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**Hint:**Use the properties of finding the solutions by comparing with the standard equations. \[{a_1}x + {b_1}y + {c_1} = 0\;And\;{a_2}x + {b_2}x + {c_2} = 0\;\;\]There are conditions found on the basis of the constants. i.e.

The ratio of ${a_1}$ and ${a_2}$ is not equal to ${b_1}\,{\text{and }}{{\text{b}}_2}$ then it gives unique solution.

The ratio of ${a_1}$ and ${a_2}$ is equal to ${b_1}\,{\text{and }}{{\text{b}}_2}$ and not equal to ${c_1}\;and\;{{\text{c}}_2}$ then it gives no solution.

The ratio of ${a_1}$ and ${a_2}$ is equal to ${b_1}\,{\text{and }}{{\text{b}}_2}$ and is equal to ${c_1}\;and\;{{\text{c}}_2}$ then it gives infinite solution.

**Complete step by step answer:**

Now, take the given equations and compare with the standard equations, \[{a_1}x + {b_1}y + {c_1} = 0\;And\;{a_2}x + {b_2}x + {c_2} = 0\;\;\]

$

5x - 8y = 0 \\

{\text{3x - }}\dfrac{{24}}{5}y + \dfrac{3}{5} = 0 \\

$

Therefore,

$

{a_1} = 5 \\

{b_1} = - 8 \\

{c_1} = 1 \\

$

$

{a_2} = 3 \\

{b_2} = - \dfrac{{24}}{5} \\

{c_2} = \dfrac{3}{5} \\

$

Now, find the ratios of the terms, put the values

$

\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{5}{3}, \\

\dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{ - 8}}{{ - 24/5}} = \dfrac{5}{3} \\

\dfrac{{{c_1}}}{{{c_2}}} = \dfrac{1}{{3/5}} = \dfrac{5}{3} \\

\Rightarrow \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}} = \dfrac{5}{3} \\

$

All three ratios are of the same values, so the given set of equations has infinitely many solutions.

Therefore, the required solution –

Hence, from the given multiple choices, option A is the correct answer.

**Note:**

The same set of equations can be solved by taking the simplest form of the equations converting all the fractions and decimals in the form of natural numbers. Such as here twenty-four by five can be converted in a simple form by taking LCM (Least common multiple).

Additional Information: The solutions of the set of the linear equations can be done by two methods- (1) Algebraic Method and (2) Graphical Method. In the algebraic method, we further can use substitute methods or elimination methods.

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