Question: Solve the following system of equations algebraically:
y, equals, x, squared, plus, 7, x, minus, 12
y= x 2 +7x−12
y, equals, minus, x, minus, 3
y= −x−3
1. **State the problem:**
We need to solve the system of equations:
$$y = x^2 + 7x - 12$$
and
$$y = -x - 3$$
2. **Set the equations equal to each other:**
Since both expressions equal $y$, set them equal:
$$x^2 + 7x - 12 = -x - 3$$
3. **Bring all terms to one side to form a quadratic equation:**
$$x^2 + 7x - 12 + x + 3 = 0$$
$$x^2 + 8x - 9 = 0$$
4. **Use the quadratic formula:**
The quadratic formula is:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
where $a=1$, $b=8$, and $c=-9$.
5. **Calculate the discriminant:**
$$b^2 - 4ac = 8^2 - 4(1)(-9) = 64 + 36 = 100$$
6. **Find the roots:**
$$x = \frac{-8 \pm \sqrt{100}}{2} = \frac{-8 \pm 10}{2}$$
7. **Calculate each solution:**
- For $+$:
$$x = \frac{-8 + 10}{2} = \frac{2}{2} = 1$$
- For $-$:
$$x = \frac{-8 - 10}{2} = \frac{-18}{2} = -9$$
8. **Find corresponding $y$ values using $y = -x - 3$:**
- For $x=1$:
$$y = -1 - 3 = -4$$
- For $x=-9$:
$$y = -(-9) - 3 = 9 - 3 = 6$$
**Final solutions:**
$$(x,y) = (1, -4) \text{ and } (-9, 6)$$