Question

Solve for x: $\dfrac{{x + 2}}{6} - \left[ {\dfrac{{11 - x}}{3} - \dfrac{1}{4}} \right] = \dfrac{{3x - 14}}{{12}}$.

Answer 1

Hint: First solve the terms in the big bracket by taking LCM. After that simplify the remaining terms on the left side by taking LCM. Then, cancel the denominator of both sides as they are equal. After that move the variable part on the left side and the constant on the right side. Then simply it to get the value of x.

Complete step-by-step answer:
Given: - $\dfrac{{x + 2}}{6} - \left[ {\dfrac{{11 - x}}{3} - \dfrac{1}{4}} \right] = \dfrac{{3x - 14}}{{12}}$
Take LCM inside the bracket on the left side,
$ \Rightarrow \dfrac{{x + 2}}{6} - \left[ {\dfrac{{4\left( {11 - x} \right) - 3}}{{12}}} \right] = \dfrac{{3x - 14}}{{12}}$
Multiply the terms inside the bracket,
$ \Rightarrow \dfrac{{x + 2}}{6} - \left[ {\dfrac{{44 - 4x - 3}}{{12}}} \right] = \dfrac{{3x - 14}}{{12}}$
Subtract 3 from 44 inside the brackets,
$ \Rightarrow \dfrac{{x + 2}}{6} - \dfrac{{41 - 4x}}{{12}} = \dfrac{{3x - 14}}{{12}}$
Now, take LCM on the left side,
$ \Rightarrow \dfrac{{2\left( {x + 2} \right) - \left( {41 - 4x} \right)}}{{12}} = \dfrac{{3x - 14}}{{12}}$
Open the brackets and change the sign accordingly,
$ \Rightarrow \dfrac{{2x + 4 - 41 + 4x}}{{12}} = \dfrac{{3x - 14}}{{12}}$
Add or subtract the like terms,
$ \Rightarrow \dfrac{{6x - 37}}{{12}} - = \dfrac{{3x - 14}}{{12}}$
Cancel out the denominator of both sides,
$ \Rightarrow 6x - 37 = 3x - 14$
Move the variable part on the left side and the constant part on the right side,
$ \Rightarrow 6x - 3x = 37 - 14$
Subtract the terms on both sides,
$ \Rightarrow 3x = 23$
Divide both sides by 3,
$\therefore x = \dfrac{{23}}{3}$

Hence, the value of x is $\dfrac{{23}}{3}$.

Note: In such problems, we first need to rearrange the given equation and then simplify it and solve it for the value of x.
If it is linear i.e. the highest power of the variable is one then we can get the value directly by performing ordinary operations like addition, subtraction, multiplication, and division.
While if the equation turns out to be quadratic then we need to use a middle-term split or quadratic formula to solve for the value of the variable.