Answer
The solution of the inequality $\frac{x+3}{4}\ge \frac{x-2}{3}+1$ in set builder notation is $\left\{ x\left| x\le 5 \right. \right\}$ and in interval notation is $\left( -\infty ,5 \right]$.
Work Step by Step
Consider the inequality, $\frac{x+3}{4}\ge \frac{x-2}{3}+1$
Multiply both sides by the least common denominator of $12$.
$\begin{align}
& 12\left( \frac{x+3}{4} \right)\ge 12\left( \frac{x-2}{3}+1 \right) \\
& 3\left( x+3 \right)\ge 4\left( x-2 \right)+12
\end{align}$
Use the distributive property.
$\begin{align}
& 3x+9\ge 4x-8+12 \\
& 3x+9\ge 4x+4
\end{align}$
Subtract $3x$ and $4$ from both sides.
$\begin{align}
& 3x+9-3x-4\ge 4x+4-3x-4 \\
& 5\ge x
\end{align}$
Therefore, the solution of the inequality $\frac{x+3}{4}\ge \frac{x-2}{3}+1$ in set builder notation is $\left\{ x\left| x\le 5 \right. \right\}$ and in interval notation it is $\left( -\infty ,5 \right]$.
The number line provided below shows the solution $\left\{ x\left| x\le 5 \right. \right\}$ of the inequality $\frac{x+3}{4}\ge \frac{x-2}{3}+1$.
