Solve the following linear equations using cross multiplication method:
$5x - 3y = 2$ and $4x + 7y = - 3$

Hint: The standard form of two linear equations in two variables is
$
  {a_1}x + {b_1}y + {c_1} = 0 \\
  {a_2}x + {b_2}y + {c_2} = 0 \\
 $
Where ${a_1},{b_1},{a_2}$and ${b_2}$ belong to the set of real numbers.
To solve the above set of linear equations using cross multiplication method, we have to find of $x$ and $y$ variables using the following formulae:
$\dfrac{x}{{{b_1}{c_2} - {b_2}{c_1}}} = \dfrac{y}{{{c_1}{a_2} - {c_2}{a_1}}} = \dfrac{1}{{{a_1}{b_2} - {a_2}{b_1}}}$
Where ${a_1}{b_2} - {a_2}{b_1} \ne 0$.

Complete step-by-step solution:
The linear equations in two variables are given,
$5x - 3y = 2$
$4x + 7y = - 3$
Convert the above linear equation in the standard form of the linear equations by taking all the terms on one side of the equality.
$5x - 3y - 2 = 0$
$4x + 7y + 3 = 0$
Now comparing the above equations with the standard form of the equations of the form ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$, we get the following values of the coefficients-
${a_1} = 5$, ${b_1} = - 3$, ${c_1} = - 2$, ${a_2} = 4$, ${b_2} = 7$, ${c_2} = 3$
Using the formulae for cross multiplication method
$\dfrac{x}{{{b_1}{c_2} - {b_2}{c_1}}} = \dfrac{y}{{{c_1}{a_2} - {c_2}{a_1}}} = \dfrac{1}{{{a_1}{b_2} - {a_2}{b_1}}}$
We have to substitute the values of the coefficients in the above formulae:
$\dfrac{x}{{( - 3) \times 3 - 7 \times ( - 2)}} = \dfrac{y}{{( - 2) \times 4 - 3 \times 5}} = \dfrac{1}{{5 \times 7 - 4 \times ( - 3)}}$
Further simplify the above equality
$\dfrac{x}{{( - 9) - ( - 14)}} = \dfrac{y}{{( - 8) - 15}} = \dfrac{1}{{35 - ( - 12)}}$
$\dfrac{x}{{ - 9 + 14}} = \dfrac{y}{{ - 8 - 15}} = \dfrac{1}{{35 + 12}}$
$\dfrac{x}{5} = \dfrac{y}{{ - 23}} = \dfrac{1}{{47}}$
Since the denominator is not $0$, so a unique solution exist.
Now separating the above into two equalities,
$\dfrac{x}{5} = \dfrac{1}{{47}}$ and \[\dfrac{y}{{ - 23}} = \dfrac{1}{{47}}\]
Cross multiply the above fractions,
$47x = 5$ and \[47y = - 23\]
Now, solve for $x$ and $y$, to get the solution
$x = \dfrac{5}{{47}}$ and $y = \dfrac{{ - 23}}{{47}}$
Convert it into decimal from
$x \approx 0.10638$ and $y \approx - 0.48936$

Note: Cross multiplication method for solving the linear equations can only be used when we have a pair of linear equations in two variables. That is for standard equations ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$, where ${a_1},{b_1},{a_2}$and ${b_2}$ belong to the set of real numbers ${a_1}{b_2} - {a_2}{b_1}$ should not be equal to $0$.
Also, if the value of ${b_1}{c_2} - {b_2}{c_1}$ is $0$ then $x$ has infinitely many solutions and if the value of ${c_1}{a_2} - {c_2}{a_1}$ is $0$ then $y$ has infinitely many solutions.