Question: Solve the following system of equations algebraically:
y= x squared plus 20 x plus 90
y= x^2 + 20x + 90
y= 2 x plus 10
y= 2x + 10
1. **State the problem:**
We need to solve the system of equations:
$$y = x^2 + 20x + 90$$
$$y = 2x + 10$$
2. **Set the equations equal to each other:**
Since both expressions equal $y$, set them equal:
$$x^2 + 20x + 90 = 2x + 10$$
3. **Bring all terms to one side to form a quadratic equation:**
$$x^2 + 20x + 90 - 2x - 10 = 0$$
Simplify:
$$x^2 + (20x - 2x) + (90 - 10) = 0$$
$$x^2 + 18x + 80 = 0$$
4. **Factor the quadratic equation:**
We look for two numbers that multiply to $80$ and add to $18$.
These are $10$ and $8$.
So,
$$x^2 + 18x + 80 = (x + 10)(x + 8) = 0$$
5. **Solve for $x$:**
Set each factor equal to zero:
$$x + 10 = 0 \Rightarrow x = -10$$
$$x + 8 = 0 \Rightarrow x = -8$$
6. **Find corresponding $y$ values:**
Use $y = 2x + 10$ for simplicity.
- For $x = -10$:
$$y = 2(-10) + 10 = -20 + 10 = -10$$
- For $x = -8$:
$$y = 2(-8) + 10 = -16 + 10 = -6$$
7. **Final solution:**
The system has two solutions:
$$(-10, -10) \text{ and } (-8, -6)$$