Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Pair of Linear Equations in Two Variables:

Exercise 3.3

Short Answer Type Questions (Q. No. 1 to 11)

Quesntion1. For which value(s) of l, do the pair of linear equations lx + y = l² and

x+ly = 1 have

(i) no solution?           

(ii) infinitely many solutions?

(iii) a unique solution?

Solution:

Given, lx + y = l² and x + ly -1                  ….(i)

Here, a₁ = l, b₁, = 1, c, = -l² and a₂ = 1, b₂ = l, c₂ = -1

(iii) Condition for a unique solution is:

Therefore, all real values of l except  ±1 equation will have unique solution.

Quesntion2. For which value (s) of k will the pair of equations

kx + 3y = k -3,

12x+ ky = k

have no solution?

Solution:

Given,

kx + 3y – (k -3) = 0 and 12x + ky – k = 0

On comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 we have         

Here, a₁ = k, b₁=3 and c₁ = -(k -3)

And a₂ = 12, b₂ = k and c₂ = -k         …..(i)

Condition for a pair of linear equations to have no solution is:

Thus, from (iii) and (iv), it’s clear that at k = - 6 the given pair of linear equations will have no solution.

Quesntion3. For which values of a and b will the following pair of linear equations has infinitely many solutions?

x + 2y = 1

(a - b) x + (a + b) y = a + b - 2

Solution:

Given equation are:

x + 2y −1 = 0 and (a - b) x + (a + b) y – (a + b - 2) = 0 ….(i)

On comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 we have:

Quesntion4. Find the values of p in (i) to (iv) and p and q in (v) for the following pair of equations

(i) 3x -y - 5 = 0 and 6x - 2y - p = 0, if the Lines represented by these

equations are parallel.

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

(iii) -3x + 5y = 7 and 2px - 3y =1

if the lines represented by these equations are intersecting at a unique point.

(iv) 2x + 3y -5 = 0 and px - 6y - 8 = 0,

if the pair of equations has a unique solution.

(v) 2x + 3y = 7 and 2px + py = 28 -qy,

if the pair of equations has infinitely many solutions.

Solution:

(i) Given, 3x - y -5 = 0 and 6x-2y - p = 0

On comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 we have:

c₁ = 5 and a₂ = 6, b₂ = -2, c₂ = -p

As, the lines represented by these equations are parallel, therefore

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

(ii) Given, -x + py -1 = 0 and px-y-1 = 0      …..(i)

We have, a₁ = -1, b₁ = p, c₁= -1and a₂ = p, b₂ = -1 and c₂ = -1

As, the pair of linear equations has no solution i.e., both lines are parallel to each other.

Therefore, the given pair of linear equations has no solution for p = 1.

(iii) Given, -3x + 5y - 7 = 0 and 2px-3y-1 = 0

Here, a₁ = -3, b₁ = 5, c₁ = -7 and a₂ = 2p, b₂ = -3, c₂ = -1

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

Therefore, the lines represented by these equations are intersecting at a unique point for all great values of p except 9/10.

(iv) Given, 2x + 3y-5 = 0 and px - 6y -8 = 0

Here,c₁ = -5 and a₂ = p, b₂ = -6, c₂ = −8

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

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

(v)   Given, 2x + 3y = 7 and 2px + py = 28 -qy

Here, a₁ = 2, b₂, = 3, c_1 = -7 and a₂ = 2p, b₂ = (p + q), c₂ = -28

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

⟹ 4 + q =12

Therefore, q = 8

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

Quesntion5. Two straight paths are represented by the equations x - 3y = 2 and -2x + 6y = 5. Check whether the paths cross each other or not.

Solution:

Hence, two straight paths represented by the given equations never cross each other because they are parallel to each other.

Quesntion6. Write a pair of linear equations which has the unique solution

x = - 1 and y 3. How many such pairs can you write?

Solution:

Hence, infinitely many pairs of linear equations are possible.

Quesntion7. If 2x + y = 23 and 4x - y = 19, then find the values of 5y -2x and

Solution:

Quesntion8. Find the values of x and y in the following rectangle

Quesntion9. Solve the following pairs of equations

Solution:

Hence, the required values of x and y are 1.2 and 2.1, respectively

Hence, the required values of x and y are 6 and 8, respectively

Hence, the required values of x and y are 3 and 2, respectively.

(v) Given pair of linear equations is

          43x + 67y = -24     ...(i)

And   67x + 43y = 24      ...(ii)

Now taking 43 × (i) + 67 ×(ii), we get

Hence, the required values of x and y are 1 and -1, respectively.

(vi) Given pair of linear equations is

Hence, the required values of x and y are a² and b², respectively.

Now, multiplying both sides of equation (i) by LCM (10, 5) = 10, we get

x + 2y – 10 = 0

Þ x + 2y = 10 ... (iii)

Again, multiplying both sides of (iv) by LCM (8, 6) = 24, we get

3x + 4y = 360 ... (iv)

Now, (iv) – 2 × (iii) gives:

Hence, the solution of the pair of equations is x = 340, y = - 165 and the required value of l is

Quesntion11. By the graphical method, find whether the following pair of equations are consistent or not. If consistent, solve them.

(i) 3x + y + 4 = 0, 6x - 2y + 4 = 0

(ii) x - 2y = 6, 3x - 6y = 0

(iii) x + y = 3, 3x + 3y = 9

Solution:

(i) Given pair of equations is:

 3x+ y + 4= 0 and 6x - 2y + 4 = 0

On comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 we have

and c₁ = 4  and a₂ = 6, b₂ = - 2 and c₂ = 4

So, the given pair of linear equations has a unique solution and thereby it is consistent.    

We have,  3x + y + 4 = 0

Þ y = - 4 - 3x                           ….(i)

If x = 0, y = - 4

x = - 1,  y = - 1

x = - 2, y = 2                    

And 6 - 2 + 4 = 0                     …..(ii)

Þ 2y = 6x + 4

Þ y = 3x + 2                                                                      

If x = 0, y = 2

x = -1, y = - 1

x = 1, y = 5

Plotting (i) and (ii) as per the respective values of x and y, we get two lines AB and PQ respectively that intersect at C (-1.-1).

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

(iii) Given pair of equations is:

x + y = 3 and 3x + 3y = 9

On comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 we have

a₁ = 1, b₁ = 1 and c₁, = - 3 and a₂ = 3, b₂ = 3 andc₂ = - 9                         

So, the given pair of lines is coincident. Therefore, these lines have. Hence, the given pair of linear equations has infinitely many solutions, i.e., consistent.

Now, x + y = 3             ….(i)

Or      y = 3 - x

If x = 0, y = 3

x = 3, y = 0

Also 3x + 3y = 9           ….(ii)

Or    3y = 9 - 3x 

Plotting (i) and (ii) for respective set of values for x and y, we get two lines AB and CD respectively, that are coincident.