Show that the following system of equations has a unique solution:\(\f

1.	Show that the following system of equations has a unique solution:\(\frac{x}3+\frac{y}2=3\),x – 2y = 2. Also, find the solution of the given system of equations.
Answer» The given system of equations is: x/3 + y/2 = 3 ⇒ 2x+3y/6 = 3 2x + 3y = 18 ⇒ 2x + 3y – 18 = 0 ….(i) and x – 2y = 2 x – 2y – 2 = 0 …..(ii) These equations are of the forms: a₁x+b₁y+c₁ = 0 and a₂x+b₂y+c₂ = 0 where, a₁ = 2, b₁= 3, c₁ = -18 and a₂ = 1, b₂ = -2, c₂ = -2 For a unique solution, we must have: a₁/a₂ ≠ b₁/b₂ , i.e., 2/1 ≠ 3/−2 Hence, the given system of equations has a unique solution. Again, the given equations are: 2x + 3y – 18 = 0 …..(iii) x – 2y – 2 = 0 …..(iv) On multiplying (i) by 2 and (ii) by 3, we get: 4x + 6y – 36 = 0 …….(v) 3x - 6y – 6 = 0 ……(vi) On adding (v) from (vi), we get: 7x = 42 ⇒x = 6 On substituting x = 6 in (iii), we get: 2(6) + 3y = 18 ⇒3y = (18 - 12) = 6 ⇒y = 2 Hence, x = 6 and y = 2 is the required solution.

Show that the following system of equations has a unique solution:\(\frac{x}3+\frac{y}2=3\),x – 2y = 2. Also, find the solution of the given system of equations.

Answer»

The given system of equations is:

x/3 + y/2 = 3

⇒ 2x+3y/6 = 3

2x + 3y = 18

⇒ 2x + 3y – 18 = 0 ….(i)

and

x – 2y = 2

x – 2y – 2 = 0 …..(ii)

These equations are of the forms:

a₁x+b₁y+c₁ = 0 and a₂x+b₂y+c₂ = 0

where, a₁ = 2, b₁= 3, c₁ = -18 and a₂ = 1, b₂ = -2, c₂ = -2

For a unique solution, we must have:

a₁/a₂ ≠ b₁/b₂ , i.e., 2/1 ≠ 3/−2

Hence, the given system of equations has a unique solution.

Again, the given equations are:

2x + 3y – 18 = 0 …..(iii)

x – 2y – 2 = 0 …..(iv)

On multiplying (i) by 2 and (ii) by 3, we get:

4x + 6y – 36 = 0 …….(v)

3x - 6y – 6 = 0 ……(vi)

On adding (v) from (vi), we get:

7x = 42

⇒x = 6

On substituting x = 6 in (iii), we get:

2(6) + 3y = 18

⇒3y = (18 - 12) = 6

⇒y = 2

Hence, x = 6 and y = 2 is the required solution.