Question: Solve the following system of equations algebraically:
y= x squared plus 15 x plus 35
y= 2 x minus 5
1. **State the problem:**
We need to solve the system of equations:
$$y = x^2 + 15x + 35$$
$$y = 2x - 5$$
2. **Set the equations equal to each other:**
Since both expressions equal $y$, set them equal:
$$x^2 + 15x + 35 = 2x - 5$$
3. **Bring all terms to one side to form a quadratic equation:**
$$x^2 + 15x + 35 - 2x + 5 = 0$$
Simplify:
$$x^2 + (15x - 2x) + (35 + 5) = 0$$
$$x^2 + 13x + 40 = 0$$
4. **Factor the quadratic equation:**
We look for two numbers that multiply to $40$ and add to $13$.
These are $8$ and $5$.
So,
$$x^2 + 13x + 40 = (x + 8)(x + 5) = 0$$
5. **Solve for $x$:**
Set each factor equal to zero:
$$x + 8 = 0 \Rightarrow x = -8$$
$$x + 5 = 0 \Rightarrow x = -5$$
6. **Find corresponding $y$ values:**
Use $y = 2x - 5$ for simplicity.
- For $x = -8$:
$$y = 2(-8) - 5 = -16 - 5 = -21$$
- For $x = -5$:
$$y = 2(-5) - 5 = -10 - 5 = -15$$
7. **Write the solution pairs:**
$$(-8, -21) \quad \text{and} \quad (-5, -15)$$
**Final answer:**
The solutions to the system are $$\boxed{(-8, -21) \text{ and } (-5, -15)}$$.