### Instruction: Solve  \[\frac{x^2+2x+3}{x+4}=x+5\]for $x$.### Response: Let's multiply both sides by $x+4$ to get rid of the fraction.
That gives $(x^2+2x+3)=(x+5)(x+4)$.
Expanding the right side gives $x^2+2x+3=x^2+9x+20$.
Subtracting $x^2$ from both sides gives $2x+3=9x+20$.
Subtracting $2x$ from both sides gives $3=7x+20$.
Finally subtracting 20 from both sides gives $-17=7x$.