Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method.
(i) \(x – 3y – 3 = 0,3x – 9y – 2 = 0\)
(ii) \(2x + y = 5,3x + 2y = 8\)
(ii) \(3x - 5y = 20,6x - 10y = 40\)
(ii) \(x - 3y – 7 = 0,3x - 3y – 15 = 0\)

Ans.(i) \(x – 3y – 3 = 0,3x – 9y – 2 = 0\)

On comparing these equations with the general form : \(ax^2 + bx + c\)

\(a_1 = 1, b_1 = -3, c_1 = -3,a_2 = 3, b_2 = -9, c_2 = -2 \)

Here \({{a_1}\over {a_2}} = {{b_1}\over {b_2}} \ne {{c_1}\over {c_2}}\)
=>\({{1}\over {3}} = {{-3}\over {-9}} \ne {{-3}\over {-2}}\)

So, these lines are parallel to each other.

Thus, the following pairs of linear equations will have no solution.

(ii) \(2x + y = 5,3x + 2y = 8\)

On comparing these equations with the general form : \(ax^2 + bx + c\)
\(a_1 = 2, b_1 = 1, c_1 = -5,a_2 = 3, b_2 = 2, c_2 = 8 \)

Here \({{a_1}\over {a_2}} \ne {{b_1}\over {b_2}} \)
=>\({{2}\over {3}} \ne {{1}\over {2}} \)

So, these lines have a unique solution.

We can find the unique solution using cross multiplication:

=>\({{x}\over {(-8)(-1)-(2)(-5)}} = {{y}\over {(-5)(3)-(-8)(2)}} = {{1}\over {(2)(2)-(3)(1)}}\)
=>\({{x}\over {-8 + 10}} = {{y}\over {-15 + 16}} = {{1}\over {4 - 3}}\)
=>\({{x}\over {2}} = {{y}\over {1}} = {{1}\over {1}}\)
=> x = 2 and y = 1.

(iii) \(3x - 5y = 20,6x - 10y = 40\)

On comparing these equations with the general form : \(ax^2 + bx + c\)

\(a_1 = 3, b_1 = -5, c_1 = -20,a_2 = 6, b_2 = -10, c_2 = -40 \)

Here \({{a_1}\over {a_2}} = {{b_1}\over {b_2}} = {{c_1}\over {c_2}}\)
=>\({{3}\over {6}} = {{-5}\over {-10}} = {{-20}\over {-40}}\)

So, these lines coincide with each other.

Hence, there are infinite many solutions.

(iv) \(x - 3y – 7 = 0,3x - 3y – 15 = 0\)

On comparing these equations with the general form : \(ax^2 + bx + c\)

\(a_1 = 1, b_1 = -3, c_1 = -7,a_2 = 3, b_2 = -3, c_2 = -15 \)

Here \({{a_1}\over {a_2}} \ne {{b_1}\over {b_2}} \)
=>\({{1}\over {3}} \ne {{-3}\over {-3}} \)

So, these lines have a unique solution.

We can find the unique solution using cross multiplication:

=>\({{x}\over {(-3)(-15)-(-3)(-7)}} = {{y}\over {(-7)(3)-(-15)(1)}} = {{1}\over {(-3)(1)-(-3)(3)}}\)
=>\({{x}\over {45 - 21}} = {{y}\over {-21 + 15}} = {{1}\over {-3 + 9}}\)
=>\({{x}\over {24}} = {{y}\over {-6}} = {{1}\over {6}}\)

=> x = 4 and y = -1.