How do you solve the inequality \[-\dfrac{1}{6}\le 4x-4<\dfrac{1}{3}\]?

Hint: In order to find the solution of the given question that is to find how to solve \[-\dfrac{1}{6}\le 4x-4< \dfrac{1}{3}\] and find the range of \[x\], apply the concepts of addition, division and multiplication to simplify the expression to get the range of variable \[x\] that is to find the value of the variable \[x\] is greater than equal to which term and less than to which term.

Complete step by step solution:
According to the question, given equation in the question is as follows:
\[-\dfrac{1}{6}\le 4x-4<\dfrac{1}{3}\]
To solve the above equation, add the term \[4\] to all the whole inequality, we will have:
\[\Rightarrow -\dfrac{1}{6}+4\le 4x-4+4<\dfrac{1}{3}+4\]
Then simplify the terms of the above equation by using addition and taking LCM, we will have:
\[\Rightarrow \dfrac{24-1}{6}\le 4x<\dfrac{1+12}{3}\]
After simplifying the above equation by solving the terms on the numerator of the equation, we will have:
\[\Rightarrow \dfrac{23}{6}\le 4x<\dfrac{13}{3}\]
Now divide \[4\] to the whole equation, we will have:
\[\Rightarrow \dfrac{23}{4\times 6}\le \dfrac{4x}{4}<\dfrac{13}{3\times 4}\]
Simplifying it further, we will get:
\[\Rightarrow \dfrac{23}{4\times 6}\le x<\dfrac{13}{3\times 4}\]
After simplifying the above equation by solving the terms on the denominator of the equation with the help of multiplication, we will have:
\[\Rightarrow \dfrac{23}{24}\le x<\dfrac{13}{12}\]
Therefore, after solving the inequality \[-\dfrac{1}{6}\le 4x-4<\dfrac{1}{3}\], the range of the variable \[x\] is \[\dfrac{23}{24}\le x<\dfrac{13}{12}\].

Note: Students make mistakes in calculations while simplifying the expressions with inequality and sometimes end changing the sign of the inequality like in the given question some student might miswrite the inequality as this \[-\dfrac{1}{6}\le 4x-4\le \dfrac{1}{3}\] instead of writing the actual given inequality which is \[-\dfrac{1}{6}\le 4x-4<\dfrac{1}{3}\]. It’s important to cross check the answer again once solved to avoid such miscalculations in this type of question.