Question: Solve and write the answer in interval notation:
$12x + 4 < 5x + 25$ AND $4 + 6x \leq 15x - 5$
1. **State the problem:**
Solve the system of inequalities:
$$12x + 4 < 5x + 25$$
AND
$$4 + 6x \leq 15x - 5$$
and express the solution in interval notation.
2. **Solve the first inequality:**
$$12x + 4 < 5x + 25$$
Subtract $5x$ from both sides:
$$12x - 5x + 4 < 25$$
Simplify:
$$7x + 4 < 25$$
Subtract $4$ from both sides:
$$7x + \cancel{4} - \cancel{4} < 25 - 4$$
$$7x < 21$$
Divide both sides by $7$ (positive number, inequality direction stays the same):
$$\frac{7x}{\cancel{7}} < \frac{21}{\cancel{7}}$$
$$x < 3$$
3. **Solve the second inequality:**
$$4 + 6x \leq 15x - 5$$
Subtract $6x$ from both sides:
$$4 + \cancel{6x} - \cancel{6x} \leq 15x - 6x - 5$$
$$4 \leq 9x - 5$$
Add $5$ to both sides:
$$4 + 5 \leq 9x - 5 + 5$$
$$9 \leq 9x$$
Divide both sides by $9$ (positive number, inequality direction stays the same):
$$\frac{9}{\cancel{9}} \leq \frac{9x}{\cancel{9}}$$
$$1 \leq x$$
or equivalently
$$x \geq 1$$
4. **Combine the two inequalities:**
$$x < 3$$
AND
$$x \geq 1$$
This means $x$ is between $1$ and $3$, including $1$ but not including $3$.
5. **Write the solution in interval notation:**
$$[1, 3)$$
**Final answer:**
$$[1, 3)$$