To solve this system of equations algebraically, we can set the two equations equal to each other and solve for $x$. Here are the steps:

Given Equations:

1. $y = -2x + 6$
2. $y = x^2 - x + 4$

Since both expressions are equal to $y$, we can set them equal to each other: $-2x + 6 = x^2 - x + 4$

Step 1: Rearrange the Equation

Move all terms to one side of the equation to set it equal to zero: $0 = x^2 - x + 4 + 2x - 6$ Simplify: $0 = x^2 + x - 2$

Step 2: Solve the Quadratic Equation

Now, we have a quadratic equation: $x^2 + x - 2 = 0$ This equation can be solved using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = 1$, and $c = -2$.

Plugging in the values: $x = \frac{-1 \pm \sqrt{(1)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1}$ $x = \frac{-1 \pm \sqrt{1 + 8}}{2}$ $x = \frac{-1 \pm \sqrt{9}}{2}$ $x = \frac{-1 \pm 3}{2}$

So, we have two solutions for $x$:

1. $x = \frac{-1 + 3}{2} = 1$
2. $x = \frac{-1 - 3}{2} = -2$

Step 3: Find Corresponding $y$ Values

Now, we substitute these $x$ values back into one of the original equations to find $y$.

For $x = 1$:

$y = -2(1) + 6 = -2 + 6 = 4$ So, one solution is $(1, 4)$.

For $x = -2$:

$y = -2(-2) + 6 = 4 + 6 = 10$ So, another solution is $(-2, 10)$.

Final Answer

The solutions to the system are: $(1, 4) \quad \text{and} \quad (-2, 10)$

Would you like more details on any of these steps?

Here are some related questions that could further your understanding:

1. How do you determine whether a system of equations is linear, quadratic, or both?
2. Can we use graphing to verify these solutions?
3. What are some alternative methods to solve a linear-quadratic system?
4. How does the discriminant help in solving quadratic equations?
5. What do the solutions represent in terms of the graphs of the equations?

Tip: When solving systems involving quadratics, try to simplify the equation to standard form $ax^2 + bx + c = 0$ and check if factoring is possible before using the quadratic formula.