NCERT Solutions for Class 9 Maths Chapter 4 (Ex 4.3)
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EXERCISE 4.3
1. Draw the Graph of Each of the Following Linear Equations in Two Variables:
(i) x+y=4
Ans. Given linear equation x+y=4 ,
⇒ y = 4 - x
By substituting different values of x, we obtain distinct values of y -
To obtain the graph of \[x+y=4\] , we plot the points obtained above, \[(0,4)\] , \[(2,2)\] and \[(4,0)\] , on a graph and join them to form a straight line that represents given equation.
The graph for above points will be constructed as follows:
(ii) x-y=2
Ans. Given linear equation x-y=2 ,
⇒ y = x - 2
By substituting different values of x, we obtain distinct values of y -
To obtain the graph of \[x-y=2\] , we plot the points obtained above, \[(0,-2)\] , \[(2,0)\] and \[(4,2)\] , on a graph and join them to form a straight line that represents given equation.
The graph for above points will be constructed as follows:
(iii) y=3x
Ans. Given linear equation y=3x.
By substituting different values of x , we obtain distinct values of y -
To obtain the graph of \[y=3x\] , we plot the points obtained above, \[(-1,-3)\] , \[(0,0)\] and \[(1,3)\] , on a graph and join them to form a straight line that represents given equation.
The graph for above points will be constructed as follows:
(iv) 3=2x+y
Ans. Given linear equation 3=2x+y ,
⇒ y = 3 - 2x
By substituting different values of x, we obtain distinct values of y -
To obtain the graph of \[3=2x+y\] , we plot the points obtained above, \[(-1,5)\] , \[(0,3)\] and \[(3,-3)\] , on a graph and join them to form a straight line that represents given equation.
The graph for above points will be constructed as follows:
2. Give the Equations of two Lines Passing Through \[(2,14)\] . How many more such lines are there, and why?
Ans. We find the equations of two lines passing through \[(2,14)\] , given that \[x=2\] and \[y=14\] :
\[x+y=2+14\]
\[\Rightarrow x+y=16\]
\[x-y=2-14\]
\[\Rightarrow x-y=-12\]
It can be observed that point \[(2,14)\] satisfies the equation \[x+y=16\] and \[x-y=-12\] . Therefore, \[x+y=16\] and \[x-y=-12\] are two lines passing through point \[(2,14)\] .
It is known that infinite number of lines can pass through one point. Therefore, there are infinite possible equations of lines passing through the given point \[(2,14)\] .
3. If the Point \[(3,4)\] Lies on the Graph of the Equation\[3y=ax+7\] , Find the Value of \[a\] .
Ans. Given equation of the line \[3y=ax+7\] , and point \[(3,4)\] lies on this line.
Substituting the value of \[x=3\] and \[y=4\] in the given equation:
\[3y=ax+7\]
\[\Rightarrow 3(4)=a(3)+7\]
\[\Rightarrow 12=3a+7\]
\[\Rightarrow 3a=5\]
We get,
\[a=\frac{5}{3}\]
4. The Taxi Fare in a City is As Follows: For the First Kilometre, the Faresis Rs. \[8\] and for the Subsequent Distance it is Rs. \[5\] per km. Taking the Distance cCvered as \[x\] km and Total Fare as Rs. \[y\] ,Write a Linear Equation for This Information, and Draw Its Graph.
Ans. We are given the following information:
• Taxi fare for the first kilometre \[=\] Rs. \[8\]
• Taxi fare for each subsequent kilometre \[=\] Rs. \[5\]
• Total distance covered \[=\] \[x\] km
• Total fare for the distance after the first kilometre \[=\] Rs. \[(x-1)5\]
• Total fare covered \[=\] Rs. \[y\]
The linear equation for the above information can be formed as follows:
\[y=8+(x-1)5\]
\[\Rightarrow y=8+5x-5\]
\[\Rightarrow y=5x+3\]
\[\Rightarrow 5x-y+3=0\]
By substituting different values of \[x\] in the given equation, we get different values for \[y\] .
When \[x=-2\] , then \[y=5(-2)+3\] . So, \[y=-7\]
When \[x=-1\] , then \[y=5(-1)+3\] . So, \[y=-2\]
When \[x=0\] , then \[y=5(0)+3\] . So, \[y=3\]
When \[x=1\] , then \[y=5(1)+3\] . So, \[y=8\]
When \[x=2\] , then \[y=5(2)+3\] . So, \[y=13\]
Thus, we have the following table with all the obtained solutions:
Plotting the points \[(-2,-7)\] , \[(-1,-2)\] , \[(0,3)\] , \[(1,8)\] , \[(2,13)\] on the graph and joining them through a line, we obtain the required graph.
The graph of the line represented by the given equation as shown:
Here, the variable \[x\] and \[y\] represent the distance covered and the taxi fare paid for that distance respectively.
Since distance and amount cannot be negative, only those values of \[x\] and \[y\] which are lying in the first quadrant will be considered.
5. From the Choices Given Below, Choose the Equation Whose Graphs are Given in the Given Figures.
(1) For the first figure:
(i) \[y=x\]
(ii) \[x+y=0\]
(iii) \[y=2x\]
(iv) \[2+3y=7x\]
Ans. Points on the given line in Figure 1 are \[(-1,1)\] , \[(0,0)\] , and \[(1,-1)\] .
It can be observed that the coordinates of the points on the given graph satisfy the equation \[x+y=0\] .
Therefore, \[x+y=0\] is the equation corresponding to the graph as shown in Figure 1.
Hence, the correct answer is (ii) \[x+y=0\] .
(2) For the second figure:
(i) \[y=x+2\]
(ii) \[y=x-2\]
(iii) \[y=-x+2\]
(iv) \[x+2y=6\]
Ans. Points on the given line in Figure 2 are \[(-1,3)\] , \[(0,2)\] , and \[(2,0)\] .
It can be observed that the coordinates of the points on the given graph satisfy the equation \[y=-x+2\] .
Therefore, \[y=-x+2\] is the equation corresponding to the graph as shown in Figure 2.
Hence, the correct answer is (iii) \[y=-x+2\] .
6. If the Work Done by a Body on Application of a Constant Force is Directly Proportional to the Distance Travelled 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 Travelled by the Body is:
(i) \[2\] units
Ans. Let the distance travelled and the work done by the body be \[x\] and \[y\] respectively.
It is given that both these quantities are directly proportional.
So, Work done \[\propto \] Distance travelled.
Hence, \[y\propto x\] :
\[y=kx\]
Where, \[k\] is a constant.
We are given that constant force applied on the body is \[5\] units.
Therefore, \[y=5x\] ......(1)
By substituting the different values of \[x\] in the equation (1) we get different values for \[y\] .
When \[x=0\] , \[y=0\]
When \[x=1\] , \[y=5\]
When \[x=-1\] , \[y=-5\]
Thus, we have the following table with all the obtained solutions:
Plotting \[(0,0)\] , \[(1,5)\] and \[(-1,-5)\] on the graph paper and drawing a line joining them, we obtain the required graph.
The graph of the line represented by the given equation is shown below.
From the graph, it can be observed that the value of \[y\] corresponding to \[x=2\] is \[10\] . This implies that the work done by the body is \[10\] units when the distance travelled by it is \[2\] units.
(ii) \[0\] units
Ans. Let the distance travelled and the work done by the body be \[x\] and \[y\] respectively.
It is given that both these quantities are directly proportional.
So, Work done \[\propto \] Distance travelled.
Hence, \[y\propto x\] :
\[y=kx\]
Where, \[k\] is a constant.
We are given that constant force applied on the body is \[5\] units.
Therefore, \[y=5x\] ......(1)
By substituting the different values of \[x\] in the equation (1) we get different values for \[y\] .
When \[x=0\] , \[y=0\]
When \[x=1\] , \[y=5\]
When \[x=-1\] , \[y=-5\]
Thus, we have the following table with all the obtained solutions:
Plotting \[(0,0)\] , \[(1,5)\] and \[(-1,-5)\] on the graph paper and drawing a line joining them, we obtain the required graph.
The graph of the line represented by the given equation is shown below.
From the graph, it can be observed that the value of \[y\] corresponding to \[x=0\] is \[0\] . This implies that the work done by the body is \[0\] unit when the distance travelled by it is \[0\] unit.
7. 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 Which Satisfies This Data. (You may take their contributions as Rs. \[x\] and Rs. \[y\] ) Draw the Graph of the Same.
Ans. Let the amount that Yamini and Fatima contributed towards the Prime Minister’s Relief Fund be Rs. \[x\] and Rs. \[y\] respectively.
Amount contributed by Yamini and Fatima together would be Rs. \[100\] .
\[x+y=100\]
\[y=100-x\]
By substituting the different values of \[x\] in the equation we get different values for \[y\]
When \[x=0\] , \[y=100\]
When \[x=50\] , \[y=50\]
When \[x=100\] , \[y=0\]
Thus, we have the following table with all the obtained solutions:
Plotting the points \[(0,100)\] , \[(50,50)\] and \[(100,0)\] on the graph paper and drawing a line to join them, we obtain the following graph:
Here, we can see that variable \[x\] and \[y\] are representing the amount contributed by Yamini and Fatima respectively and these quantities cannot be negative. Hence, only those values of \[x\] and \[y\] which are lying in the first quadrant will be considered.
8. IN Countries Like USA and Canada, Temperature is Measured in Fahrenheit, Whereas in Countries Like India, It is Measured in Celsius. Here is a Linear Equation That Converts Fahrenheit to Celsius:
\[F=\left( \frac{9}{5} \right)C+32\]
(i) Draw the graph of the linear equation above using Celsius for \[x\] -axis and Fahrenheit for \[y\] -axis.
Ans. Given the equation \[F=\left( \frac{9}{5} \right)C+32\] .
By substituting the different values of \[x\] in the equation we get different values for \[y\] .
When \[C=0\] , \[F=32\]
When \[C=10\] , \[F=50\]
When \[C=-40\] , \[F=-40\]
Thus, we have the following table with all the obtained solutions:
Plotting the points \[(0,32)\] , \[(10,50)\] and \[(-40,-40)\] on the graph and joining these points by a line segment, we obtain the required graph.
(ii) If the temperature is \[{{30}^{\circ }}\] C, what is the temperature in Fahrenheit?
Ans. Given temperature \[{{30}^{\circ }}\] C, and the formula \[F=\left( \frac{9}{5} \right)C+32\] , we simply substitute the value:
\[\Rightarrow F=\left( \frac{9}{5} \right)(30)+32\]
\[\Rightarrow F=54+32\]
\[\Rightarrow {{86}^{\circ }}\] F
Therefore, the temperature \[{{30}^{\circ }}\] C in Fahrenheit is \[{{86}^{\circ }}\] F.
(iii) If the temperature is \[{{95}^{\circ }}\] F, what is the temperature in Celsius?
Ans. Given temperature \[{{95}^{\circ }}\] F, and the formula \[F=\left( \frac{9}{5} \right)C+32\] , we simply substitute the value:
\[\Rightarrow 95=\left( \frac{9}{5} \right)C+32\]
\[\Rightarrow 63=(\frac{9}{5})C\]
\[\Rightarrow {{35}^{\circ }}\] C
Therefore, the temperature \[{{95}^{\circ }}\] F in Celsius is \[{{35}^{\circ }}\] C.
(iv) If the temperature is \[{{0}^{\circ }}\] C, what is the temperature in Fahrenheit and if the temperature is \[{{0}^{\circ }}\] F, what is the temperature in Celsius?
Ans. We know that \[F=\left( \frac{9}{5} \right)C+32\] . So, if \[C={{0}^{\circ }}\] , then substituting this value in the former equation, we get:
\[F=\left( \frac{9}{5} \right)(0)+32\]
\[\Rightarrow {{32}^{\circ }}\] F
Similarly, if \[F={{0}^{\circ }}\] , then substituting this value we get:
\[0=\left( \frac{9}{5} \right)C+32\]
\[\Rightarrow C=\left( \frac{5}{9} \right)(-32)\]
\[\Rightarrow -{{17.77}^{\circ }}\] C
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.
Ans. We know that \[F=\left( \frac{9}{5} \right)C+32\] .
Next, we consider \[F=C\] , and substitute in above equation:
\[F=\left( \frac{9}{5} \right)F+32\]
\[\Rightarrow -\left( \frac{4}{5} \right)F=32\]
\[\Rightarrow \frac{1}{5}F=-8\]
\[\Rightarrow F=-{{40}^{\circ }}\]
Hence, there exists a temperature which is numerically same in both Fahrenheit and Celsius.
It is at \[-{{40}^{\circ }}\] .
NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Exercise 4.3
Opting for the NCERT solutions for Ex 4.3 Class 9 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 4.3 Class 9 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.
Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 9 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 9 Maths Chapter 4 Exercise 4.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.
Besides these NCERT solutions for Class 9 Maths Chapter 4 Exercise 4.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.
Do not delay any more. Download the NCERT solutions for Class 9 Maths Chapter 4 Exercise 4.3 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.
Important Topics Covered in Exercise 4.3 of Class 9 Maths NCERT Solutions
Exercise 4.3 of Class 9 Maths NCERT Solutions is mainly based on the concept of graphical representation of the linear equations in two variables. These solutions are very helpful to have a deeper understanding of graphs and plot the solutions of the linear equations on a coordinate plane.
The concepts covered in this exercise will help to build a solid base for the core algebra concepts which is required for advanced Math studies. Students are advised to make a note of the key concepts and formulas present in Exercise 4.3 of the NCERT Solutions for Class 9 Maths which will benefit the students to memorize them fast. By doing this practice, students will be well prepared for the exam.
NCERT Solutions for Class 9 Maths Chapterwise PDF
CBSE Class 9 Maths Chapter 4 Other Exercises
FAQs on NCERT Solutions for Class 9 Maths Chapter 4: Linear Equations in Two Variables - Exercise 4.3
1. What is the linear equation in two variables?
The linear expressions in this chapter have two variables and are in the form of aX+bY+c=0. Students are required to solve the equation and find values for the variables x and y. A linear equation has an infinite number of solutions as there can be many possible combinations of values of x and y. This chapter also teaches how to plot variables or coordinates on a graph which is a straight line.
All solutions of linear equations are points on the graph of the linear equation. In this chapter, students learn how to form an equation of two variables of the format:
ax + by + c = 0, where a, b and c are any real number except zero.
2. How many questions are there in this exercise?
There are a total of 8 questions in this exercise. Here, in question 1 there are 3 subparts, in which you will be given the equation where you have to represent it over the graph. In question 2, you will be given a point through which 2 lines will be passing, you will have to find out how many lines pass through that particular point.
Question 3 asks you to find the value of a from the given equation. Question 4, 6, 7 and 8 are scenario-based questions. In which, a particular situation will be given and you will have to solve it through the scenario. Question 5 is a graphical representation question, in which from the given figure, you have to mark the exact positions of the points.
3. What are the main topics which are covered in chapter 4?
This is one of the vast chapters in class 9 maths and it is also interesting for the students who like graphs related problems. The chapter consists of 5 main topics in it. They are:
Introduction to the Linear Equations.
Linear equations.
A solution of a linear equation.
Graph of a Linear equation in two variables.
Equations of lines parallel to the X-axis and Y-axis.
You will also learn three types of variables which are independent, dependent and controlled. Basically variable is referred to factors, traits and conditions.
4. Where can I find all the solutions of the Class 9 Maths NCERT book?
Vedantu covers all the solutions asked in the NCERT solutions book. All our NCERT Solutions are framed by our experienced faculty and academic experts in an easy and understandable way. These solutions also will help you in gaining a better understanding of various crucial concepts covered in this chapter with ease.
All the solutions to the exercise questions have been organised in a structured way, only through this, the main concept will be clear to you as they are connected to each other. The solutions provided by us will help you in improving your knowledge about one of the important parts of the body.
5. Is NCERT Solutions for Class 9 Chapter 4 enough to prepare for exams?
The NCERT Class 9 Solutions for Chapter 4 provides stepwise solutions for all the exercises and examples in chapter 4. They are explained in an easy manner so that students irrespective of their calibre can understand the concepts easily. Apart from the detailed understanding of the concepts, the solutions PDF also provides extra exercises with solutions for the students to practice and prepare from. Thorough practice can help them ace the exam.
6. What is an ideal study plan for Class 9 Maths Chapter 4?
Class 9 Maths Chapter 4 introduces important concepts like linear equations that are helpful in higher classes for students. It is important that the students understand these concepts well and form a strong base. It is essential that while preparing the students practice each question and example from the NCERT textbook. They should write and memorise all the formulas and also practice everything that is taught on a daily basis. Practice is key to score good marks.
7. How many exercises are there in the Chapter 4 of Class 9 Maths textbook?
Chapter 9 of the CBSE textbook is Linear Equations in Two Variables. The chapter focuses on the introduction to linear equations, graphs and interpreting values of variables. The NCERT Solutions for Class 9 Maths Chapter 4 available for download on the Vedantu website provides solutions for all the exercises in the textbook. There are a total of four exercises- 4.1, 4.2, 4.3 and 4.4. All the solutions are given in an easy to understand step by step format.
8. What are the two methods of solving the linear equations?
The linear equations with two variables can be solved with the help of two methods- the Graphical Method and the Linear Method. In the graphical method, the given equations are plotted on the graph and the point at which both the lines intersect is the final answer. In the linear method, the solution is determined by substituting the value of one variable in terms of the other.
9. Is Class 9 Maths Chapter 4 tough?
No, it is not tough. Being focused and thorough in your studies throughout the year will help you achieve good marks. To help you with this, Vedantu provides the students with NCERT textbook solutions available for free download from the website and from the Vedantu app. The solutions are formulated by experts and are aimed at making the topics easier for students to understand.