Question: Solve the following system of equations algebraically:
y, equals, x, squared, plus, 2
y= x 2 +2
y, equals, minus, 3, x, plus, 2
y= −3x+2
1. **State the problem:**
We need to solve the system of equations:
$$y = x^2 + 2$$
$$y = -3x + 2$$
2. **Set the equations equal to each other:**
Since both expressions equal $y$, set them equal:
$$x^2 + 2 = -3x + 2$$
3. **Simplify the equation:**
Subtract $2$ from both sides:
$$x^2 + 2 - 2 = -3x + 2 - 2$$
$$x^2 = -3x$$
4. **Rewrite the equation:**
Bring all terms to one side:
$$x^2 + 3x = 0$$
5. **Factor the equation:**
$$x(x + 3) = 0$$
6. **Solve for $x$:**
Set each factor equal to zero:
$$x = 0$$
$$x + 3 = 0 \Rightarrow x = -3$$
7. **Find corresponding $y$ values:**
Use $y = -3x + 2$ to find $y$ for each $x$:
- For $x=0$:
$$y = -3(0) + 2 = 2$$
- For $x=-3$:
$$y = -3(-3) + 2 = 9 + 2 = 11$$
8. **Final solution:**
The system has two solutions:
$$(0, 2) \text{ and } (-3, 11)$$