Solving Systems of Quadratic Equations Worksheet | Algebra II PDF ... - Free Printable

To solve the given systems of quadratic equations, we will use algebraic methods such as substitution and elimination. 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: Equate the two expressions for \( y \)
\[
x + 2 = x^2 - 4
\]

#### Step 2: Rearrange into 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: Equate the two expressions for \( y \)
\[
x^2 + 3x - 1 = x + 14
\]

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

#### 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: Equate the two expressions for \( y \)
\[
2x^2 + x - 7 = 2x + 8
\]

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

#### 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 the first equation
\[
y = x^2 + 3x
\]

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

#### Step 3: Rearrange into 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: Solve the second equation for \( y \)
\[
y = -2x - 22
\]

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

#### Step 3: Rearrange into standard quadratic form
\[
-3x^2 - 2x + 5 + 2x + 22 = 0
\]
\[
-3x^2 + 27 = 0
\]

#### Step 4: Solve for \( x \)
\[
-3x^2 = -27
\]
\[
x^2 = 9
\]
\[
x = \pm 3
\]

#### Step 5: Find corresponding \( y \)-values using \( y = -2x - 22 \)
- For \( x = 3 \):
\[
y = -2(3) - 22 = -6 - 22 = -28
\]
- For \( x = -3 \):
\[
y = -2(-3) - 22 = 6 - 22 = -16
\]

#### 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 the first equation
\[
(10 - 3x) + 6 = 2x^2 + x
\]
\[
16 - 3x = 2x^2 + x
\]

#### Step 3: Rearrange into 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:

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