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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.

Solve the linear-quadratic system of equations algebraically. Express your values as radical expressions, if necessary.

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

Learn how to solve a linear-quadratic system of equations involving y = -2x + 6 and y = x^2 - x + 4. This step-by-step solution covers key algebraic concepts, including using the quadratic formula to find the points of intersection.

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