Chapter 3: Pair of Linear Equations in Two Variables

The given pair of linear equations is

λx + y = λ² and x + λy = 1

a₁ = λ, b₁= 1, c₁ = – λ²

a₂ =1, b₂=λ, c₂=-1

The given equations are;

λ x + y – λ² = 0

x + λ y – 1 = 0

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

We get,

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) For no solution,

a₁/a₂ = b₁/b₂≠ c₁/c₂

i.e. λ = 1/ λ ≠ λ²

so, λ ² = 1;

and λ ² ≠ λ

Here, we take only λ = – 1,

Since the system of linear equations has no solution.

(ii) For infinitely many solutions,

a₁/a₂ = b₁/b₂ = c₁/c₂

i.e. λ = 1/ λ  = λ²

so λ = 1/ λ gives λ = + 1;

λ = λ ² gives λ = 1,0;

Hence satisfying both the equations

λ = 1 is the answer.

(iii) For a unique solution,

a₁/a₂ ≠ b₁/b₂

so λ ≠1/ λ

hence, λ² ≠ 1;

λ ≠ + 1;

So, all real values of λ except +1.

The given pair of linear equations is

kx + 3y = k – 3 …(i)

12x + ky = k …(ii)

On comparing the equations (i) and (ii) with ax + by = c = 0,

We get,

a₁ = k, b₁ = 3, c₁ = -(k – 3)

a₂ = 12, b₂ = k, c₂ = – k

Then,

a₁ /a₂ = k/12

b₁ /b₂ = 3/k

c₁ /c₂ = (k-3)/k

For no solution of the pair of linear equations,

a₁/a₂ = b₁/b₂≠ c₁/c₂

k/12 = 3/k ≠ (k-3)/k

Taking the first two parts, we get

k/12 = 3/k

k² = 36

k = + 6

Taking the last two parts, we get

3/k ≠ (k-3)/k

3k ≠ k(k – 3)

k² – 6k ≠ 0

so, k ≠ 0,6

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

The given pair of linear equations are:

x + 2y = 1 …(i)

(a-b)x + (a + b)y = a + b – 2 …(ii)

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

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 of the pair of linear equations,

a₁/a₂ = b₁/b₂=c₁/c₂(coincident lines)

so, 1/(a-b) = 2/ (a+b) = 1/(a+b-2)

Taking the first two parts,

1/(a-b) = 2/ (a+b)

a + b = 2(a – b)

a = 3b …(iii)

Taking the last two parts,

2/ (a+b) = 1/(a+b-2)

2(a + b – 2) = (a + b)

a + b = 4 …(iv)

Now, put the value of a from Eq. (iii) in Eq. (iv), and we get

3b + b = 4

4b = 4

b = 1

Put the value of b in Eq. (iii), and we get

a = 3

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

Given pair of linear equations is

3x – y – 5 = 0 …(i)

6x – 2y – p = 0 …(ii)

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

We get,

a₁ = 3, b₁ = – 1, c₁ = – 5;

a₂ = 6, b₂ = – 2, c₂ = – p;

a₁ /a₂ = 3/6 = ½

b₁ /b₂ = ½

c₁ /c₂ = 5/p

Since the lines represented by these equations are parallel, then

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Taking the last two parts, we get ½ ≠ 5/p

So, p ≠ 10

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

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

Given pair of linear equations is

– x + py = 1 …(i)

px – y – 1 = 0 …(ii)

On comparing with ax + by + c = 0,

We get,

a₁ = -1, b₁ = p, c₁ =- 1;

a₂ = p, b₂ = – 1, c₂ =- 1;

a₁ /a₂ = -1/p

b₁ /b₂ = – p

c₁ /c₂ = 1

Since the lines equations have no solution, i.e., both lines are parallel to each other,

a₁/a₂ = b₁/b₂≠ c₁/c₂

-1/p = – p ≠ 1

Taking the last two parts, we get

p ≠ -1

Taking the first two parts, we get

p² = 1

p = + 1

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

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

Given, pair of linear equations is

– 3x + 5y = 7

2px – 3y = 1

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

Here, a₁ = -3, b₁ = 5, c₁ = – 7;

And a₂ = 2p, b₂ = – 3, c₂ = – 1;

a₁ /a₂ = -3/ 2p

b₁ /b₂ = – 5/3

c₁ /c₂ = 7

Since the lines intersect at a unique point, i.e., it has a unique solution

a₁/a₂ ≠ b₁/b₂

-3/2p ≠ -5/3

p ≠ 9/10

Hence, the lines represented by these equations intersect at a unique point for all real values of p except 9/10.

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

Given, pair of linear equations is

2x + 3y – 5 = 0

px – 6y – 8 = 0

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

Here, 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

Since the pair of linear equations has a unique solution,

a₁/a₂ ≠ b₁/b₂

so 2/p ≠ – ½

p ≠ – 4

Hence, 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 has infinitely many solutions.

Given pair of linear equations is

2x + 3y = 7

2px + py = 28 – qy

or 2px + (p + q)y – 28 = 0

On comparing with ax + by + c = 0,

We get,

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

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

a₁/a₂ = b₁/b₂ = c₁/c₂

1/p = 3/(p+q) = ¼

Taking the first and third parts, we get

p = 4

Again, taking the last two parts, we get

3/(p+q) = ¼

p + q = 12

Since p = 4

So, q = 8

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

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

Given linear equations are

x – 3y – 2 = 0 …(i)

-2x + 6y – 5 = 0 …(ii)

On comparing with ax + by c=0,

We get

a₁ =1, b₁ =-3, c₁ =- 2;

a₂ = -2, b₂ =6, c₂ =- 5;

a₁/a₂ = – ½

b₁/b₂ = – 3/6 = – ½

c₁/c₂ = 2/5

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

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

Condition for the pair of system to have a unique solution

a₁/a₂ ≠ b₁/b₂

Let the equations be,

a₁x + b₁y + c₁ = 0

a₂x + b₂y + c₂ = 0

Since, x = – 1 and y = 3 is the unique solution of these two equations, then

It must satisfy the equations –

a₁(-1) + b₁(3) + c₁ = 0

– a₁ + 3b₁ + c₁ = 0 …(i)

and a₂(- 1) + b₂(3) + c₂ = 0

– a₂ + 3b₂ + c₂ = 0 …(ii)

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

Hence, infinitely many pairs of linear equations are possible.

Given equations are

2x + y = 23 …(i)

4x – y = 19 …(ii)

On adding both equations, we get

6x = 42

So, x = 7

Put the value of x in Eq. (i), and we get

2(7) + y = 23

y = 23 – 14

so, y = 9

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

y/x – 2 = 9/7 -2 = -5/7

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

Using the property of a rectangle,

We know that,

Lengths are equal,

i.e., CD = AB

Hence, x + 3y = 13 …(i)

Breadth are equal,

i.e., AD = BC

Hence, 3x + y = 7 …(ii)

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

We get,

8x = 8

So, x = 1

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

We get,

y = 4

Therefore, the required values of x and y are 1 and 4, respectively.