How do you solve $\dfrac{{3x}}{4} + \dfrac{{5x}}{2} = 13$?

Hint: The above expression is the equation having fractions in it.
In order to solve the given equation we have to find the LCM (lowest common factor) of the given fraction and after that we will perform the basic mathematical operations such as addition, multiplication etc.
LCM is general is the common multiples contained in two or more given numbers.
Using the operation of LCM we will calculate the problem.

Complete step-by-step solution:
Let’s learn LCM first and then we will perform the calculations.
LCM is the lowest common factor, when two numbers are given they are expanded in their prime factors (prime numbers are the one which are divided by 1 and the number itself). We will learn LCM by an example;
4 and 6 are the two numbers, for which we have to find the LCM. First we will expand the two given numbers;
$ \Rightarrow 4 = 2 \times 2$
$ \Rightarrow 6 = 3 \times 2$
Now, we will take 2 as common to both and then multiply the two uncommon numbers with 2.
$ \Rightarrow 2 \times 2 \times 3 = 12$ (We got 12 as LCM)
Now, we will solve the given equation;
$ \Rightarrow \dfrac{{3x}}{4} + \dfrac{{5x}}{2} = 13$(We will take the LCM of 4 and 2)
$ \Rightarrow \dfrac{{3x + 10x}}{4} = 13$ (LCM comes out to be 4; we will divide the LCM by the denominator of the fraction and then multiply with the numerator)
$ \Rightarrow 13x = 13 \times 4$ (We have added the numerator and cross multiplied the denominator with the term at RHS)
$ \Rightarrow x = \dfrac{{13 \times 4}}{{13}}$ (We will cancel the common term)
$ \Rightarrow x = 4$ (Required value)

Therefore the value of x is equal to 4.

Note: The equation given to us in the question is a monomial equation which has the degree 1 of the variable. We use to make monomial equations in order to find the variable value such as in solving many problems of physics like projectile motion, in mathematics problems of ages, average, time and work, even graphs can use equations to be drawn like parabolic graph, exponential graph.