Question: Solve the following system of equations algebraically:
y, equals, x, squared, plus, 18, x, plus, 78
y= x 2 +18x+78
y, equals, x, plus, 8
y= x+8
1. **State the problem:**
We need to solve the system of equations:
$$y = x^2 + 18x + 78$$
$$y = x + 8$$
2. **Set the equations equal to each other:**
Since both expressions equal $y$, set them equal:
$$x^2 + 18x + 78 = x + 8$$
3. **Bring all terms to one side to form a quadratic equation:**
$$x^2 + 18x + 78 - x - 8 = 0$$
Simplify:
$$x^2 + (18x - x) + (78 - 8) = 0$$
$$x^2 + 17x + 70 = 0$$
4. **Factor the quadratic equation:**
We look for two numbers that multiply to $70$ and add to $17$.
These are $7$ and $10$.
So,
$$x^2 + 17x + 70 = (x + 7)(x + 10) = 0$$
5. **Solve for $x$:**
Set each factor equal to zero:
$$x + 7 = 0 \Rightarrow x = -7$$
$$x + 10 = 0 \Rightarrow x = -10$$
6. **Find corresponding $y$ values:**
Use $y = x + 8$:
- For $x = -7$:
$$y = -7 + 8 = 1$$
- For $x = -10$:
$$y = -10 + 8 = -2$$
7. **Final solution:**
The system has two solutions:
$$(x, y) = (-7, 1) \text{ and } (-10, -2)$$