Official Class 10 Mathematics Revision Material: CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Notes Set 04
Review targeted revision notes for Class 10 Mathematics with the CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Notes Set 04. Built according to official educational guidelines for the 2026-27 academic year, these downloadable summaries for Chapter 03 Pair of Linear Equations in Two Variables support daily study and last-minute exam readiness.
Chapter-wise Concept Summaries: Chapter 03 Pair of Linear Equations in Two Variables
Access the complete concept summary PDF for Chapter 03 Pair of Linear Equations in Two Variables below. Regular review of these targeted notes builds familiarity with complex Class 10 Mathematics themes and helps secure higher marks in final school evaluations.
Introduction
In mathematics, an equation is a statement that asserts the equality of two expressions which are connected by the equal sign “=”. Example: \( 2x = 10 \), \( 5x + 5 = 40 \) etc.
The variables for which the equation has to be solved are also called unknowns, and the values of the unknowns that satisfy the equation are known as the solutions of the equation. A linear equation is an algebraic equation in which all variables involved have the highest power 1. For example: \( 3x - 4 = 0 \).
Linear Equation
A linear equation is an algebraic equation in which all variables involved have the highest power as 1. For example: \( 3x - 4 = 0 \), \( 2x + y = 6 \).
Linear equations in one variable. An equation which contains only one variable with degree 1 is known as a linear equation in one variable. Example: \( 4y = 20 \), \( 13x = 52 \) etc.
Linear equations in two variables. An equation which contains two unknown variables is known as a linear equation in two variables. Example: \( 2x - y = 10 \), \( 10x + 4y = 3 \) etc. If an equation is written in the form \( ax + by = c \), where a, b, c are real numbers and the coefficients of x and y, i.e., a and b, are not equal to zero, it is known as a linear equation in two variables x and y.
For a linear equation \( ax + by = c \), the graph is a straight line. \( 6x + 4y = 36 \) is an example of a linear equation in two variables.
Note.
1. A straight line parallel to the y-axis at a distance of k units from the y-axis is the graph of an equation of the type \( x = k \) (where k is a constant).2. A straight line parallel to the x-axis at a distance of k units from the x-axis is the graph of an equation of the type \( y = k \) (where k is a constant).3. The solution of the two equations is given by the point of intersection of the two lines.
4. There are infinitely many solutions for a single linear equation in two variables.
Pair of Linear Equations in Two Variables
Two or more linear equations in the same two variables are known as a pair of linear equations in two variables. For example,
\[ a_1x + b_1y + c_1 = 0,\qquad a_2x + b_2y + c_2 = 0 \]is the general form of a pair of linear equations in two variables. Here \( a_1, a_2, b_1, b_2, c_1, c_2 \) are real numbers such that \( a_1^2 + b_1^2 \neq 0 \) and \( a_2^2 + b_2^2 \neq 0 \). A system of simultaneous linear equations is formed by a pair of linear equations in two variables. For example, \( 2x + 3y = 7 \) and \( 9x - 2y = -8 \) are a pair of linear equations.
Pair of Linear Equations are of Two Types
1. Consistent pair of linear equations. A pair of linear equations in two variables which has at least one solution is called a consistent pair of linear equations. A consistent pair is of two kinds:
(a) Independent pair of linear equations: a pair which has exactly one solution.
(b) Dependent pair of linear equations: a pair which has infinitely many distinct common solutions. A dependent pair of linear equations is always consistent.
2. Inconsistent pair of linear equations. A pair of linear equations which has no solution is called an inconsistent pair of linear equations.
Note.
1. Dependent equations are equations of the type \( ax + by = c \) and \( kax + kby = kc \). For example, \( 2x + 3y = 9 \) ...(i) and \( 4x + 6y = 18 \) ...(ii). Equation (ii) is a multiple of equation (i); both equations are basically the same, that is, equation (ii) depends on equation (i). A dependent pair of equations has infinitely many solutions.
2. Inconsistent equations are equations of the type \( ax + by = c \) and \( kax + kby = mc \) (\( m \neq k \)). For example, \( x + 2y = 4 \) and \( 2x + 4y = 12 \).
Solution of a Pair of Linear Equations
The pair of linear equations in two variables can be solved by the following two methods: (i) Graphical method, (ii) Algebraic method.
Graphical Method of Solution of a Pair of Linear Equations
The graph of a pair of linear equations in two variables is shown by two straight lines. The behaviour of the lines representing a pair of linear equations in two variables and the existence of solutions can be summarised as follows:
(i) Lines are intersecting, if the system has a unique solution.
(ii) Lines are coincident, if the system has an infinite number of solutions.
(iii) Lines are parallel, if the system has no solution.
Graphical Representation of Simultaneous Equations
(i) Intersecting lines (unique solution)
\( x + y = 4 \) ...(i) and \( 3x + 2y = 11 \) ...(ii)
To obtain the equivalent geometric representation, we find two points on the line representing each equation. That is, we find two solutions of each equation, given in the tables below.
| x | 1 | 3 |
| y | 3 | 1 |
| x | 3 | 1 |
| y | 1 | 4 |
For \( x + y = 4 \) (left table) and \( 3x + 2y = 11 \) (right table). The two lines intersect at the unique point \( (3, 1) \). Hence the equations are consistent with a unique solution.
(ii) Coincident lines (infinite number of solutions)
\( x + 2y = 8 \) ...(i) and \( 2x + 4y = 16 \) ...(ii)
Solutions of each equation are given in the tables below.
| x | 0 | 8 |
| y | 4 | 0 |
| x | 2 | 4 |
| y | 3 | 2 |
For \( x + 2y = 8 \) (left table) and \( 2x + 4y = 16 \) (right table). Clearly, the graphs of the given equations are coincident lines. Hence the given system of equations has infinitely many solutions.
(iii) Parallel lines (no solution)
\( x + y = 4 \) ...(i) and \( 2x + 2y = 12 \) ...(ii)
Solutions of each equation are given in the tables below.
| x | 0 | 4 |
| y | 4 | 0 |
| x | 0 | 6 |
| y | 6 | 0 |
For \( x + y = 4 \) (left table) and \( 2x + 2y = 12 \) (right table). Clearly, the graphs of the given equations are parallel lines. As they have no common point, there is no common solution. Hence the given system of equations has no solution.
Conditions for simultaneous equations \( a_1x + b_1y + c_1 = 0 \), \( a_2x + b_2y + c_2 = 0 \)
| Algebraic condition | Graphical representation | Algebraic interpretation | Type of pair |
|---|---|---|---|
| \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) | Intersecting lines | Exactly one solution (unique solution) | Consistent (independent) |
| \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) | Coincident lines | Infinitely many solutions | Consistent (dependent) |
| \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \) | Pair of parallel lines | No solution | Inconsistent |
Solved Examples
(a) How many solutions are possible, if the system is consistent?
(b) If the coefficient of y in the second equation is replaced by 10, will there be any change in the number of solutions? Solve the following system of simultaneous linear equations graphically and explain your answer.
Answer:
(a) Method I. Here \( a_1 = 2, b_1 = 5, a_2 = 4, b_2 = 3 \). Now \( \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} \) and \( \frac{b_1}{b_2} = \frac{5}{3} \), that is, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \). Hence the given system is consistent and has exactly one solution.
Method II. \( 2x + 5y = 0 \Rightarrow y = \frac{-2x}{5} \) and \( 4x + 3y = 0 \Rightarrow y = \frac{-4x}{3} \).
| x | 0 | -5 | 5 |
| y | 0 | 2 | -2 |
| x | 0 | 3 | -3 |
| y | 0 | -4 | 4 |
Hence the points are \( (0,0), (-5,2), (5,-2) \) for the first line and \( (0,0), (3,-4), (-3,4) \) for the second line. From the graph, the two lines intersect at the point \( (0, 0) \). Hence the solution of the pair of linear equations is \( x = 0, y = 0 \).
(b) If the coefficient of y in the second equation is replaced by 10, then \( 2x + 5y = 0 \) ...(iii) and \( 4x + 10y = 0 \) ...(iv).
| x | 0 | 5 | -5 |
| y | 0 | -2 | 2 |
| x | 0 | -5 | 5 |
| y | 0 | 2 | -2 |
Hence the points are \( (0,0), (-5,2), (5,-2) \) for both equations. From the graph the two lines are coincident. Hence the system has infinitely many solutions.
Answer:
Let the number of violins of type A be x and of type B be y. Total number of violins sold = 7; price of type A = Rs. 200, price of type B = Rs. 300; total sales = Rs. 1500.
Algebraically: \( x + y = 7 \) ...(i) and \( 200x + 300y = 1500 \), that is, \( 2x + 3y = 15 \) ...(ii).
Graphically: for (i), \( y = 7 - x \):
| x | 0 | 3 | 7 |
| y | 7 | 4 | 0 |
For (ii), \( y = \frac{15 - 2x}{3} \):
| x | 0 | -3 | 3 |
| y | 5 | 7 | 3 |
The two lines intersect at \( (6, 1) \), so the lines are intersecting in nature (6 violins of type A and 1 of type B).
Answer:
For \( x + 2y - 6 = 0 \), \( y = \frac{6 - x}{2} \):
| x | 0 | 2 | 6 |
| y | 3 | 2 | 0 |
For \( x + 2y - 4 = 0 \), \( y = \frac{4 - x}{2} \):
| x | 0 | 2 | 4 |
| y | 2 | 1 | 0 |
Hence the points are \( (0,3), (2,2), (6,0) \) and \( (0,2), (2,1), (4,0) \). From the graph we can see that the two lines are parallel to each other.
Answer:
For \( 3x + y + 1 = 0 \), \( y = -(3x + 1) \):
| x | -2 | -1 | 1 |
| y | 5 | 2 | -4 |
For \( 2x - 3y + 8 = 0 \), \( y = \frac{2x + 8}{3} \):
| x | 2 | -1 | 5 |
| y | 4 | 2 | 6 |
The two straight lines intersect each other at the point \( P(-1, 2) \). Therefore \( x = -1, y = 2 \) is the solution of the given system. The area of the region bounded by the two lines and the x-axis is shaded in the graph.
Answer:
For \( 2x + y = 6 \), \( y = 6 - 2x \):
| x | 1 | 2 | 4 |
| y | 4 | 2 | -2 |
For \( x - 2y + 2 = 0 \), \( y = \frac{x + 2}{2} \):
| x | 0 | 2 | 4 |
| y | 1 | 2 | 3 |
The lines intersect at \( P(2, 2) \), so \( x = 2, y = 2 \) is the solution. The line \( 2x + y = 6 \) meets the x-axis at \( B(3, 0) \) and the line \( x - 2y + 2 = 0 \) meets the x-axis at \( C(-2, 0) \). The vertices of the triangle are \( P(2,2), B(3,0) \) and \( C(-2,0) \).
With \( PL \perp BC \): \( BC = 5 \) units and \( PL = 2 \) units. Therefore the area of \( \triangle PBC = \frac{1}{2}(5 \times 2) = 5 \) sq. units.
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Study Material and Revision Notes for Class 10 Mathematics Chapter 03 Pair of Linear Equations in Two Variables
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