This worksheet provides six practice problems for solving systems of linear and quadratic equations algebraically.

To solve the given systems of quadratic equations, we will use substitution and algebraic manipulation. Let's solve each system step by step.

---

Problem 1:

\[
\begin{aligned}
1) & \quad y = x + 2 \\
& \quad y = x^2 - 4
\end{aligned}
\]

#### Step 1: Substitute \( y = x + 2 \) into \( y = x^2 - 4 \)
\[
x + 2 = x^2 - 4
\]

#### Step 2: Rearrange the equation to standard quadratic form
\[
x^2 - x - 6 = 0
\]

#### Step 3: Factor the quadratic equation
\[
(x - 3)(x + 2) = 0
\]

#### Step 4: Solve for \( x \)
\[
x - 3 = 0 \quad \text{or} \quad x + 2 = 0
\]
\[
x = 3 \quad \text{or} \quad x = -2
\]

#### Step 5: Find corresponding \( y \)-values using \( y = x + 2 \)
- For \( x = 3 \):
\[
y = 3 + 2 = 5
\]
- For \( x = -2 \):
\[
y = -2 + 2 = 0
\]

#### Solutions:
\[
(x, y) = (3, 5) \quad \text{and} \quad (x, y) = (-2, 0)
\]

---

Problem 2:

\[
\begin{aligned}
2) & \quad y = x^2 + 3x - 1 \\
& \quad y = x + 14
\end{aligned}
\]

#### Step 1: Substitute \( y = x + 14 \) into \( y = x^2 + 3x - 1 \)
\[
x + 14 = x^2 + 3x - 1
\]

#### Step 2: Rearrange the equation to standard quadratic form
\[
x^2 + 3x - x - 14 - 1 = 0
\]
\[
x^2 + 2x - 15 = 0
\]

#### Step 3: Factor the quadratic equation
\[
(x + 5)(x - 3) = 0
\]

#### Step 4: Solve for \( x \)
\[
x + 5 = 0 \quad \text{or} \quad x - 3 = 0
\]
\[
x = -5 \quad \text{or} \quad x = 3
\]

#### Step 5: Find corresponding \( y \)-values using \( y = x + 14 \)
- For \( x = -5 \):
\[
y = -5 + 14 = 9
\]
- For \( x = 3 \):
\[
y = 3 + 14 = 17
\]

#### Solutions:
\[
(x, y) = (-5, 9) \quad \text{and} \quad (x, y) = (3, 17)
\]

---

Problem 3:

\[
\begin{aligned}
3) & \quad y = 2x^2 + x - 7 \\
& \quad y = 2x + 8
\end{aligned}
\]

#### Step 1: Substitute \( y = 2x + 8 \) into \( y = 2x^2 + x - 7 \)
\[
2x + 8 = 2x^2 + x - 7
\]

#### Step 2: Rearrange the equation to standard quadratic form
\[
2x^2 + x - 2x - 8 - 7 = 0
\]
\[
2x^2 - x - 15 = 0
\]

#### Step 3: Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Here, \( a = 2 \), \( b = -1 \), and \( c = -15 \):
\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-15)}}{2(2)}
\]
\[
x = \frac{1 \pm \sqrt{1 + 120}}{4}
\]
\[
x = \frac{1 \pm \sqrt{121}}{4}
\]
\[
x = \frac{1 \pm 11}{4}
\]

#### Step 4: Solve for \( x \)
\[
x = \frac{1 + 11}{4} = \frac{12}{4} = 3
\]
\[
x = \frac{1 - 11}{4} = \frac{-10}{4} = -\frac{5}{2}
\]

#### Step 5: Find corresponding \( y \)-values using \( y = 2x + 8 \)
- For \( x = 3 \):
\[
y = 2(3) + 8 = 6 + 8 = 14
\]
- For \( x = -\frac{5}{2} \):
\[
y = 2\left(-\frac{5}{2}\right) + 8 = -5 + 8 = 3
\]

#### Solutions:
\[
(x, y) = (3, 14) \quad \text{and} \quad \left(x, y\right) = \left(-\frac{5}{2}, 3\right)
\]

---

Problem 4:

\[
\begin{aligned}
4) & \quad y = x(x + 3) \\
& \quad y - x = 8
\end{aligned}
\]

#### Step 1: Simplify \( y = x(x + 3) \)
\[
y = x^2 + 3x
\]

#### Step 2: Substitute \( y = x^2 + 3x \) into \( y - x = 8 \)
\[
x^2 + 3x - x = 8
\]
\[
x^2 + 2x = 8
\]

#### Step 3: Rearrange the equation to standard quadratic form
\[
x^2 + 2x - 8 = 0
\]

#### Step 4: Factor the quadratic equation
\[
(x + 4)(x - 2) = 0
\]

#### Step 5: Solve for \( x \)
\[
x + 4 = 0 \quad \text{or} \quad x - 2 = 0
\]
\[
x = -4 \quad \text{or} \quad x = 2
\]

#### Step 6: Find corresponding \( y \)-values using \( y = x^2 + 3x \)
- For \( x = -4 \):
\[
y = (-4)^2 + 3(-4) = 16 - 12 = 4
\]
- For \( x = 2 \):
\[
y = (2)^2 + 3(2) = 4 + 6 = 10
\]

#### Solutions:
\[
(x, y) = (-4, 4) \quad \text{and} \quad (x, y) = (2, 10)
\]

---

Problem 5:

\[
\begin{aligned}
5) & \quad y = -3x^2 - 2x + 5 \\
& \quad y + 2x + 22 = 0
\end{aligned}
\]

#### Step 1: Substitute \( y = -3x^2 - 2x + 5 \) into \( y + 2x + 22 = 0 \)
\[
-3x^2 - 2x + 5 + 2x + 22 = 0
\]

#### Step 2: Simplify the equation
\[
-3x^2 + 27 = 0
\]

#### Step 3: Solve for \( x \)
\[
-3x^2 = -27
\]
\[
x^2 = 9
\]
\[
x = \pm 3
\]

#### Step 4: Find corresponding \( y \)-values using \( y = -3x^2 - 2x + 5 \)
- For \( x = 3 \):
\[
y = -3(3)^2 - 2(3) + 5 = -27 - 6 + 5 = -28
\]
- For \( x = -3 \):
\[
y = -3(-3)^2 - 2(-3) + 5 = -27 + 6 + 5 = -16
\]

#### Solutions:
\[
(x, y) = (3, -28) \quad \text{and} \quad (x, y) = (-3, -16)
\]

---

Problem 6:

\[
\begin{aligned}
6) & \quad y + 6 = 2x^2 + x \\
& \quad y + 3x = 10
\end{aligned}
\]

#### Step 1: Solve the second equation for \( y \)
\[
y = 10 - 3x
\]

#### Step 2: Substitute \( y = 10 - 3x \) into \( y + 6 = 2x^2 + x \)
\[
10 - 3x + 6 = 2x^2 + x
\]
\[
16 - 3x = 2x^2 + x
\]

#### Step 3: Rearrange the equation to standard quadratic form
\[
2x^2 + x + 3x - 16 = 0
\]
\[
2x^2 + 4x - 16 = 0
\]

#### Step 4: Simplify by dividing by 2
\[
x^2 + 2x - 8 = 0
\]

#### Step 5: Factor the quadratic equation
\[
(x + 4)(x - 2) = 0
\]

#### Step 6: Solve for \( x \)
\[
x + 4 = 0 \quad \text{or} \quad x - 2 = 0
\]
\[
x = -4 \quad \text{or} \quad x = 2
\]

#### Step 7: Find corresponding \( y \)-values using \( y = 10 - 3x \)
- For \( x = -4 \):
\[
y = 10 - 3(-4) = 10 + 12 = 22
\]
- For \( x = 2 \):
\[
y = 10 - 3(2) = 10 - 6 = 4
\]

#### Solutions:
\[
(x, y) = (-4, 22) \quad \text{and} \quad (x, y) = (2, 4)
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
1) & \quad (3, 5), (-2, 0) \\
2) & \quad (-5, 9), (3, 17) \\
3) & \quad (3, 14), \left(-\frac{5}{2}, 3\right) \\
4) & \quad (-4, 4), (2, 10) \\
5) & \quad (3, -28), (-3, -16) \\
6) & \quad (-4, 22), (2, 4)
\end{aligned}
}
\]