Question: 4x-3 divided by x+3 plus 1 equals 3x+1 divided x+3
1. **State the problem:** Solve the equation $$\frac{4x-3}{x+3} + 1 = \frac{3x+1}{x+3}$$.
2. **Rewrite the equation:** To work with a common denominator, express 1 as $$\frac{x+3}{x+3}$$:
$$\frac{4x-3}{x+3} + \frac{x+3}{x+3} = \frac{3x+1}{x+3}$$
3. **Combine the left side fractions:**
$$\frac{4x-3 + x + 3}{x+3} = \frac{3x+1}{x+3}$$
Simplify numerator:
$$\frac{(4x + x) + (-3 + 3)}{x+3} = \frac{5x}{x+3}$$
4. **Rewrite the equation:**
$$\frac{5x}{x+3} = \frac{3x+1}{x+3}$$
5. **Since denominators are equal and not zero (note: $x \neq -3$), set numerators equal:**
$$5x = 3x + 1$$
6. **Solve for $x$:**
$$5x - 3x = 1$$
$$2x = 1$$
$$x = \frac{1}{2}$$
7. **Check for restrictions:**
$x \neq -3$ to avoid division by zero, and $x=\frac{1}{2}$ is valid.
**Final answer:**
$$x = \frac{1}{2}$$