Find all $x$ such that $1-3x < x + 8 \le 7-x$. Give your answer in interval notation.
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 $\boxed{\left( -\frac{7}{4}, -\frac{1}{2} \right]}$.
To find the values of $x$ that satisfy the given inequalities $1-3x < x + 8 \le 7-x$, we will break it down into two separate inequalities and solve each one.
First, let's solve $1-3x < x + 8$. We can simplify this inequality by moving all the terms involving $x$ to one side:
$1 - 3x - x < 8$
$1 - 4x < 8$
Next, let's solve $x + 8 \le 7-x$:
$x + 8 - 8 \le 7 - x - 8$
$x \le -1 - x$
Now, let's solve each inequality separately:
$1 - 4x < 8$
Subtracting $1$ from both sides gives:
$-4x < 7$
Dividing both sides by $-4$, remembering to flip the inequality sign since we are dividing by a negative number:
$x > -\frac{7}{4}$
Now, let's solve $x \le -1 - x$:
Adding $x$ to both sides gives:
$2x \le -1$
Dividing both sides by $2$ gives:
$x \le -\frac{1}{2}$
Putting the solutions together, we have:
$x > -\frac{7}{4}$ and $x \le -\frac{1}{2}$
The solution in interval notation is:
$\boxed{\left( -\frac{7}{4}, -\frac{1}{2} \right]}$
To find the values of $x$ that satisfy the inequalities $1-3x < x + 8$ and $x + 8 \le 7-x$, we can solve each inequality separately and then find the intersection of the solution sets.
Let's start with the first inequality, $1-3x < x + 8$:
1. Subtract $x$ from both sides: $1-4x < 8$.
2. Subtract $1$ from both sides: $-4x < 7$.
3. Divide both sides by $-4$ (remember to reverse the inequality sign when dividing by a negative number): $x > - \frac{7}{4}$.
Now, let's move on to the second inequality, $x + 8 \le 7 - x$:
1. Add $x$ to both sides: $2x + 8 \le 7$.
2. Subtract $8$ from both sides: $2x \le -1$.
3. Divide both sides by $2$: $x \le - \frac{1}{2}$.
So, we have found the two solution sets:
For the first inequality: $x > - \frac{7}{4}$
For the second inequality: $x \le - \frac{1}{2}$
To find the intersection of these solution sets, we need to find the values of $x$ that satisfy both inequalities. Thus, the values of $x$ that satisfy both inequalities are those that are greater than $- \frac{7}{4}$ and less than or equal to $- \frac{1}{2}$.
In interval notation, we can express the solution as $\left(-\frac{7}{4}, -\frac{1}{2} \right]$.