NCERT Solutions for Exercise 4.4 Class 9 Maths Chapter 4 - Linear Equations in Two Variables

NCERT Solutions for Exercise 4.4 Class 9 Maths Chapter 4 - Linear Equations in Two Variables

Edited By Safeer PP | Updated on Jul 29, 2022 06:15 PM IST

NCERT Solutions for Class 9 Maths exercise 4.4 which is based on the geometrical depiction of equations of lines parallel to the x-axes and y-axes and covers the concept of linear equations in two variables. Linear equations in two variables are a system of equations that can have a single solution, no solutions, or an unlimited number of solutions and are of the type ax+by+c=0, where a, b, c are real numbers. To identify the set of solutions to the linear equations with two variables, we must first determine the value of variables. The graphical depiction of equations of lines parallel to x-axes and y-axes was the topic of exercise 4.4 Class 9 Maths. A line parallel to the y-axis will have the form x=k, which means that the value of x is the same for any value of y , and k is any constant value. Similarly, A line parallel to the x-axis will be of the form y=k, which means that the value of y is the same for any value of x, and k is any constant value.

This Story also Contains
  1. Linear Equations In Two Variables Class 9 Maths Chapter 4 Exercise: 4.4
  2. More About Ncert Solutions For Class 9 Maths Exercise 4.4
  3. Benefits of NCERT Solutions for Class 9 Maths Exercise 4.4:
  4. NCERT Solutions of Class 10 Subject Wise
  5. Subject Wise NCERT Exemplar Solutions

NCERT solutions for Class 9 Maths chapter 4 exercise 4.4 include two questions, one of which is the main question with two subsections and the other is the main question with two subsections. This NCERT book Class 9 Maths chapter 4 exercise 4.4 thoroughly explains the fundamentals of linear equations in two variables. The following activities are included along with NCERT syllabus Class 9 Maths chapter 4 exercise 4.4.

Linear Equations In Two Variables Class 9 Maths Chapter 4 Exercise: 4.4

Q1 (i) Give the geometric representations of y = 3 as an equation

in one variable

Answer:

Equation y = 3 can be represented in one variable on the number line.

1640075843993

Q1 (ii) Give the geometric representations of y = 3 as an equation: in two variables

Answer:

For 2 variable representations of y = 3 .

Equation : 0.x+y=3

For graph,

x=0,we have y=3

x=1,we have y=3

x=2,we have y=3

Hence, (0,3) ,(1,3) and (2,3) are solutions of equation.

1640075877910

Q2 (i) Give the geometric representations of 2x + 9 = 0 as an equation: in one variable

Answer:

Equation 2x + 9 = 0 can be represented in one variable on the number line.

\Rightarrow x=\frac{-9}{2}

The yellow mark represents x= - 4.5.

1640075916094

Q2 (ii) Give the geometric representations of 2x + 9 = 0 as an equation: in two variables

Answer:

For 2 variable representations of 2x + 9 = 0 .

Equation : 2x +0.y =-9

For graph,

y=0,we have x=\frac{-9}{2}

y=1,we have x=\frac{-9}{2}

y=2,we have x=\frac{-9}{2}

Hence, (\frac{-9}{2},0) , (\frac{-9}{2},1) and (\frac{-9}{2},2) are solutions of equation.

More About Ncert Solutions For Class 9 Maths Exercise 4.4

To solve linear equations in two variables visually in Class 9 Maths chapter 4 exercise 4.4, we must first construct the graph of the equations by constructing a table for various x and y values. After that, we must use the values to plot the points on the graph for each equation. We'll be able to connect all of the locations at that moment. As a result, the graph is created. We can describe the graph of linear equations in two ways in exercise 4.4 Class 9 Maths.

• A one-variable linear equation with a solution represented by a point that can be plotted on a number line.

• The solution to a two-variable linear equation that can be plotted on the cartesian plane is represented by a line parallel to the x or y-axis.

If the constant term ‘c’ in the linear equation ax+by+c=0 is zero, the line will always pass through the origin.

Also Read| Linear Equations In Two Variables Class 9 Notes

Benefits of NCERT Solutions for Class 9 Maths Exercise 4.4:

• NCERT solutions for Class 9 Maths exercise 4.4 are very accurate and clear so that we may get a clear idea about the concept and topic and these solutions can be helpful for us not only for the first term examination but also in solving homework and assignments in higher education.

• By solving the NCERT solution for Class 9 Maths chapter 4 exercise 4.4, we may study for the exams based on their knowledge gaps, which will also be useful in Class 10 Maths chapter 3 exercise 3.2 pair of linear equations with two variables.

• On solving the questions of exercise 4.4 Class 9 Maths, the questions will help us to self evaluate our performance in examinations and we can also solve the questions from other textbooks prescribed by the CBSE Board.

Also see-

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. Line of the equation y=m is __________ (where m is constant).

The line of the equation y=m is parallel to the x-axis. 

Reason : 

A line that is parallel to the x-axis will be in the form y=k where k is a constant. 

2. Quadrant I is _______.

Quadrant I is positive 

Reason: In quadrant I, the x-coordinates are positive also the y-coordinates are positive.

3. Number of solutions of the equation x-8=0 are _______.

Since the given equation is a linear equation in one variable x-8=0 has only one solution.

4. (0, 8) lies on _____ axis.

(0, 8) lies on the y- axis. Since the x coordinate is 0, the point lies in the y axis. 

5. Equation of horizontal line above the x axis at 5 units from x axis is ________.

The horizontal line above the x-axis is parallel to the x-axis and it is above the x-axis at 5 units, so the equation of the line will be y=5 

6. Equation of vertical line to the right of the y axis at 8 units from y axis is ________.

The vertical line to the right of the y axis is parallel to the x-axis at 8 units from the y-axis, so the equation is x=8

7. What is the equation of the x-axis , as per NCERT solutions for Class 9 Maths chapter 4 exercise 4.4 ?

 The equation of the x-axis is y=0, as per NCERT solutions for Class 9 Maths chapter 4 exercise 4.4

8. As per NCERT solutions for Class 9 Maths chapter 4 exercise 4.4, What is the equation of line parallel to the y-axis?

The equation of a line parallel to the y-axis is x=k, where k is any constant real number, as per NCERT solutions for Class 9 Maths chapter 4 exercise 4.4.

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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

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

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

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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

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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

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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

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less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

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more than 9

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