Important Questions Class 10 Maths Chapter 3

Important Questions Class 10 Mathematics Chapter 3 – Pair of Linear Equations in Two Variables

Mathematics is based on calculations and it requires a lot of practice. Linear equations in two variables is a bit of a lengthy topic and involves a wide  variety of problems. 

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Questions covered in Chapter 3 Class 10 Mathematics Important Question will be based on a variety of concepts included in the chapter. It has  been curated to bring conceptual clarity among the students and enhance their knowledge with concise answers that can help them in their higher classes as well.

• Pair of Linear Equations in Two Variables
• Graphical Method of Solution of a Pair of Linear Equations
• Algebraic Methods of Solving a Pair of Linear Equations
• Equations Reducible to a Pair of Linear Equations in Two Variables

By solving a lot of questions given in our Important Questions Class 10 Mathematics Chapter 3 can make students confident enough  to face any  questions during board examinations.

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Important Questions Class 10 Mathematics Chapter 3 – With Solutions

The following are the  list of  questionnaires and their solutions which are included in the Class 10 Mathematics Chapter 3 Important Questions:

Question 1. Graphically, the pair of equations

6x – 3y + 10 = 0

2x – y + 9 = 0

are represented by two lines that are

(A) Intersecting at exactly one point. (B) Intersecting at exactly two points.

(C) Coincident (D) parallel.

Answer 1: (D) Parallel

Explanation:  We have given equations as,

6x-3y+10 = 0

On dividing with 3

2x-y+ 10/3= 0… (i)

Also, 2x-y+9=0…(ii)

Table for 2x-y+ 10/3 = 0 is given below,

Table for 2x-y+9=0 is given as,

Thus, the pair of the equations represents two parallel lines.

Question 2. The pair of the equations x + 2y + 5 = 0 as well as –3x – 6y + 1 = 0 have

(A) unique solution (B) exactly two solutions

(C) infinitely many solutions (D) no solution

Answer 2: (D) No solution

Explanation:

The given equations are:

x + 2y + 5 = 0

–3x – 6y + 1 = 0

a1 = 1; b1 = 2; c1 = 5

a2 = -3; b2 = -6; c2 = 1

a1/a2 = -1/3

b1/b2 = -2/6 = -1/3

c1/c2 = 5/1 = 5

Where,

a1/a2 = b1/b2 ≠ c1/c2

Hence, the pair of equations has no solution.

Question 3. When the pair of linear equations is consistent, then the lines will be given as;

(A) parallel (B) always coincident

(C) intersecting or coincident (D) always intersecting

Answer 3: (C) intersecting or coincident

Explanation: Condition for the pair of the linear equations to be consistent can be:

The intersecting lines have unique solution if,

a1/a2 ≠ b1/b2

Coincident and dependent

a1/a2 = b1/b2 = c1/c2

Question 4. The pair of the equations y = 0 as well as y = –7 have

(A) one solution (B) two solutions

(C) infinitely many solutions (D) no solution

Answer 4: (D) no solution

Explanation: The given pair of the equations are y = 0 as well as y = – 7.

Graphically, both the given lines are parallel as well as have no solution

Question 5. The pair of the equations x = a as well as y = b graphically shows lines that are

(A) parallel (B) intersecting at (b, a)

(C) coincident (D) intersecting at (a, b)

Answer 5: (D) intersecting at (a, b)

Explanation: Graphically for every condition,

a, b>>0

a, b< 0

a>0, b< 0

a<0, b>0 and a = b≠ 0.

The pair of the equations x = a as well as y = b graphically represents the given lines that are intersecting

at (a, b).

Therefore, the cases of  the given two lines intersect at (a, b).

Question 6. Do the following pair of the given linear equations have no solution? Explain your answer.

(i) 2x + 4y = 3

12y + 6x = 6

(ii) x = 2y

y = 2x

(iii) 3x + y – 3 = 0

2x + 2/3y = 2

Answer 6: The given condition for no solution = a1/a2 = b1/b2 ≠ c1/c2 (parallel lines)

(i) Yes.

Given that the pair of the equations are,

2x+4y – 3 = 0 as well as 6x + 12y – 6 = 0

On comparing the equations with ax+ by +c = 0;

We have,

a1 = 2, b1 = 4, c1 = – 3;

a2 = 6, b2 = 12, c2 = – 6;

a1 /a2 = 2/6 = 1/3

b1 /b2 = 4/12 = 1/3

c1 /c2 = – 3/ – 6 = ½

Where, a1/a2 = b1/b2 ≠ c1/c2, that is parallel lines

Therefore, the given pair of the linear equations has no solution.

(ii) No.

Given that the pair of the equations,

x = 2y or x – 2y = 0

y = 2x or 2x – y = 0;

On comparing the equations with ax+ by +c = 0;

We have,,

a1 = 1, b1 = – 2, c1 = 0;

a2 = 2, b2 = – 1, c2 = 0;

a1 /a2 = ½

b1 /b2 = -2/-1 = 2

Where, a1/a2 ≠ b1/b2.

Therefore, the given pair of the linear equations has a unique solution.

(iii) No.

Given that the pair of the equations,

3x + y – 3 = 0

2x + 2/3 y = 2

On comparing the equations with ax+ by +c = 0;

We have,

a1 = 3, b1 = 1, c1 = – 3;

a2 = 2, b2 = 2/3, c2 = – 2;

a1 /a2 = 2/6 = 3/2

b1 /b2 = 4/12 = 3/2

c1 /c2 = – 3/-2 = 3/2

Where, a1/a2 = b1/b2 = c1/c2, that is coincident lines

Question 7. Do the following equations represent the pair of coincident lines? Justify   your answer.

(i) 3x + 1/7y = 3

7x + 3y = 7

(ii) –2x – 3y = 1

6y + 4x = – 2

(iii) x/2 + y + 2/5 = 0

4x + 8y + 5/16 = 0

Answer 7: Condition for coincident lines,

a1/a2 = b1/b2 = c1/c2;

(i) No.

as Given pair of linear equations are:

3x + 1/7y = 3

7x + 3y = 7

Comparing the above equations ax + by + c = 0;

where, a1 = 3, b1 = 1/7, c1 = – 3;

as well as a2 = 7, b2 = 3, c2 = – 7;

a1 /a2 = 3/7

b1 /b2 = 1/21

c1 /c2 = – 3/ – 7 = 3/7

where, a1/a2 ≠ b1/b2.

Thus, the given pair of linear equations has a unique solution.

(ii) Yes,

Given a pair of the linear equations.

– 2x – 3y – 1 = 0 and 4x + 6y + 2 = 0;

as Comparing the above equations ax + by + c = 0;

where, a1 = – 2, b1 = – 3, c1 = – 1;

And a2 = 4, b2 = 6, c2 = 2;

a1 /a2 = – 2/4 = – ½

b1 /b2 = – 3/6 = – ½

c1 /c2 = – ½

where, a1/a2 = b1/b2 = c1/c2, i.e. coincident lines

Therefore, the given pair of linear equations is coincident.

(iii) No,

Given pair of the linear equations are

x/2 + y + 2/5 = 0

4x + 8y + 5/16 = 0

 by   comparing the above equations ax + by + c = 0;

where, a1 = ½, b1 = 1, c1 = 2/5;

as well as a2 = 4, b2 = 8, c2 = 5/16;

a1 /a2 = 1/8

b1 /b2 = 1/8

c1 /c2 = 32/25

where, a1/a2 = b1/b2 ≠ c1/c2, i.e. parallel lines

 Thus, the given pair of linear equations has no solution.

Question 8. For which value(s) for λ , do the pair of the given linear equations

λx + y = λ2  as well as x + λy = 1 have

(i) no solution?

(ii) infinitely many solutions?

(iii) unique solution?

Answer 8:

The given pair of the linear equations can be;

λx + y = λ2 and x + λy = 1

a1 = λ, b1= 1, c1 = – λ2

a2 =1, b2=λ, c2=-1

The given equations can be;

λ x + y – λ2 = 0

x + λ y – 1 = 0

On comparing the above given equations with ax + by + c = 0;

We have,

a1 = λ, b1 = 1, c1 = – λ 2;

a2 = 1, b2 = λ, c2 = – 1;

a1 /a2 = λ/1

b1 /b2 = 1/λ

c1 /c2 = λ2

(i) In case of no solution,

a1/a2 = b1/b2 ≠ c1/c2

λ = 1/ λ ≠ λ2

Thus, λ2 = 1;

also, λ2 ≠ λ

Where, we take only λ = – 1,

As the system of the linear equations has no solution.

(ii) For infinitely many solutions,

a1/a2 = b1/b2 = c1/c2

 λ = 1/ λ  = λ2

Thus, λ = 1/ λ gives λ = + 1;

λ = λ2 gives λ = 1,0;

On satisfying both the given equations

λ = 1 is the answer.

(iii) For unique solution,

a1/a2 ≠ b1/b2

so λ ≠1/ λ

Thus, λ2 ≠ 1;

λ ≠ + 1;

Hence, all the real values of λ except +1

Question 9. For which value(s) of k would the pair of given equations

kx + 3y = k – 3

12x + ky = k

have no solution?

Answer 9: 

The given pair of the linear equations is as follows:

kx + 3y = k – 3 …(i)

12x + ky = k …(ii)

When we compare the equations (i) as well as (ii) with ax + by = c = 0,

We 

observe,

a2 = 12, b2 = k, c2 = – k

a1 = k, b1 = 3, and c1 = -(k – 3)

Thus,

a1 /a2 = k/12

b1 /b2 = 3/k

c1 /c2 = (k-3)/k

For no solution of the pair of the given linear equations,

a1/a2 = b1/b2 ≠ c1/c2

k/12 = 3/k ≠ (k-3)/k

Taking the given first two parts, we have,

k/12 = 3/k

k2 = 36

k = + 6

Taking the given last two parts, we have,

3/k ≠ (k-3)/k

3k ≠ k(k – 3)

k2 – 6k ≠ 0

so, k ≠ 0,6

Thus, the value of k for which the given pair of the linear equations has no solution is k = – 6.

Question 10. For which values of a and b, would the following pair of the given linear equations consist of infinitely many solutions?

x + 2y = 1

(a – b)x + (a + b)y = a + b – 2

Answer 10:

The given pair of the linear equations are as follows:

x + 2y = 1 ……(1)

(a-b)x + (a + b)y = a + b – 2 …..(2)

When we compare with ax + by = c = 0 we have,

a1 = 1, b1 = 2, c1 = – 1

a2 = (a – b), b2 = (a + b), c2 = – (a + b – 2)

a1 /a2 = 1/(a-b)

b1 /b2 = 2/(a+b)

c1 /c2 = 1/(a+b-2)

For infinitely many solutions, the pair of the given linear equations will,

a1/a2 = b1/b2=c1/c2 (here, coincident lines)

Thus, 1/(a-b) = 2/ (a+b) = 1/(a+b-2)

Taking the given first two parts,

1/(a-b) = 2/ (a+b)

a + b = 2(a – b)

a = 3b …(iii)

Taking the given last two parts,

2/ (a+b) = 1/(a+b-2)

2(a + b – 2) = (a + b)

a + b = 4 …(iv)

Also, putting the value of a from Eq. (iii) in Eq. (iv), we have,

3b + b = 4

4b = 4

b = 1

Putting the value for b in Eq. (iii), we get

a = 3

Thus, the values for (a,b) = (3,1) satisfies all the given parts. Thus, the required values of a and b are 3 and 1 respectively, and the given pair of the linear equations has infinitely many solutions.

Question 11. Find out the value(s) of p in (i) to (iv) and p and q in (v) for the following pair of the equations:

(i) 3x – y – 5 = 0 as well as 6x – 2y – p = 0, if the lines presented by these equations are  parallel.

Answer 11: Given pair of linear equations is

3x – y – 5 = 0 …(i)

6x – 2y – p = 0 …(ii)

On comparing with ax + by + c = 0 

We observe,

a1 = 3, b1 = – 1, c1 = – 5;

a2 = 6, b2 = – 2, c2 = – p;

a1 /a2 = 3/6 = ½

b1 /b2 = ½

c1 /c2 = 5/p

hence, the lines presented by these equations are as parallel, then

a1/a2 = b1/b2 ≠ c1/c2

Taking last two parts, we observe½ ≠ 5/p

So, p ≠ 10

Thus, the given pair of linear equations are  parallel for all real values of p  except 10.

(ii) – x + py = 1 as well as px – y = 1, if the pair of the equations has no solution.

Answer: Given pair of linear equations is

– x + py = 1 …(i)

px – y – 1 = 0 …(ii)

On comparing with ax + by + c = 0,

We observe,

a1 = -1, b1 = p, c1 =- 1;

a2 = p, b2 = – 1, c2 =- 1;

a1 /a2 = -1/p

b1 /b2 = – p

c1 /c2 = 1

Hence, the line equation has no solution i.e., both lines are parallel to each other.

a1/a2 = b1/b2≠ c1/c2

-1/p = – p ≠ 1

Taking last two parts, we get

p ≠ -1

Taking first two parts, we get

p2 = 1

p = + 1

Thus, the given pair of the linear equations has not any solution for p = 1.

(iii) – 3x + 5y = 7 as well as 2px – 3y = 1, if the lines presented by these equations are intersecting at a unique point.

Answer : Given, pair of the linear equations is

– 3x + 5y = 7

2px – 3y = 1

On comparing with ax + by + c = 0, we get

where, a1 = -3, b1 = 5, c1 = – 7;

And a2 = 2p, b2 = – 3, c2 = – 1;

a1 /a2 = -3/ 2p

b1 /b2 = – 5/3

c1 /c2 = 7

Hence, the lines are intersecting at the unique point i.e., it has a unique solution

a1/a2 ≠ b1/b2

-3/2p ≠ -5/3

p ≠ 9/10

Thus, the lines presented by these equations are intersecting at the unique point for all real values of p are except 9/10

(iv) 2x + 3y – 5 = 0 as well as px – 6y – 8 = 0, if the pair of the equations has a unique solution.

Answer : Given, pair of the linear equations is

2x + 3y – 5 = 0

px – 6y – 8 = 0

with comparing  ax + by + c = 0 we observe

where, a1 = 2, b1 = 3, c1 = – 5;

And a2 = p, b2 = – 6, c2 = – 8;

a1 /a2 = 2/p

b1 /b2 = – 3/6 = – ½

c1 /c2 = 5/8

Hence, the pair of the linear equations has a unique solution.

a1/a2 ≠ b1/b2

so 2/p ≠ – ½

p ≠ – 4

Thus, the pair of linear equations has a unique solution for all values of p except – 4.

(v) 2x + 3y = 7 and 2px + py = 28 – qy, if the pair of equations have infinitely many solutions.

Answer : Given pair of the linear equations 

2x + 3y = 7

2px + py = 28 – qy

or 2px + (p + q)y – 28 = 0

with comparing  the equation ax + by + c = 0,

We observe,

Here, a1 = 2, b1 = 3, c1 = – 7;

And a2 = 2p, b2 = (p + q), c2 = – 28;

a1/a2 = 2/2p

b1/b2 = 3/ (p+q)

c1/c2 = ¼

Hence, the pair of the equations has infinitely many solutions i.e., both lines are  coincident.

a1/a2 = b1/b2 = c1/c2

1/p = 3/(p+q) = ¼

Taking first and third parts, we observe

p = 4

Again, taking last two parts, we observe

3/(p+q) = ¼

p + q = 12

Since p = 4

So, q = 8

where, we see that the values of p = 4 and q = 8 satisfies all three parts.

Thus, the pair of equations has infinitely many solutions for all values of p = 4 and q = 8.

Question 12. Two given straight paths are shown by the equations x – 3y = 2 and –2x + 6y = 5.Check if the paths cross each other or not.

Answer 12: Given that the linear equations are

x – 3y – 2 = 0 …(i)

-2x + 6y – 5 = 0 …(ii)

When we compare with ax + by c=0,

We have,

a1 =1, b1 =-3, c1 =- 2;

a2 = -2, b2 =6, c2 =- 5;

a1/a2 = – ½

b1/b2 = – 3/6 = – ½

c1/c2 = 2/5

that is, a1/a2 = b1/b2 ≠ c1/c2 [parallel lines]

Thus, the two straight paths that are  represented by the given equations never cross each other, as they are parallel to each other.

Question 13. Write the pair of the linear equations that have the unique solution x = – 1, y =3. How many pairs of this kind can you write?

Answer 13: The condition for the pair of the system to have unique solution

a1/a2 ≠ b1/b2

Assume the equations as,

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

As, x = – 1 and y = 3 is the unique solution of the two given equations, then we get,

It must satisfy the given equations –

a1(-1) + b1(3) + c1 = 0

– a1 + 3b1 + c1 = 0 …(i)

and a2(- 1) + b2(3) + c2 = 0

– a2 + 3b2 + c2 = 0 …(ii)

As for the different values of a1, b1, c1 as well as a2, b2, c2 satisfy the Eqs. (i) and (ii).

Therefore, infinitely many pairs of the linear equations are possible.

Question 14. When 2x + y = 23 as well as 4x – y = 19, find the values of 5y – 2x as well as y/x – 2.

Answer 14: Given that the equations are

2x + y = 23 …(i)

4x – y = 19 …(ii)

On adding both the equations, we have,

6x = 42

So, x = 7

Putting the value for x in Eq. (i), we have,

2(7) + y = 23

y = 23 – 14

Thus, y = 9

Therefore, 5y – 2x = 5(9) – 2(7) = 45 – 14 = 31

y/x – 2 = 9/7 -2 = -5/7

Therefore, the values of (5y – 2x) and y/x – 2 are 31 as well as -5/7 respectively.

Question 15. Find out the values for x and y in the following rectangle

Answer 15: Using the property of the rectangle,

We have,

Lengths are same,

So, CD = AB

Therefore, x + 3y = 13 …(i)

Breadth are same,

So, AD = BC

Therefore, 3x + y = 7 …(ii)

On multiplying Eq. (ii) by 3 and then by subtracting Eq. (i),

We have,

8x = 8

So, x = 1

On Putting x = 1 in Eq. (i),

We have,

y = 4

Thus, the required values for x and y are 1 and 4, respectively.

Question 16. Aftab explains to his daughter, “Seven years ago, I was seven times as old as you were then. Moreover, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent the given situation algebraically and graphically.

Answer 16: Assume the present age of Aftab be ‘x’.

Also, the present age of his daughter is ‘y’.

It can be written as; seven years ago,

For the Age of Aftab = x-7

For the Age of his daughter = y-7

As per the question,

x−7 = 7(y−7)

⇒x−7 = 7y−49

⇒x−7y = −42         ………………………(i)

And, three years from now or after three years,

Age of Aftab will be as= x+3.

Age of his daughter will be as = y+3

As per the situation given,

x+3 = 3(y+3)

x+3 = 3y+9

x−3y = 6       …………..…………………(ii)

Subtracting equation (i) from the equation (ii) we get,

(x−3y)−(x−7y) = 6−(−42)

⇒−3y+7y = 6+42

⇒4y = 48

⇒y = 12

The algebraic equation is represented in the form as

x−7y = −42

x−3y = 6

For, x−7y = −42 or x = −42+7y

The solution table can be

For the case of,  x−3y = 6   or     x = 6+3y

The solution table can be

The graphical representation is:

Question 17. The coach of the cricket team buys 3 bats as well as 6 balls for Rs.3900. Later on, she buys another bat and 3 more balls of the same kind for Rs.1300. Represent the given situation algebraically as well as geometrically.

Answer 17:  Assume the cost of the bat be ‘Rs x’

Also, the cost of a ball be ‘Rs y’

As per the question, the algebraic representation can be

3x+6y = 3900

Also, x+3y = 1300

For, 3x+6y = 3900

Or x = (3900-6y)/3

The solution table can be

For the case of, x+3y = 1300

And x = 1300-3y

The solution table can be

The graphical representation can be as follows.

Question 18. The cost of 2 kg of apples as well as 1kg of grapes on the day was found to be Rs.160. A month later, the cost of 4 kg of apples as well as 2 kg of grapes is Rs.300. Represent the following situation algebraically and geometrically.

Answer 18: Assume the cost of 1 kg of apples be ‘Rs. x’

Also, the cost of 1 kg of grapes be ‘Rs. y’

As per the question, the algebraic form is

2x+y = 160

Also, 4x+2y = 300

For the case of 2x+y = 160 or y = 160−2x, 

The solution table is;

For the case of 4x+2y = 300 or y = (300-4x)/2, 

The solution table is;

The graphical representation can be as follows;

Question 19. Are the following pair of given linear equations consistent? Explain your answer.

(i) –3x– 4y = 12

4y + 3x = 12

(ii) (3/5)x – y = ½

(1/5)x – 3y= 1/6

(iii) 2ax + by = a

1. ax + 2by – 2a = 0; a, b ≠ 0

(iv) x + 3y = 11

2 (2x + 6y) = 22

Answer 19:

Conditions for the pair of linear equations to be consistent are:

For a1/a2 ≠ b1/b2. [unique solution]

For a1/a2 = b1/b2 = c1/c2 [coincident or infinitely many solutions]

(i) No.

The given pair of the linear equations

– 3x – 4y – 12 = 0 or 4y + 3x – 12 = 0

On comparing the above equations with ax + by + c = 0;

We have,

a1 = – 3, b1 = – 4, c1 = – 12;

a2 = 3, b2 = 4, c2 = – 12;

a1 /a2 = – 3/3 = – 1

b1 /b2 = – 4/4 = – 1

c1 /c2 = – 12/ – 12 = 1

Where, a1/a2 = b1/b2 ≠ c1/c2

So, the pair of the linear equations has no solution, that is inconsistent.

(ii) Yes.

The given pair of the linear equations

(3/5)x – y = ½

(1/5)x – 3y= 1/6

On comparing the above equations with ax + by + c = 0;

We have,

a1 = 3/5, b1 = – 1, c1 = – ½;

a2 = 1/5, b2 = 3, c2 = – 1/6;

a1 /a2 = 3

b1 /b2 = – 1/ – 3 = 1/3

c1 /c2 = 3

Where, a1/a2 ≠ b1/b2.

So, the given pair of the linear equations has a unique solution that is consistent.

(iii) Yes.

The given pair of the linear equations –

2ax + by –a = 0 or 4ax + 2by – 2a = 0

On comparing the above equations with ax + by + c = 0;

We have,

a1 = 2a, b1 = b, c1 = – a;

a2 = 4a, b2 = 2b, c2 = – 2a;

a1 /a2 = ½

b1 /b2 = ½

c1 /c2 = ½

Where, a1/a2 = b1/b2 = c1/c2

So, the given pair of the linear equations has infinitely many solutions, that is consistent

(iv) No.

The given pair of the linear equations

x + 3y = 11 or 2x + 6y = 11

On comparing the above equations with ax + by + c = 0;

We have,

a1 = 1, b1 = 3, c1 = 11

a2 = 2, b2 = 6, c2 = 11

a1 /a2 = ½

b1 /b2 = ½

c1 /c2 = 1

Where, a1/a2 = b1/b2 ≠ c1/c2.

So, the given pair of the linear equations has no solution.

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• In case students want to practice with additional study material to check their preparation, they   can access Important Questions Class 10 Mathematics Chapter 3 from the Extramarks website to ensure 100% score in the exam results. Trust Extramarks , it is one stop solution to all your problems.

Extramarks provides a repository of  resources like NCERT textbooks, NCERT revision notes, CBSE sample papers, CBSE past year question papers, and CBSE additional questions, Important Questions Class 10 Mathematics Chapter 3 and mock tests for the students of Classes 1 to 12. Students must click on the links given below to access some of these resources:

• NCERT books
• CBSE Revision Notes
• CBSE syllabus
• CBSE sample papers
• CBSE past year question papers
• Important formulas 
• CBSE extra questions