NCERT Solutions for class 9 Maths Chapter 4- Linear Equations In Two Variables Exercise 4.3

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 \[x+y=4\] , 

\[\Rightarrow y=4-x\]

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:

(Image will be uploaded soon)

(ii) \[x-y=2\] 

Ans. Given linear equation \[x-y=2\] , 

\[\Rightarrow y=x-2\]

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:

(Image will be uploaded soon)

(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:

(Image will be uploaded soon)

(iv) \[3=2x+y\] 

Ans. Given linear equation \[3=2x+y\] , 

\[\Rightarrow y=3-2x\]

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:

(Image will be uploaded soon)

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: 

(Image will be uploaded soon)

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) \[2+3y=7x\]

(Image will be uploaded soon)

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) \[x+2y=6\]

(Image will be uploaded soon)

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. 

(Image will be uploaded soon)

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. 

(Image will be uploaded soon)

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:

(Image will be uploaded soon)

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.

(Image will be uploaded soon)

(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.

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