Solve the following pair of linear equations by the substitution and cross-multiplication methods:
\(8x + 5y = 9\)
\(3x + 2y = 4\)

Ans.(x + y = 14\)...........(i)
=> \(x – y = 4\)..............(ii)

Using equation (ii)
=> \(x = 4 + y\)......................(iii)

On substituting (iii) in (i):
=>\(4 + 2y = 14\)
=>\(2y = 10\)
=>\(y = 5\)...........(iv)

On substituting (iv) in (i):
=>\(x + 5 = 14\)
=>\(x = 14 - 5\)
=>\(x = 9\)

So, the value of \(y = 5\) and \(x = 9\).
\(x - 3y – 7 = 0,3x - 3y – 15 = 0\)

On comparing quations with general form of equation \(ax^2 + bx + c\)
\(a_1 = 1, \)
\( b_1 = -3, \)
\( c_1 = -7, \)
\( a_2 = 3, \)
\(b_2 = -3,\)
\(c_2 = -15 \)

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.