Sure, let's go through the steps to solve the equation:
[tex]\[
\frac{1}{3}(2x + 1) - \frac{1}{4}(2x + 1) = \frac{1}{12}(2 - x)
\][/tex]
Step 1: Simplify the terms within the parentheses.
[tex]\[
\frac{1}{3}(2x + 1) \quad \text{and} \quad \frac{1}{4}(2x + 1)
\][/tex]
Step 2: Multiply the terms outside the parentheses.
[tex]\[
\frac{1}{3}(2x + 1) = \frac{2x + 1}{3}
\][/tex]
[tex]\[
\frac{1}{4}(2x + 1) = \frac{2x + 1}{4}
\][/tex]
Step 3: Rewrite the equation using the simplified terms.
[tex]\[
\frac{2x + 1}{3} - \frac{2x + 1}{4} = \frac{2 - x}{12}
\][/tex]
Step 4: Find a common denominator for the fractions on the left side. The common denominator for 3 and 4 is 12.
[tex]\[
\frac{8(2x + 1)}{24} - \frac{6(2x + 1)}{24} = \frac{2 - x}{12}
\][/tex]
Step 5: Simplify the fractions on the left side.
[tex]\[
\frac{8(2x + 1) - 6(2x + 1)}{24} = \frac{2 - x}{12}
\][/tex]
[tex]\[
\frac{2(2x + 1)}{24} = \frac{2 - x}{12}
\][/tex]
[tex]\[
\frac{(2x + 1)}{12} = \frac{2 - x}{12}
\][/tex]
Step 6: Since the denominators are equal, we can equate the numerators.
[tex]\[
2x + 1 = 2 - x
\][/tex]
Step 7: Solve for [tex]\( x \)[/tex].
[tex]\[
2x + x = 2 - 1
\][/tex]
[tex]\[
3x = 1
\][/tex]
[tex]\[
x = \frac{1}{3}
\][/tex]
Thus, the solution to the equation [tex]\(\frac{1}{3}(2 x+1)-\frac{1}{4}(2 x+1)=\frac{1}{12}(2-x)\)[/tex] is:
[tex]\[
x = \frac{1}{3}
\][/tex]
