SEBA Class 9 Mathematics Chapter 4 Linear Equations in Two Variables

SEBA Class 9 Mathematics Chapter 4 Linear Equations in Two Variables Solutions, SEBA Class 9 Maths Textbook Notes in English Medium, SEBA Class 9 Mathematics Chapter 4 Linear Equations in Two Variables Solutions in English to each chapter is provided in the list so that you can easily browse throughout different chapter Assam Board SEBA Class 9 Mathematics Chapter 4 Linear Equations in Two Variables Notes and select needs one.

SEBA Class 9 Mathematics Chapter 4 Linear Equations in Two Variables

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Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. SEBA Class 9 Mathematics Chapter 4 Linear Equations in Two Variables Question Answer. These solutions are part of SCERT All Subject Solutions. Here we have given SEBA Class 9 Mathematics Chapter 4 Linear Equations in Two Variables Solutions for All Subject, You can practice these here.

Linear Equations in Two Variables

Chapter – 4

Q. 1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.

Ans: Let the cost of a notebook is Rs. x and cost of a pen is Rs. y.

Therefore a linear equation in two variables (x and y) representing the given statement is x = 2y.

Q.2. Express the following linear equations in the form a x + by + c = 0 and indicate the values of a, b and c in each case.

Ans: 

On comparing the coefficients of x and y and also constant terms we observe that 

a=2,b=3 and c = – 

Ans: 

(iii) -2x+3y= 6

Ans: -2x+3y =6 in the form of

ax + by +c=0 is written as

-2x+3y-6=0 [Changing 6 to L.H.S.]

⇒ -2x+3y+(-6)=0

On comparing the coefficients of x and y and also constant terms we observe that

a=-2, b=3 and c=0

(iv) x = 3y

Ans: x=3y in the form of ax + by + c= 0 is written as

x-3y = 0 ⇒1.x+(-3)y-+0=0

On comparing the coefficients of x and y and also constant terms

we observe that a=1,b=-3 and c=0

(v) 2x = -5y

Ans: 2x=-5y in the form of ax + by + c= 0 is written

as 2x+5y =0 [Changing -5y to L.H.S.]

⇒ 2x+5y+0=0

On comparing the coefficients of x and y and also constant terms

we observe that a =2, b=5 and c= 0

(vi) 3x + 2 = 0

Ans: 3x + 2 = 0 in the form of ax + by + c = 0 is written as

3x+0.y +2 = 0

On comparing the coefficients of x and y and also constant terms

we observe that a =3, b=0 and c= 0

(vii) y – 2 = 0

Ans: y -2 = 0 in the form of ax + by + c = 0 is written as

0.x+1.y+(-2)=0

On comparing the coefficients of x and y and also constant terms

we observe that a =0, b=1 and c= -2

(viii) 5 = 2x

Ans: 5= 2x in the form of ax + by + c = 0 is written as 5-2.x=0

⇒ -2x+0.y+5=0

On comparing the coefficients of x and y and also constant terms

we observe that a =2, b = 0 and c=5

Q. 3. The equation 3x + 2y =12 has-

(a) A unique solution.

(b) Two solutions.

(c) Infinite solutions.

(d) No solutions.

Ans: (c) Infinite solutions.

Q.1. Which one of the following statements is true and why?

y = 3x + 5 has

(i) a unique solution.

Ans: We are given the equation y=3x+5 Put x = 0, then y=5

Therefore, (0,5) is the solution of this equation. Again, put x=1 then y =8

(ii) Only two solutions.

Ans: Therefore, (1, 8) is also the solution 

of this equation. Again, put x = 2 

then y=11.

(iii) Infinitely many solutions.

Ans: Therefore, (2, 11) is also the solution of this equation. Again x=3, y=8 and so on.

Now, it is clear that this equation has infinitely many solutions.

Remember. [A linear equation in two variable has infinitely many solution.]

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

(i) 2x + y = 7 

Ans: (i) We have 2x + y = 7 put x = 0, then y = 7

Therefore, (0, 7) is the solution of this equation.

Again put x = 1, then x = 5

Therefore (1, 5) is also the solution of this equation. Again put x= 2, 

then y = 3.

Therefore (2, 3) is also the solution of this equation. Again put x = 3

then y = 1

Therefore (3, 1) is also the solution of this equation. 

Therefore, (0,7), (1,5), (2, 3) and (3, 1) are

the four solutions of the equation 2x + y = 7

(ii) лx + y = 9

Ans: We have given πx+=9 or,

Therefore (0,9) is the solution of this equation.

Again put x = 7 they y = – 13

Therefore (7,-13) is also the solution of this equation. 

Again put x = 14, then y = – 35

Therefore (14,-35) is also the solution of this equation.

of equation πx+y=9

(iii) x+4y

Ans: We are given that x = 4y put y=0 then x=0

Therefore, (0, 0) is the solution of this equation.

Again y =1, then x=4.

Therefore (4, 1) is also part of this equation.

Again, put y = 2 then x= 8

Therefore, (8, 2) is also the solution of this equation.

Again put y = 3, x = 12

Therefore, (12, 3) is also the solution of this equation.

Hence, (0, 0), (4, 1), (8, 2) and (12, 3) are the four solutions of equation

x=4y

Q. 3. Check which of the following are solutions of the equation

x-2y=4 and which are not.

(i) (0, 2) 

Ans: We are given that x-2y=4

put x=0 and y = 2

then, 0-2×2=4,⇒-4=4

Here, L.H.S.≠R.H.S

Therefore, (0, 2) is not the solution of equation x-2y=4

(ii) (2, 0) 

Ans: Put x=2 and y = 0

Then, 2-2×0=4⇒2=4

Here, L.H.S.≠R.H.S.

Therefore (2, 0) is not the solution of equation x-2y=4

(iii) (4,0) 

Ans: Put x=2 and y = 0

Then 4-2×0=4⇒4=4

Here, L.H.S.=R.H.S.

Therefore, (4, 0) is the solution of equation x-2y=4

(iv) (√2, 4√2) 

Ans: Put x=√2 and y = 4√2 in equation x-2y=4

Then √2-2×4√2=4 ⇒√2-8√2=4⇒7√2=4

Here, L.H.S.≠R.H.S.

Therefore, (√2, 4√2) is not the solution of equation x-2y=4

(v) (1, 1)

Ans: Put x=1 and y = 1 in equation x-2y=4

Then 1-2×1=4⇒1-2=4

L.H.S.≠R.H.S

Therefore, (1, 1) is not the solution of equation x – 2y = 4

2x + 2y = k.

Ans: We are given that x=2 and y = 1 is the solution of the equation

2x+3y=k

Put x=2 and y = 1 in equation 2x + 3y=k

2×2+3×1=k

4+3=k∴k=7

Hence, the value of k in equation 2x + 3y = k is 7

Q.1. Draw the graph of each of the following linear equations in two variables:

(i) x+y=4 

Ans: We are given the equation x+y=4

To draw the graph, we need at least two solutions of the equation. 

Put x = 0 then Y = 4 and if x = 4 then y = 0 are solutions of the given equation. 

So we use the following table to draw the graph.

(ii) x-y=2 

Ans: We have given the equation x – y = 2 To draw the graph, we need at least two solutions of the equation. 

Put x = 0 then y = – 2 and if x = 2 then y = 0 are solutions of the given equation. 

So we use the following table to draw the graph.

(iii) y = 3x 

Ans: We are given the equation y = 3x 

To draw the graph, we need at least two solutions of the equa- tion. 

Put x = 0 then y = 0 and if x = 1 then y = 3 are the solu- tions of the given equation. 

So we use the following table to draw the graph.

(iv) 3 = 2x+y

Ans: We have given the equation 3 = 2x + y

To draw a graph. We need at least two solutions of the equation: 

Put x = 0 the y = 3 , and if x = 1 then y = 1 are the solutions of the given equation. 

So we use the following table to draw the graph:

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

Ans: Here (2, 14) is the solution of a linear equation. The equation of two lines passing through (2, 14) are 7x = y and x + 12 = y.

There are infinitely many linear equations which are satisfied by the co- ordinate of the point (2, 14).

2x + 10 = y 3x + 8 = y 4x + 6 = y and so on. 

It is because we know that from a given point these are infinitely many straight lines.

Q.3. 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 (3, 4) lies on the graph of the equation 3y = ax + 7. 

It is clear that (3, 4) is the solution of this equation. 

Therefore we put x = 3 and Y = 4 in this equation.

 3 × 4 = a ×3 + 7

Q. 4. The taxi 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 dis- tance covered as x km and total fare as Rs. y, write a linear equation for this information and draw its graph.

Ans: We are given that total distance covered is x km and total fare is Rs.y. 

But, we are given that the rate of first km is Rs. 8. 

Therefore the fare of remaining distance (x – 1) km is Rs. 5 per km. 

Therefore, according to question, we can say 

(x – 1) ×5 × 8 = y 

or, 5x – 5 + 8 = y or, 5x + 3 = y 

Therefore, 5x + 3 = y is required for this information.

We use the following table to represent the graph.

Q. 5. From the choices given below, choose the equation whose graph

are given in Fig. 4.22 and Fig. 4.23 (A) For Fig. 4.22

(i) y = x

(ii) x + y = 0

(iii) y = 2x

(iv) 2 + 3y = 7x

Ans:  In Fig. 4.22 The points on the line are (-1, 1), (0, 0) and (1,-1). 

By inspection x + y = 0 is the equation corresponding to this graph. We find that the y coordinate in each case is equal but opposite sign of the x coor- dinate.

(B) For Fig. 4.23

(i) y = x + 2

(ii) y = x – 2

(iii) y = – x + 2

(iv) x + 2y = 6

Ans: In Fig. 4.23 The points on the line are (0, 2) and (2, 0). 

By inspec- tion y = – x + 2 is the equation corresponding to this graph. 

We find that when we put x = 0 then the value of y = 2 and when we put x = 2 then the value of y = 0

Q. 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 real from the graph the work done when the distance travelled by the body is

(i) 2 units

(ii) 0 units

Ans: Let us assume,

The work done by a constant force is y-units and the distance travelled by the body is x-units. Therefore, According to the question. 

Force y is directly proportional to distance travelled x 

y∝x

∴ y = k x       (where k is constant)

Again according to question, we have given that constance force

(k) = 5 units.

∴ The required two variable equation is y = 5x

(i) According to the graph, when the distance travelled by the body is 2 unit, then work done is 10 units.

(ii) According to the graph, when the distance travelled by the body is 0 unit, then work done is 0 unit.

To draw the graph of y = 5x we use the following table.”

Q. 7. Yamini and Fatima, two students of class IX of a school, together contributed Rs. 100 towards 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 Yamini contributed towards Prime Minister Relief Fund is Rs. x 

and Fatima contributed Rs. y

Therefore, according to question, x + y = 100 

∴ The linear equation which satisfies this data is x + y = 100 

To draw the graph we use following table 

Q. 8. In countries like the 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:

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

Ans: We have given the linear equation. 

To draw the graph, we need at least two solutions of the equation. 

Put C = 0 then F = 32 and if C = 5 then F = 41 are the solutions of the given equation. So we use the following table to draw the graph.

Taking C on x-axis and F on y-axis.

(ii) If the temperature is 30°C, what is the temperature in Fahren- heit?

Ans: We have given that 

C = 30⁰ 

Then by the linear equation

Therefore, if the temperature is 30°C then the temperature in Fahren- heit is 86⁰.

(iii) If the temperature is 95°F, what is the temperature? 

Ans: We have given that F = 95⁰

Then by the linear equation 

Put F=95⁰

The temperature of 95⁰F 35⁰ Celsius

(iv) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0⁰F, what is the temperature in Celsius?

Ans: We have given that 

C=0⁰

Then by the linear equation 

Put C=0⁰

or, F=0+32

F=32

∴ The temperature of 0⁰C is 32 Fahrenheit. Again in case II.

We have given that 

F=0⁰

Then by linear equation 

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

Ans: Yes, there is a temperature which is numerically the same in both Fahrenheit and Celsius. 

This is -40°C 

At – 40⁰C the value of Fahrenheit scale is also – 40⁰F.

Q. 1. Give the geometric representation of y = 3 as an equation

(i) In one variable

Ans: We have given that y=3

It is treated as an equation in one variable, y only. It has a unique solution y = 3, which is a point on the graph.

(ii) In two variables.

Ans: We have given that y = 3

It can be also expressed as 0.x + y = 3

It is treated as an equation in two variables, x and y.

It has infinitely many solutions which are in the form of (r, 3) where r is any real number. The solutions of this equation are represented by a line. We use following table to draw the graph

Q. 2. Give the geometric representations of 2x + 9 = 0 as an equation.

(i) In one variable. 

Ans: We have given that 2x + 9 = 0

 or, x = – 9/2

It is treated as an equation in one variable, x. only. It has a unique solution x = – 9/2 which is a point graph. 

(ii) In two variables.

Ans: We have given that 2x + 9 = 0 It can be also expresses as 2x + 0,

y + 9 = 0 

It is treated as an equation in two variables, x and y

It has infinitely many solutions which are in the form of (- 9/2, r) where r is any real number.

The solutions of this equation are represented by a line. To draw the graph we use the following table.