Question: Which of the following pairs of linear equations has a unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by cross multiplication method?
2x + y = 5, 3x + 2y = 8
A.Unique solutions, x = 2 and y = 1
B.No solutions
C.Infinitely many solutions
D.Data insufficient

Hint:Before attempting this question, one must have prior knowledge of conditions under which the linear equation shows the solution such as unique solution, no solution and infinitely many solution and how to apply the cross multiplication method i.e. \[\dfrac{x}{{{b_1}{c_2} - {b_2}{c_1}}} = \dfrac{y}{{{c_1}{a_2} - {c_2}{a_1}}} = \dfrac{1}{{{a_1}{b_2} - {a_2}{b_1}}}\], utilize this information to solve the problem.

Complete step-by-step answer:
According to the given information we have pair of linear equations 2x + y = 5 and 3x + 2y = 8
Comparing the pair of given linear equation with general linear equations $ {a_1}x + {b_1}y + {c_1} = 0 $ and $ {a_2}x + {b_2}y + {c_2} = 0 $ to find the values of $ {a_1} $ , $ {b_1} $ , $ {c_1} $ , \[{a_2}\], \[{b_2}\] and \[{c_2}\]
We get $ {a_1} $ = 2, $ {b_1} $ = 1, $ {c_1} $ = - 5, \[{a_2}\] = 3, \[{b_2}\] = 2 and \[{c_2}\] = - 8
Let substitute the given values in the formula $ \dfrac{{{a_1}}}{{{a_2}}} $ , $ \dfrac{{{b_1}}}{{{b_2}}} $ and $ \dfrac{{{c_1}}}{{{c_2}}} $
\[\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{2}{3}\], $ \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{1}{2} $ and $ \dfrac{{{c_1}}}{{{c_2}}} = \dfrac{{ - 5}}{{ - 8}} = \dfrac{5}{8} $
Since after observing the values we get $ \dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}} $
Therefore the pair of linear equations shows a unique solution
Now using the method of cross multiplication to find the solution of our equation
According to the cross multiplication method i.e. \[\dfrac{x}{{{b_1}{c_2} - {b_2}{c_1}}} = \dfrac{y}{{{c_1}{a_2} - {c_2}{a_1}}} = \dfrac{1}{{{a_1}{b_2} - {a_2}{b_1}}}\]
Substituting the given values in the above formula
\[\dfrac{x}{{1 \times \left( { - 8} \right) - 2 \times \left( { - 5} \right)}} = \dfrac{y}{{\left( { - 5} \right) \times 3 - \left( { - 8} \right) \times 2}} = \dfrac{1}{{2 \times 2 - 3 \times 1}}\]
 $ \Rightarrow $ \[\dfrac{x}{{ - 8 + 10}} = \dfrac{y}{{ - 15 + 16}} = \dfrac{1}{{4 - 3}}\]
 $ \Rightarrow $ \[\dfrac{x}{2} = \dfrac{y}{1} = \dfrac{1}{1}\]

 $ \Rightarrow $ \[\dfrac{x}{2} = \dfrac{1}{1}\] and \[\dfrac{y}{1} = \dfrac{1}{1}\]
Therefore x = 2 and y = 1 is the solution of the given equation
Hence option A is the correct option

Note: In the above questions we came across the terms has unique solution, no solution, or infinitely many solutions let’s discuss about these terms when 2 lines coincide together shows an infinite many solution due to infinite sets of common values for lines which are intersect each other, shows a unique solution because of a common set of values but when 2 lines are parallel to each other due to no common set of values it shows no solution. In the system where equations of linear pairs show infinitely many solutions and unique solutions is called a consistent system whereas the system which consists of equations of linear pairs which shows infinitely many solutions is called inconsistent system.