NCERT Solutions of class 9 maths exercise 4.3 of chapter 4- Linear equations in two variables - Future Study Point

NCERT Solutions of class 9 maths exercise 4.3 of chapter 4- Linear equations in two variables

Linear equation in two variables

NCERT Solutions of class 9 maths exercise 4.3-Linear equations in two variables

Linear equation in two variables

NCERT Solutions of exercise 4.3 are the solutions of unsolved questions of class 9 maths NCERT textbook prescribed by CBSE for 9 class students. All questions are solved by the maths expert by a step by step method. Exercise 4.3 is good for the preparation of exams because every year one or two questions of 3 to 5 marks are compulsorily asked from this exercise.

Exercise 4.1 & 4.2 -Linear Equations in Two Variables

Class 9-Sample papers and question papers

Mid point theorem

Solutions of important question papers of last year’s question papers

Mean,mode median

Solutions of specific questions

How to write linear equation in two variable

Technics of achieving hundred percent marks in Maths

Maths assignments for class 9 and 10

9 class maths assignment for SA-1

Lines,angles and Triangles

Addition, subtraction,multiplication and division of polynomial

Number System

NCERT Solutions of class 9 maths exercise 4.3 of chapter 4- Linear equations in two variables

Click for online shopping

Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc

Q1. Draw the graph of each of the following linear equations in two variables.

(i) x + y =4    (ii) x – y = 2   (iii) y = 3x   (iv) 3 = 2x + y

Ans.(i)  We are given the equation x + y =4

For x = 0, y= 4, for x = 1, y= 3, for x =2, y= 2

So, we have the following solutions of the equation

x012
y432

Drawing the graph of the equations as follows.

x +y=4

 

 

 

 

 

 

 

(ii) The given equation is x – y = 2

For x =0, y= -2,for x = 2, y = 0,for x = 3, y=1

The solutions of the given equation is shown as follows

x023
y-201

Drawing the graph of the given equation as follows.

x - y = 2  graph

 

 

 

 

 

 

 

(iv) We are given the equation 3 = 2x + y

When x = 0, y = 3, x =2, y= -1, x =3, y= -3

The table of solutions of the equation is given below.

x023
y3-1-3

3 = 2x + y graph

 

 

 

 

 

 

 

Q2. Give the equations of two lines passing through (2,14), How many more such lines are there, and why?

Ans. The given coordinates of the point are =(2,14)

Observing the coordinates of the point we find that the ratio between x-coordinate and the y coordinate is 1: 7, so we have

y- 7x = 0

We also observe that the difference between both coordinates is 2-14 = -12

So,x – y = -12

x -y + 12 = 0

Both of these equation satisfies the coordinates (2,14)

Hence the required two-equation passes through (2,14) are y- 7x = 0 and x -y + 12 = 0

From one point infinite lines can be drawn, so through the point (2,14), there are such of infinite lines pass out.

Q3. If the point (3,4) lies on the graph of the equation 3y = ax +7, find the value of a.

Ans. We are given that the point (3,4) lies on the graph of the  equation 3y = ax +7,so the coordinates of the given point will satisfy the equation.

Putting x = 3 and y = 4 in the equation 3y = ax +7

3 × 4 = a× 3 + 7

3a = 12 – 7

3a = 5

Therefore the value of a is 5/3.

Q4. The taxi fare in a city is as follows: for the first kilometer, the fare is Rs 8 and for the subsequent distance it is Rs 5 per kilometer. Taking the distance covered as x km and total fare as Rs y . Write the linear equation for this information, and draw its graph.

Ans. The distance covered is = x km

The fare for the first 1 kilometer is = Rs 8

After the taxi covered 1 km the rest of the distance is = x -1

The rate of fare for the rest of the distance is = Rs 5/km

Fare for the distance (x -1) km = Rs 5( x- 1)

The total fare is given = y

So, the linear equation as per the question is given by

y = 8 + 5(x – 1)

y = 8 + 5x -5

5x -y +3 = 0

So, the required linear equation is 5x -y +3 = 0

Solving the linear equation

for x = 0, we get y = 3, when x =1, y = 8, when x =-1, y = 2

The solutions of the given equation are shown as follows

x01-1
y38-2

Drawing the graph of the given equation as follows.

5x -y +3 = 0 graph

 

 

 

 

 

 

 

 

 

 

 

 

Q5. From the choices given below, choose the equation whose graphs are given in the given figure.

For the first figure

(i) y = x

(ii)x + y=0

(iii) y = 2x

(iv) 2 + 3y = 7x

For the second figure

(i) y = x+ 2

(ii) y= x -2

(iii)y = -x + 2

(iv)x + 2y = 6

Q5 fig 1 and fig 2

 

 

 

 

 

 

 

Ans. In the first figure, the graph of the equation is passing through (0,0), (-1.1) and (1.-1)

The equation x + y=0 is satisfied by the coordinates (0,0), (-1.1) and (1.-1), so for the first fig. the answer is (ii)

In the second figure, the graph is passing through the points (-1,3), (0,2), and (2,0)

These coordinates satisfy the equation y = -x + 2, so for the second figure, the answer is (iii).

Q6. If the work done by a body an application of a constant force is directly proportional to the distance traveled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance traveled by the body is

(i) 2 units  (ii) 0 units

Ans. Let the work done is y ,the constant force is k and the distance travelled is x

So, we have

Work done = force × displacement  (i.e distance travelled in the same direction ,x)

y = kx

k, constant force = 5 units

The equation become

y = 5x

The solutions of the given equation y = 5x, are given as follows

When x = 0, y = 0,for x = 1, y= 5, for x= -1, y = -5

The table of solutions of the equation is as follows

x01-1
y05-5

 

Drawing the graph of the equation

Linear equation in two variables

(i) Observing the graph, the corresponding value of work done when the distance, x = 2 units is = 10 units

(ii) When distance is 0, the corres[ponding value of work done is = 0 units.

Q7. Yamini and Fatima , two students of class ix of a school, together contributed Rs 100 towards the Prime Minister’s relief fund to help the earthquake victims. Write a linear equation that satisfies this data .(You may take their contributions as Rs x and Rs y). Draw the graph of the same.

Ans.Let the Yamini contributed Rs x and Fatima contributed Rs y towards Prime Minister’s relief fund

Both of them contributed total sum of Rs 100

So, x + y = 100

Solving the equation for different values of x,we get different values of y

When x = 50, y = 50, when x = 30, y = 70, when x =60, y = 40

The solutions of the table is shown below

x503060
y507040

 

Drawing the graph of the equation

x +y =100 graph

 

 

 

 

 

 

 

 

Q8. In countries like the USA and Canada, the temperature is measured in Faherignhite, whereas in countries like India it is measured in celsius. Here is a linear equation that converts Fahreinhite to celsius:

(i) Draw the graph of the linear equation above using Celcius for the x-axis and Faherenhite for the y-axis.

(ii) If the temperature is 30°C, what is the temperature in Fahrenheit.

(iii) If the temperature is 95°F, what is the temperature in Celcius.

(iv) If the temperature is 0°C , what is the temperature in Fahrenheit and if the temperature in 0°F, what is the temperature in Celcius.

(v) Is there a temperature that is numerically the same in both Fahrenheit and Celcius? If yes, find it.

Ans. (i) The given equation is

Solving the equation by putting different values of C for getting different values of F

When C = 0, F = 32, When C=-20, F= =-4, When C = -40,F =-40

C0-20-40
F32-4-40

Drawing the graph as follows

Graph between F and C

 

 

 

 

 

 

 

 

(ii) We have the equation

When C = 30°C, the temperature in Fahrenheit is calculated as follows

F = 54 + 32 = 86

Hence in Fahrenheit, the temperature is 86°F

(ii) If the temperature in Fahrenheit is 95°F, then the temperature in Celcius is calculated as follows.

The given equation is

Therefore the temperature in Celsius is 35°C

(iv) If the temperature in Celcius = 0°C, then value in Fahrenhit is calculated as follows

The given equation is

 

F = 32

When the temperature in Fahrenheit is =0°F, then the temperature in Celcius is calculated as follows.

Hence,the temperature in Celcius is =-17.77°C

So,when C = 0, F= 32 and when F =0, C= -17.77

(v) The given equation is

Putting F =C

Hence,at -40°C, both Fahrenhit and Celcius are equal.

Study science notes 

Archimedes Principle: Complete detail

Average Speed and Average velocity

Three equation of Motions

Recoil velocity of the gun

Mole concept

The second law of motion

The universal law of gravitational force

NCERT Solutions Class 10 Science from chapter 1 to 16

NCERT Solutions of all chapters of Maths for Class 10 from Chapters 1 to 15

Class 11 Chemistry NCERT Solutions

Chapter 1-Some basic concepts of chemistry

Class 11 Physics NCERT Solutions

Chapter 1- Physical World

Science and Maths NCERT solutions for Class 9 ,10 and 11 classes

Previous year maths question paper of class 11

Maths question paper 2015-16

Maths question paper 2014-15

Maths question paper 2019-20

NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for class 9 maths

Chapter 1- Number SystemChapter 9-Areas of parallelogram and triangles
Chapter 2-PolynomialChapter 10-Circles
Chapter 3- Coordinate GeometryChapter 11-Construction
Chapter 4- Linear equations in two variablesChapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s GeometryChapter 13-Surface Areas and Volumes
Chapter 6-Lines and AnglesChapter 14-Statistics
Chapter 7-TrianglesChapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions for class 9 science 

Chapter 1-Matter in our surroundingsChapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?Chapter 10- Gravitation
Chapter3- Atoms and MoleculesChapter 11- Work and Energy
Chapter 4-Structure of the AtomChapter 12- Sound
Chapter 5-Fundamental unit of lifeChapter 13-Why do we fall ill ?
Chapter 6- TissuesChapter 14- Natural Resources
Chapter 7- Diversity in living organismChapter 15-Improvement in food resources
Chapter 8- MotionLast years question papers & sample papers

NCERT Solutions for class 10 maths

Chapter 1-Real numberChapter 9-Some application of Trigonometry
Chapter 2-PolynomialChapter 10-Circles
Chapter 3-Linear equationsChapter 11- Construction
Chapter 4- Quadratic equationsChapter 12-Area related to circle
Chapter 5-Arithmetic ProgressionChapter 13-Surface areas and Volume
Chapter 6-TriangleChapter 14-Statistics
Chapter 7- Co-ordinate geometryChapter 15-Probability
Chapter 8-Trigonometry

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

NCERT Solutions for Class 10 Science

Chapter 1- Chemical reactions and equationsChapter 9- Heredity and Evolution
Chapter 2- Acid, Base and SaltChapter 10- Light reflection and refraction
Chapter 3- Metals and Non-MetalsChapter 11- Human eye and colorful world
Chapter 4- Carbon and its CompoundsChapter 12- Electricity
Chapter 5-Periodic classification of elementsChapter 13-Magnetic effect of electric current
Chapter 6- Life ProcessChapter 14-Sources of Energy
Chapter 7-Control and CoordinationChapter 15-Environment
Chapter 8- How do organisms reproduce?Chapter 16-Management of Natural Resources

NCERT Solutions for class 11 maths

Chapter 1-SetsChapter 9-Sequences and Series
Chapter 2- Relations and functionsChapter 10- Straight Lines
Chapter 3- TrigonometryChapter 11-Conic Sections
Chapter 4-Principle of mathematical inductionChapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbersChapter 13- Limits and Derivatives
Chapter 6- Linear InequalitiesChapter 14-Mathematical Reasoning
Chapter 7- Permutations and CombinationsChapter 15- Statistics
Chapter 8- Binomial Theorem Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

NCERT solutions for class 12 maths

Chapter 1-Relations and FunctionsChapter 9-Differential Equations
Chapter 2-Inverse Trigonometric FunctionsChapter 10-Vector Algebra
Chapter 3-MatricesChapter 11 – Three Dimensional Geometry
Chapter 4-DeterminantsChapter 12-Linear Programming
Chapter 5- Continuity and DifferentiabilityChapter 13-Probability
Chapter 6- Application of DerivationCBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals

Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2

Class 12 Maths Important Questions-Application of Integrals

Class 12 Maths Important questions on Chapter 7 Integral with Solutions for term 2 CBSE Board 2021-22

Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22

Solutions of class 12  maths question paper 2021 preboard exam CBSE Solution

 

 

 

 

 

 

 

Scroll to Top