Find all $x$ such that $1-3x < x + 8 \le 7-x$. Give your answer in interval notation.

Answer 1

We start by solving $1-3x < x + 8$. Adding $-x$ to both sides gives us $1-4x < 8$. Subtracting 1 from both sides gives $-4x < 7$, and dividing by $-4$ (and flipping the inequality sign) gives $x > -\frac{7}{4}$. Now we solve the second inequality, $x + 8 \le 7-x$. Adding $x$ to both sides gives us $2x + 8 \le 7$. Subtracting 8 from both sides gives $2x \le -1$, and dividing by 2 gives $x\le -\frac{1}{2}$. Therefore, the solutions to the given inequalities are $x > -\frac{7}{4}$ and $x \le -\frac{1}{2}$. Combining these gives us $\boxed{\left( -\frac{7}{4}, -\frac{1}{2} \right]}$.