Find the range of values of x for which\(\frac{x+2}{4}-\frac{x+1}{3}>\frac{1}{2}\)
Answer Details
We can start solving the inequality by first finding a common denominator for the fractions:
\begin{align*}
\frac{x+2}{4}-\frac{x+1}{3}&>\frac{1}{2}\\
\frac{3(x+2)}{12}-\frac{4(x+1)}{12}&>\frac{6}{12}\\
\frac{3x+6-4x-4}{12}&>\frac{1}{2}\\
-\frac{x-2}{12}&>\frac{1}{2}\\
\end{align*}
Multiplying both sides by $-1$ changes the direction of the inequality:
\begin{align*}
\frac{x-2}{12}&<-\frac{1}{2}\\
\end{align*}
Multiplying both sides by $12$ gives:
\begin{align*}
x-2&<-6\\
x&< -4\\
\end{align*}
Therefore, the answer is:
- x < -4