### Exercise 4.1

**Question 1. 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 Rs. x and that of a pen to be Rs.y).**Solution:**

Let the cost of a notebook = Rs. x

and the cost of a pen = Rs. y

According to the condition, we have

[Cost of a notebook] =2 x [Cost of a pen]

i. e„ (x) = 2 x (y) or, x = 2y

or, x – 2y = 0

Thus, the required linear equation is x – 2y = 0.

**Question 2. Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:**

(i) 2x + 3y =

(ii)

(iii) – 2x + 3y = 6

(iv) x = 3y

(v) 2x = -5y

(vi) 3x + 2 = 0

(vii) y – 2 = 0

(viii) 5 = 2x**Solution:**

(i) We have 2x + 3y =

or (2)x + (3)y + ( ) = 0

Comparing it with ax + by +c= 0, we geta = 2,

b = 3 and c= – .

(ii) We have

or x + (- ) y + (10) = 0

Comparing it with ax + by + c = 0, we get

a =1, b =- and c= -10

(iii) Wehave -2x + 3y = 6 or (-2)x + (3)y + (-6) = 0

Comparing it with ax – 4 – by + c = 0,we get a = -2, b = 3 and c = -6.

(iv) We have x = 3y or (1)x + (-3)y + (0) = 0 Comparing it with ax + by + c = 0, we get a = 1, b = -3 and c = 0.

(v) We have 2x = -5y or (2)x + (5)y + (0) = 0 Comparing it with ax + by + c = 0, we get a = 2, b = 5 and c = 0.

(vi) We have 3x + 2 = 0 or (3)x + (0)y + (2) = 0 Comparing it with ax + by + c = 0, we get a = 3, b = 0 and c = 2.

(vii) We have y – 2 = 0 or (0)x + (1)y + (-2) = 0 Comparing it with ax + by + c = 0, we get a = 0, b = 1 and c = -2.

(viii) We have 5 = 2x ⇒ 5 – 2x = 0

or -2x + 0y + 5 = 0

or (-2)x + (0)y + (5) = 0

Comparing it with ax + by + c = 0, we get a = -2, b = 0 and c = 5.

### NCERT Solutions for Class 9 Maths Exercise 4.2

**Question 1. Which one of the following options is true, and why?**

has

(i) a unique solution,

(ii) only two solutions,

(iii) infinitely many solutions

**Solution:**

We need to the number of solutions of the linear equation.

We know that any linear equation has infinitely many solutions.

Justification:

If then

If then

If then = -1

Similarly, we can find infinite many solutions by putting the values of

**Question 2. Write four solutions for each of the following equations:**

(i)

(ii)

(iii)

**Solution:**

We know that any linear equation has infinitely many solutions.

Let us putin the linear equation, to get

Thus, we get first pair of solution as.

Let us putin the linear equation, to get

Thus, we get second pair of solution as.

Let us putin the linear equation, to get

Thus, we get third pair of solution as.

Let us putin the linear equation, to get

Thus, we get fourth pair of solution as.

Therefore, we can conclude that four solutions for the linear equationare.

(ii)

We know that any linear equation has infinitely many solutions.

Let us putin the linear equation, to get

Thus, we get first pair of solution as.

Let us putin the linear equation, to get

Thus, we get second pair of solution as.

Let us putin the linear equation, to get

Thus, we get third pair of solution as.

Let us putin the linear equation, to get

Thus, we get fourth pair of solution as.

Therefore, we can conclude that four solutions for the linear equationare

.

(iii)

We know that any linear equation has infinitely many solutions.

Let us putin the linear equation, to get

Thus, we get first pair of solution as.

Let us putin the linear equation, to get

Thus, we get second pair of solution as.

Let us putin the linear equation, to get

Thus, we get third pair of solution as.

Let us putin the linear equation, to get

Thus, we get fourth pair of solution as.

Therefore, we can conclude that four solutions for the linear equationare.

**Question 3. Check which of the following are solutions of the equationand which are not :**

(i)

(ii)

(iii)

(iv)

(v)

**Solution:**

(i)

We need to putin the L.H.S. of linear equation, to get

L.H.S. R.H.S.

Therefore, we can conclude thatis not a solution of the linear equation .

(ii)

We need to putin the L.H.S. of linear equation, to get

L.H.S. R.H.S.

Therefore, we can conclude thatis not a solution of the linear equation .

(iii)

We need to putin the linear equation, to get

L.H.S. R.H.S.

Therefore, we can conclude thatis a solution of the linear equation .

(iv)

We need to putin the linear equation, to get

L.H.S. R.H.S.

Therefore, we can conclude thatis not a solution of the linear equation .

(v)

We need to putin the linear equation, to get

L.H.S. R.H.S.

Therefore, we can conclude thatis not a solution of the linear equation .

**Question 4. Find the value ofk, if x = 2, y = 1 is a solution of the equation.**

**Solution: We know that, if is a solution of the linear equation, then on substituting the respective values of x and y in the linear equation , the LHS and RHS of the given linear equation will not be effected.**

**Therefore, we can conclude that the value of k, for which the linear equation has as one of its solutions is 7.**

**NCERT Solutions for Class 9 Maths Exercise 4.3**

**Question 1. 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 + ySolution:(i) x + y = 4 ⇒ y = 4 – xIf we have x = 0, then y = 4 – 0 = 4x = 1, then y =4 – 1 = 3x = 2, then y = 4 – 2 = 2∴ We get the following table:Plot the ordered pairs (0, 4), (1,3) and (2,2) on the graph paper. Joining these points, we get a straight line AB as shown.Thus, the line AB is the required graph of x + y = 4**

**(ii) x – y = 2 ⇒ y = x – 2If we have x = 0, then y = 0 – 2 = -2x = 1, then y = 1 – 2 = -1x = 2, then y = 2 – 2 = 0∴ We get the following table:Plot the ordered pairs (0, -2), (1, -1) and (2, 0) on the graph paper. Joining these points, we get a straight line PQ as shown.Thus, the ime is the required graph of x – y = 2**

**(iii) y = 3xIf we have x = 0,then y = 3(0) ⇒ y = 0x = 1, then y = 3(1) = 3x= -1, then y = 3(-1) = -3∴ We get the following table:Plot the ordered pairs (0, 0), (1, 3) and (-1, -3) on the graph paper. Joining these points, we get a straight line LM as shown.Thus, the line LM is the required graph of y = 3x.**

**(iv) 3 = 2x + y ⇒ y = 3 – 2xIf we have x = 0, then y = 3 – 2(0) = 3x = 1,then y = 3 – 2(1) = 3 – 2 = 1x = 2,then y = 3 – 2(2) = 3 – 4 = -1∴ We get the following table:Plot the ordered pairs (0, 3), (1, 1) and (2, – 1) on the graph paper. Joining these points, we get a straight line CD as shown.Thus, the line CD is the required graph of 3 = 2x + y.**

**Question 2. Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?Solution:(2, 14) means x = 2 and y = 14Equations which have (2,14) as the solution are (i) x + y = 16, (ii) 7x – y = 0There are infinite number of lines which passes through the point (2, 14), because infinite number of lines can be drawn through a point.**

**Question 3. If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.Solution:The equation of the given line is 3y = ax + 7∵ (3, 4) lies on the given line.∴ It must satisfy the equation 3y = ax + 7We have, (3, 4) ⇒ x = 3 and y = 4.Putting these values in given equation, we get3 x 4 = a x 3 + 7⇒ 12 = 3a + 7⇒ 3a = 12 – 7 = 5 ⇒ a = Thus, the required value of a is **

**Question 4. The taxi fare In a city Is as follows: For the first kilometre, the fare Is Rs. 8 and for the subsequent distance it is Rs. 5 per km. Taking the distance covered as x km and total fare as Rs.y, write a linear equation for this Information, and draw Its graph.Solution:Here, total distance covered = x km and total taxi fare = Rs. yFare for 1km = Rs. 8Remaining distance = (x – 1) km∴ Fare for (x – 1)km = Rs.5 x(x – 1)Total taxi fare = Rs. 8 + Rs. 5(x – 1)According to question,y = 8 + 5(x – 1) = y = 8 + 5x – 5⇒ y = 5x + 3,which is the required linear equation representing the given information.Graph: We have y = 5x + 3Wben x = 0, then y = 5(0) + 3 ⇒ y = 3x = -1, then y = 5(-1) + 3 ⇒ y = -2x = -2, then y = 5(-2) + 3 ⇒ y = -7∴ We get the following table:Now, plotting the ordered pairs (0, 3), (-1, -2) and (-2, -7) on a graph paper and joining them, we get a straight line PQ as shown.Thus, the line PQ is the required graph of the linear equation y = 5x + 3.**

**Question 5. From the choices given below, choose the equation whose graphs are given ¡n Fig. (1) and Fig. (2).For Fig. (1)(i) y = x(ii) x + y = 0(iii) y = 2x(iv) 2 + 3y = 7x**

**For Fig. (2)(i) y = x + 2(ii) y = x – 2(iii) y = -x + 2(iv) x + 2y = 6Solution:For Fig. (1), the correct linear equation is x + y = 0[As (-1, 1) = -1 + 1 = 0 and (1,-1) = 1 + (-1) = 0]For Fig.(2), the correct linear equation is y = -x + 2[As(-1,3) 3 = -1(-1) + 2 = 3 = 3 and (0,2)⇒ 2 = -(0) + 2 ⇒ 2 = 2]**

**Question 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(ii) 0 unitSolution:Constant force is 5 units.Let the distance travelled = x units and work done = y units.Work done = Force x Distance⇒ y = 5 x x ⇒ y = 5xFor drawing the graph, we have y = 5xWhen x = 0, then y = 5(0) = 0x = 1, then y = 5(1) = 5x = -1, then y = 5(-1) = -5∴ We get the following table:Ploffing the ordered pairs (0, 0), (1, 5) and (-1, -5) on the graph paper and joining the points, we get a straight line AB as shown.From the graph, we get(i) Distance travelled =2 units i.e., x = 2∴ If x = 2, then y = 5(2) = 10⇒ Work done = 10 units.**

**(ii) Distance travelled = 0 unit i.e., x = 0∴ If x = 0 ⇒ y = 5(0) – 0⇒ Work done = 0 unit.**

**Question 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.xand Rs.y.) Draw the graph of the same.Solution:Let the contribution of Yamini = Rs. xand the contribution of Fatima Rs. y∴ We have x + y = 100 ⇒ y = 100 – xNow, when x = 0, y = 100 – 0 = 100x = 50, y = 100 – 50 = 50x = 100, y = 100 – 100 = 0∴ We get the following table:Plotting the ordered pairs (0,100), (50,50) and (100, 0) on a graph paper using proper scale and joining these points, we get a straight line PQ as shown.Thus, the line PQ is the required graph of the linear equation x + y = 100.**

**Question 8. In countries like USA and Canada, temperature is measured In Fahrenheit, whereas in countries like India, it is measured in Celsius. Here Is alinear equation that converts Fahrenheit to Celsius:F = ( )C + 32(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for 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 Celsius?(iv) If the temperature is 0°C, what Is the temperature In Fahrenheit and If the temperature is 0°F, what Is the temperature In Celsius?(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find It.Solution:(i) We haveF = ( )C + 32When C = 0 , F = ( ) x 0 + 32 = 32When C = 15, F = ( )(-15) + 32= -27 + 32 = 5When C = -10, F = (-10)+32 = -18 + 32 = 14We have the following table:Plotting the ordered pairs (0, 32), (-15, 5) and (-10,14) on a graph paper. Joining these points, we get a straight line AB.**

**(ii) From the graph, we have 86°F corresponds to 30°C.(iii) From the graph, we have 95°F corresponds 35°C.(iv) We have, C = 0From (1), we getF = ()0 + 32 = 32Also, F = 0From (1), we get0 = ()C + 32 ⇒ = C ⇒ C = -17.8(V) When F = C (numerically)From (1), we getF = F + 32 ⇒ F – F = 32⇒ F = 32 ⇒ F = -40∴ Temperature is – 40° both in F and C.**

**NCERT Solutions for Class 9 Maths Exercise 4.4**

**Question 1. Give the geometric representation ofas an equation**

**(i) In one variable**

**(ii) In two variables**

**Solution:(i) We need to represent the linear equationgeometrically in one variable.**

**We can conclude that in one variable, the geometric representation of the linear equationwill be same as representing the number 3 on a number line.**

**Given below is the representation of number 3 on the number line.**

**(ii) We need to represent the linear equationgeometrically in two variables.**

**We know that the linear equationcan also be written as.**

**We can conclude that in two variables, the geometric representation of the linear equationwill be same as representing the graph of linear equation.**

**Given below is the representation of the linear equationon a graph.**

**We can optionally consider the given below table for plotting the linear equationon the graph.**

X | 1 | 0 |

y | 3 | 3 |

**Question 2. Give the geometric representations of as an equation**

**(i) In one variable**

**(ii) In two variables**

**Solution:(i) We need to represent the linear equationgeometrically in one variable.**

**We know that the linear equationcan also be written as.**

**We can conclude that in one variable, the geometric representation of the linear equationwill be same as representing the numberon a number line.**

**Given below is the representation of number 3 on the number line.**

**(ii) We need to represent the linear equationgeometrically in two variables.**

**We know that the linear equationcan also be written as.**

**We can conclude that in two variables, the geometric representation of the linear equationwill be same as representing the graph of linear equation.**

**Given below is the representation of the linear equationon a graph.**

**We can optionally consider the given below table for plotting the linear equationon the graph.**

X | 1 | 0 |

y | 4.5 | 4.5 |