Linear Equations In Two Variables Class 9th Notes - Free NCERT Class 9th Maths Chapter 4 Notes - Download PDF

Linear Equations In Two Variables Class 9th Notes - Free NCERT Class 9th Maths Chapter 4 Notes - Download PDF

Edited By Ramraj Saini | Updated on Apr 21, 2022 02:17 PM IST

Introduction to Linear Equations in Two Variables:-

The class 9 maths chapter 4 notes are one of the important chapters in the Ncert book. Linear Equations in Two Variables class 9 notes is the summary of the chapter. The primary topics covered in cbse class 9 maths chapter 4 notes are Linear Equations, Solution of a Linear Equation, Graph of a Linear Equation in Two Variables, and Equations of Lines Parallel to the x-axis and y-axis. These are the important topics in class 9th maths chapter 4 notes. Some examples and short notes are also covered in ncert class 9 maths chapter 4 notes. Given notes for class 9 maths chapter 4 or ncert notes for class 9 maths chapter 4 are very helpful for the revision. Class 9 Linear Equations in Two Variables notes are an important chapter from the perspective of the cbse exam and can be downloaded from Linear Equations in Two Variables class 9 notes pdf download or from class 9 maths chapter 4 notes pdf download.

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NCERT Class 9 Maths Chapter 4 Notes

Linear Equations

A polynomial equation (or algebraic equation) of the first degree is known as a linear equation. For example: 2x+y=5 or 2x+5=0 are linear equations.

Linear equations that are in the form ax + by + c = 0 or ax + b = 0 are also known as two-variable linear equations, or linear equations in one variable.

Example 1: Using the equation ax + by + c = 0, write the following equations as follows: (i) 2x + 3y = 4 (ii) 2x = y

Solution :

  1. 2x+3y=4 can be written as 2x + 3y – 4 = 0. Here a = 2, b = 3 and c = – 4

  1. The equation 2x = y can be written as 2x – y + 0 = 0. Here a = 2, b = –1 and c = 0.

Example 2: The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement. (Take the cost of a notebook to be x and that of a pen to be y).

Solution: Consider the cost of a notebook is x and the cost of a pen is y.

Given that the cost of a notebook is twice the cost of a pen. Then:

2x=y

2x-y=0

This is an example of a linear equation in two variables.

Solution of A Linear Equation

Let us consider the equation 2x+3y=12.

Suppose we put x = 3 and y = 2 in the given equation, then we get 2x + 3y = (2 × 3) + (3 × 2) = 12.

So, x = 3 and y = 2 is a solution.

In this case, the solution is written as an ordered pair (3, 2).

Similarly, (0, 4) is also a solution for the given equation.

On the other hand, (1, 4) is not a solution of 2x + 3y = 12, because on putting x = 1 and y = 4 we get 2x + 3y = ( 2 x 1) + ( 3 x 4) =14, which is not 12. So, the solution is (0,4).

Thus we can have at least two solutions for 2x + 3y = 12, i.e., (3, 2) and (0, 4).

Therefore, we can say that a linear equation in two variables has infinitely many solutions.

Example 3 : Find four different solutions of the equation x + 2y = 6.

Solution : (i) Put x = 2, y = 2 in L.H.S and we get

2 + (2 x 2) = 6 = R.H.S. So one solution is (2,2)

(ii) Put x = 0 and y = 3, we get

0 + (2 x 3) = 6. Second solution is (0,3)

(iii) Put x = 6 and y = 0, we get

6 + ( 2 x 0) = 6. Third solution is (6,0).

(iv) Put x = 4, and y = 1, we get

4 + (2 x 1) = 6. Fourth solution is (4,1).

Example 4: Write the four solutions of equation x = 4y.

Solution: Given equation can be written as x - 4y = 0

(i) Put x = 4 and y = 1, we get

4 - ( 4 x 1) = 0. One solution is (4,1).

(ii) Put x = 8 and y = 2, we get

8 - ( 4 x 2) = 0. The second solution is (8,2).

(iii) Put x = 12 and y = 3

12 - ( 4 x 3) = 0. Third solution is (12,3)

(iv) Put x = 16 and y = 4,

16 - ( 4 x 4) = 0. The fourth solution is (16,4).

Graph of A Linear Equation In Two Variables

We can represent the solution of a linear equation in two variables in a cartesian plane.

Let an equation be x + 2y = 6

Solutions of this equation are (2,2); (0,3); (6,0);(4,1).

1648024985892

Expressed these solutions in tabular form

x

2

0

6

4

y

2

3

0

1

Plot all these points in the cartesian plane. Connect any two of these points to obtain a line. This line will be AB.

Now, pick another point on this line, say (8, –1). In fact, 8 + 2(–1) = 6. So, (8, –1) is a solution. If we pick any other point on this line AB then it satisfies the equation.

1. On line AB, every point lies whose coordinates satisfy Equation (1).

2. Every point (a, b) on the line AB corresponds to a solution of Equation x = a, y = b. (1).

3. Any point that is not on the AB line is not a solution of Equation (1).

As a consequence, each point on the line satisfies the line's equation, and each solution to the equation is a point on the line.

Linear equations in two variables are represented geometrically by lines whose points are the solutions to the equations. This is called the graph of the linear equation.

Example 6: Draw the graph of x + y = 7.

Solution: For the graph to be drawn, we need at least two solutions to the equation. In the given equation, x = 0, y = 7, and x = 7, y = 0 are solutions.

So, you can use the following table to draw the graph:

x

0

7

y

7

0

So, we can draw the graph by plotting the two points and then joining these points as shown in fig.

1648026581454

Equations of Lines Parallel To The x-Axis And y-Axis

Equation of line Parallel To The x-Axis

Here y=0 is the equation of the x-axis so, the equation of the line parallel to the x-axis is y = K, where K is the real number. It is the generalized form of the line equation parallel to the x-axis. For example, a line equation parallel to the x-axis is y=2. It can also be written as y-2 = 0.

1648026614183

Equation of line Parallel To The y-Axis.

Here x=0 is the equation of the y-axis so, the equation of the line parallel to the y-axis is x = K, where K is the real number. It represents the distance from the x-axis to the line y is K. It is the common form of the line equation parallel to the y-axis. For example,x=6 is a line equation parallel to the y-axis. So, we can written as x-6 = 0.

1648026644986

Example 1: What does the equation 2x + 3 = - 1 represent in a linear equation of two variables?

Solving this equation as a linear equation in two variables results in a line parallel to the y-axis, as shown in the figure below.

the equation, 2x + 3 = -1, when solved, we get,

2x+3 = -1

2x = -1 -3

2x = -4

x = -4/2

x = -2

So we can see that 2x + 3 = -1 shows a line parallel to the y-axis, which is x = -2. The figure below illustrates this line.

Line Parallel to Axes - Solved Example 1

From The NCERT Book, A Summary of This Chapter

1. A linear equation in two variables is defined as an equation in the form ax + by + c = 0, where a, b, and c are all real values and a and b are not both zero.

2. There are an infinite number of solutions in a two-variable linear equation.

3. Linear equations in two variables always have a straight line graph.

2. The y-axis equation is x = 0, and the x-axis equation is y = 0.

5. A straight line parallel to the y-axis is the graph of x = a.

6. A straight line parallel to the x-axis is the graph of y = a.

7. A line going through the origin is represented by the equation y = mx.

8. A solution of a linear equation in two variables is any point on the graph of the equation. Furthermore, every linear equation solution is a point on the graph of the linear equation.

Significance of NCERT class 9 maths chapter 2 notes

Linear Equations in Two Variables class 9th notes can be used to go over the chapter and just have a clearer understanding of the relevant points. These ncert class 9 mathematics chapter 4 notes can also be used to cover the concepts in the CBSE Maths exam for class 9. For offline use, download it from Two Variables class 9 notes pdf download or Class 9 maths chapter 4 notes pdf download.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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