Chapter 3: Pair of Linear Equations in Two Variables

The Condition for no solution = a₁/a₂ = b₁/b₂ ≠ c₁/c₂ (parallel lines)

(i) Yes.

Given pair of equations are,

2x+4y – 3 = 0 and 6x + 12y – 6 = 0

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

We get,

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

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

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

(ii) No.

Given pair of equations,

x = 2y or x – 2y = 0

y = 2x or 2x – y = 0;

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

We get,

a₁ = 1, b₁ = – 2, c₁ = 0;

a₂ = 2, b₂ = – 1, c₂ = 0;

a₁ /a₂ = ½

b₁ /b₂ = -2/-1 = 2

Here, a₁/a₂ ≠ b₁/b_2.

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

(iii) No.

Given pair of equations,

3x + y – 3 = 0

2x + 2/3 y = 2

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

We get,

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

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

Condition for coincident lines,

a₁/a₂ = b₁/b₂ = c₁/c₂;

(i) No.

Given pair of linear equations are:

3x + 1/7y = 3

7x + 3y = 7

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

Here, a₁ = 3, b₁ = 1/7, c₁ = – 3;

And a₂ = 7, b₂ = 3, c₂ = – 7;

a₁ /a₂ = 3/7

b₁ /b₂ = 1/21

c₁ /c₂ = – 3/ – 7 = 3/7

Here, a₁/a₂ ≠ b₁/b_2.

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

(ii) Yes.

Given pair of linear equations.

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

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

Here, 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

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

Hence, the given pair of linear equations is coincident.

(iii) No.

Given pair of linear equations are

x/2 + y + 2/5 = 0

4x + 8y + 5/16 = 0

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

Here, a₁ = ½, b₁ = 1, c₁ = 2/5;

And a₂ = 4, b₂ = 8, c₂ = 5/16;

a₁ /a₂ = 1/8

b₁ /b₂ = 1/8

c₁ /c₂ = 32/25

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

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

Conditions for pair of linear equations to be consistent are:

a₁/a₂ ≠ b₁/b_2. [unique solution]

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

(i) No.

The given pair of linear equations

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

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

We get,

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

Here, a₁/a₂ = b₁/b₂ ≠ c₁/c₂

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

(ii) Yes.

The given pair of linear equations

(3/5)x – y = ½

(1/5)x – 3y= 1/6

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

We get,

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

Here, a₁/a₂ ≠ b₁/b_2.

Hence, the given pair of linear equations has a unique solution, i.e., consistent.

(iii) Yes.

The given pair of linear equations –

2ax + by –a = 0 and 4ax + 2by – 2a = 0

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

We get,

a₁ = 2a, b₁ = b, c₁ = – a;

a₂ = 4a, b₂ = 2b, c₂ = – 2a;

a₁ /a₂ = ½

b₁ /b₂ = ½

c₁ /c₂ = ½

Here, a₁/a₂ = b₁/b₂ = c₁/c₂

Hence, the given pair of linear equations has infinitely many solutions, i.e., consistent

(iv) No.

The given pair of linear equations

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

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

We get,

a₁ = 1, b₁ = 3, c₁ = 11

a₂ = 2, b₂ = 6, c₂ = 11

a₁ /a₂ = ½

b₁ /b₂ = ½

c₁ /c₂ = 1

Here, a₁/a₂ = b₁/b₂ ≠ c₁/c₂.

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