How do solve \[\left| {\dfrac{{8y}}{3} - \dfrac{{2y}}{4}} \right| < - 13\]?

Hint: Here the equation is an algebraic equation that is a combination of constant and variables. we have to solve the given equation for variable x. Since the equation involves the modulus, by using the definition of modulus or absolute value and simple arithmetic operation we determine the value of y.

Complete step-by-step answer:
The absolute value or modulus of a real function f(x), it is denoted as |f(x)|, is the non-negative value of f(x) without considering its sign.
Now consider the given question \[\left| {\dfrac{{8y}}{3} - \dfrac{{2y}}{4}} \right| < - 13\]
By the definition of absolute number, we are determined the unknown variable and
By definition the modulus, separate \[\left| {\dfrac{{8y}}{3} - \dfrac{{2y}}{4}} \right| < - 13\] into two equations:
\[\dfrac{{8y}}{3} - \dfrac{{2y}}{4} < - 13\] (1)
 and
\[ - \left( {\dfrac{{8y}}{3} - \dfrac{{2y}}{4}} \right) < - 13\] (2)
Consider the equation (1)
\[ \Rightarrow \dfrac{{8y}}{3} - \dfrac{{2y}}{4} < - 13\]
Take LCM for 3 and 4. The LCM for 3 and 4 is 12.
\[ \Rightarrow \dfrac{{\dfrac{{8y}}{3} \times 12 - \dfrac{{2y}}{4} \times 12}}{{12}} < - 13\]
On simplification, we get
\[ \Rightarrow \dfrac{{32y - 6y}}{{12}} < - 13\]
\[ \Rightarrow \dfrac{{26y}}{{12}} < - 13\]
\[ \Rightarrow \dfrac{{13y}}{6} < - 13\]
Cancelling the 13 on both sides
\[ \Rightarrow \dfrac{y}{6} < - 1\]
Multiplying 6 on both sides
\[ \Rightarrow y < - 6\]
Now consider the equation (2)
\[ - \left( {\dfrac{{8y}}{3} - \dfrac{{2y}}{4}} \right) < - 13\]
First multiply the -ve sign inside to the parenthesis on LHS.
\[\left( { - \dfrac{{8y}}{3} + \dfrac{{2y}}{4}} \right) < - 13\]
Take LCM for 3 and 4. The LCM for 3 and 4 is 12.
\[ \Rightarrow \dfrac{{ - \dfrac{{8y}}{3} \times 12 + \dfrac{{2y}}{4} \times 12}}{{12}} < - 13\]
On simplification, we get
\[ \Rightarrow \dfrac{{ - 32y + 6y}}{{12}} < - 13\]
\[ \Rightarrow \dfrac{{ - 26y}}{{12}} < - 13\]
\[ \Rightarrow \dfrac{{ - 13y}}{6} < - 13\]
Cancelling the 13 on both sides
\[ \Rightarrow \dfrac{{ - y}}{6} < - 1\]
Multiplying -6 on both sides
\[ \Rightarrow y > 6\]
Hence the value of y is determined.

Note: The algebraic equation or an expression is a combination of variables and constants, it also contains the coefficient. The alphabets are known as variables. The x, y, z etc., are called as variables. The numerals are known as constants. The numeral of a variable is known as co-efficient. we must know about the modulus definition.