs on Pair of Linear Equations in Two Variables

Understanding the concepts can be reinforced by practicing multiple-choice questions (MCQs), which often appear in exams. Here are a few sample MCQs:

1. If two lines are parallel, the system of equations will have:

a. One solution

b. No solution

c. Infinitely many solutions

d. None of these

2. The pair of equations 2x + 3y = 5 and 4x + 6y = 10 represents:

a. Intersecting lines

b. Parallel lines

c. Coincident lines

d. None of these

3. In the substitution method, if one equation is x = 4y – 5, what is the next step?

a. Substitute x = 4y – 5 into the second equation.

b. Solve for y in the first equation.

c. Add the two equations.

d. Multiply the second equation by a constant.

7.0Extra Questions on Pair of Linear Equations in Two Variables

To fully grasp the topic, students should solve extra questions beyond the textbook. Some example problems include:

1. Use the elimination method to solve the following system of equations:

3x + 4y = 12 and 5x - 6y = –3 

2. Use the substitution method to solve:

x + 2y = 7 and 3x - y = 5

3. Solve using the graphical method:

2x - y = 1 and x + y = 3

8.0Practice Questions on Pair of Linear Equations in Two Variables 

Practice is key to mastering this topic. Here are some questions to practice:

1. Solve the system of equations:

4x + 5y = 9 and 7x – 2y = 5

2. For what values of k will the following system have no solution?

kx + 3y = 7 and 4x + 6y = 5 

3. Find the solution using the cross-multiplication method:

2x - 3y = 4 and 5x + y = 1

9.0Sample Question on Pair of Linear Equations in Two Variables

1. What is a pair of linear equations in two variables?

Ans: A pair of linear equations in two variables is a set of two linear equations that contain 2 unknowns or variables, typically written as:

$a_{1}x+b_{1}y+c_1=0$

$a_{2}x+b_{2}y+c_2=0$

Here, x and y are the variables, while a₁, b₁, c₁, a₂, b₂, and c₂ are constants.

10.0CBSE Class 10 Math Chapter 3 - Key Notes

Simultaneous linear equations in two variables

A pair of linear equations in two variables is said to form a system of simultaneous linear equations.

General Form:

a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, where a₁, a₂, b₁, b₂, c₁ and c₂ are real non zero numbers;
a₁² + b₁² ≠ 0 and a₂² + b₂² ≠ 0 and x, y are variables.

Investigating graphs of a system of equations

The graphical representation of a pair of simultaneous linear equations in two variables will be in one of the forms.

Types of System of Equation

• Consistent system : A system of equations with at least one solution is called a consistent system (Unique and infinite many solutions).
• Inconsistent system : A system of equations with no solution is called an inconsistent system (No solutions).
• Dependent system : A system of equations with an infinite number of solutions is said to be dependent (Infinite many solutions).
• Independent system : A system of equations with only one solution is said to be independent (Unique solutions).

Algebraic conditions for consistency / inconsistency of the system

Algebraic solution by substitution method

Ex. Solve x + y = 5, 3x + y = 11 and find x, y by substitution method.

Sol.

x + y = 5 …………… (i)
3x + y = 11 …………… (ii)

From 1st eq. we find that y = 5 − x. Substituting value of y in eq. (ii), we get

3x + 5 − x = 11
2x = 11 − 5
2x = 6
x = 6/2
x = 3

Substituting x = 3 in y = 5 − x

y = 5 − 3
y = 2

Hence, x = 3 and y = 2.

Algebraic solution by elimination method

Ex. Solve the equations by elimination method
3x + 2y = 11
2x + 3y = 4

Sol.

3x + 2y = 11 ........(i)
2x + 3y = 4 ..........(ii)

Multiplying equation (i) by 2 and equation (ii) by 3 and subtracting

(6x + 4y) − (6x + 9y) = 22 − 12
−5y = 10
y = −2

Substitute the value of y in equation (i)

3x + 2 × (−2) = 11
3x = 11 + 4
3x = 15
x = 15/3
x = 5

∴ Solution of system of equations is x = 5, y = −2.

Algebraic solution by cross multiplication method

Ex. Solve the equations by cross multiplication method:

3x + 2y = 10 ....(1)
4x − 2y = 4 ....(2)

Using formula for cross multiplication method.

So, from equation (1) and (2) we can write the value of a, b and c.

−14x = −28
x = 2

Again,

−14y = −28
y = 2

Therefore, x = 2, y = 2

Interchanged Coefficients Method

To solve the equations of the form:

ax + by = c ...(1)
bx + ay = d ...(2)

where a ≠ b, we follow the following steps:

Step-1 : Add (1) and (2) and obtain (a + b)x + (b + a)y = c + d

i.e., x + y = ... (3)

Step-2 : Subtract (2) from (1) and obtain (a − b)x − (a − b)y = c − d

i.e., x − y = ... (4)

Step-3 : Solve (3) and (4) to get x and y.

Example

Solve for x and y:

47x + 31y = 63 ...(1)
31x + 47y = 15 ...(2)

Adding (1) and (2), we get

78x + 78y = 78
x + y = 1 ...(3)

Subtracting (2) from (1)

16x − 16y = 48
x − y = 3 ...(4)

Now adding (3) and (4)

2x = 4
x = 2

Putting x = 2 in (3)

2 + y = 1
y = −1

Hence, the solution is
x = 2 and y = −1

Equations reducible to linear equations in two variables

Ex. Solve for x and y : 2/x + 3/y = 2; 1/x − 1/(2y) = 1/3

Sol.

The given equations are

2/x + 3/y = 2 ....eq (1)
1/x − 1/(2y) = 1/3 ....eq (2)

Put 1/x = u and 1/y = v in eq (1) and (2)

2u + 3v = 2 ....eq (3)

u − 1/2 v = 1/3

6u − 3v = 2 ....eq (4)

Subtract eq (3) and eq (4)

8u = 4

u = 1/2

put the value of u in eq (3)

2 × 1/2 + 3v = 2
1 + 3v = 2
3v = 1
v = 1/3

Hence,

1/x = u = 1/2
x = 2

1/y = v = 1/3
y = 3

Word Problems based on time, distance and speed

In solving problems based on time, distance and speed, we use the following formulae: Distance = Speed × Time

• Time = Distance / Speed
• Speed = Distance / Time

Upstream & Downstream concept

If Speed of a boat in still water = u km/hr

Speed of the current = v km/hr

Then,

Speed in upstream = (u − v) km/hr

Speed in downstream = (u + v) km/hr

Example

The speed of a boat in still water is 10 Km per hours. If it can travel 26 Km downstream and 14 Km upstream in the same time, find the speed of the current.

Sol.

Let x be the speed of the current

speed downstream = (10 + x) Kmph
speed upstream = (10 − x) Kmph

We know that, time = distance/speed

So,

time downstream = 26/(10 + x)
time upstream = 14/(10 − x)

26/(10 + x) = 14/(10 − x)

Take reciprocal on both sides.

(10 + x)/26 = (10 − x)/14

14(10 + x) = 26(10 − x)

140 + 14x = 260 − 26x

40x = 120

x = 3 Kmph