Solved! Worksheet 4 - Nonlinear Equations Name: Grade & Section ... - Free Printable

Problem: Solve the following systems of nonlinear equations using either elimination or substitution.

We will solve each system step by step. Let's start with the first system and proceed systematically.

---

#### System 1:
\[
\begin{cases}
x^2 - y^2 = 81 \\
x = 2y
\end{cases}
\]

1. Substitute \( x = 2y \) into the first equation:
\[
(2y)^2 - y^2 = 81
\]
Simplify:
\[
4y^2 - y^2 = 81
\]
\[
3y^2 = 81
\]
\[
y^2 = 27
\]
\[
y = \pm \sqrt{27} = \pm 3\sqrt{3}
\]

2. Find \( x \) using \( x = 2y \):
- If \( y = 3\sqrt{3} \):
\[
x = 2(3\sqrt{3}) = 6\sqrt{3}
\]
- If \( y = -3\sqrt{3} \):
\[
x = 2(-3\sqrt{3}) = -6\sqrt{3}
\]

3. Solutions:
\[
(x, y) = (6\sqrt{3}, 3\sqrt{3}) \quad \text{and} \quad (x, y) = (-6\sqrt{3}, -3\sqrt{3})
\]

---

#### System 2:
\[
\begin{cases}
4x^2 + y^2 = 16 \\
x^2 = -2y
\end{cases}
\]

1. Substitute \( x^2 = -2y \) into the first equation:
\[
4(-2y) + y^2 = 16
\]
Simplify:
\[
-8y + y^2 = 16
\]
Rearrange into standard quadratic form:
\[
y^2 - 8y - 16 = 0
\]

2. Solve the quadratic equation using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = -8 \), and \( c = -16 \):
\[
y = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-16)}}{2(1)}
\]
\[
y = \frac{8 \pm \sqrt{64 + 64}}{2}
\]
\[
y = \frac{8 \pm \sqrt{128}}{2}
\]
\[
y = \frac{8 \pm 8\sqrt{2}}{2}
\]
\[
y = 4 \pm 4\sqrt{2}
\]

3. Find \( x^2 \) using \( x^2 = -2y \):
- If \( y = 4 + 4\sqrt{2} \):
\[
x^2 = -2(4 + 4\sqrt{2}) = -8 - 8\sqrt{2}
\]
Since \( x^2 \) cannot be negative, this solution is invalid.
- If \( y = 4 - 4\sqrt{2} \):
\[
x^2 = -2(4 - 4\sqrt{2}) = -8 + 8\sqrt{2}
\]
Similarly, this is also invalid because \( x^2 \) cannot be negative.

4. Conclusion:
There are no real solutions for this system.

---

#### System 3:
\[
\begin{cases}
y = x^2 - 9 \\
y = x - 3
\end{cases}
\]

1. Set the two expressions for \( y \) equal to each other:
\[
x^2 - 9 = x - 3
\]
Rearrange into standard quadratic form:
\[
x^2 - x - 6 = 0
\]

2. Factor the quadratic equation:
\[
(x - 3)(x + 2) = 0
\]
So, \( x = 3 \) or \( x = -2 \).

3. Find \( y \) using \( y = x - 3 \):
- If \( x = 3 \):
\[
y = 3 - 3 = 0
\]
- If \( x = -2 \):
\[
y = -2 - 3 = -5
\]

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

---

#### System 4:
\[
\begin{cases}
y = -x^2 - 5 \\
y = 2x^2 + 2x + 3
\end{cases}
\]

1. Set the two expressions for \( y \) equal to each other:
\[
-x^2 - 5 = 2x^2 + 2x + 3
\]
Rearrange into standard quadratic form:
\[
-x^2 - 2x^2 - 2x - 5 - 3 = 0
\]
\[
-3x^2 - 2x - 8 = 0
\]
Multiply through by \(-1\) to simplify:
\[
3x^2 + 2x + 8 = 0
\]

2. Solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3 \), \( b = 2 \), and \( c = 8 \):
\[
x = \frac{-2 \pm \sqrt{2^2 - 4(3)(8)}}{2(3)}
\]
\[
x = \frac{-2 \pm \sqrt{4 - 96}}{6}
\]
\[
x = \frac{-2 \pm \sqrt{-92}}{6}
\]
Since the discriminant is negative (\( -92 \)), there are no real solutions.

---

#### System 5:
\[
\begin{cases}
y = x^2 + 4x + 5 \\
y = -x^2 - x + 3
\end{cases}
\]

1. Set the two expressions for \( y \) equal to each other:
\[
x^2 + 4x + 5 = -x^2 - x + 3
\]
Rearrange into standard quadratic form:
\[
x^2 + x^2 + 4x + x + 5 - 3 = 0
\]
\[
2x^2 + 5x + 2 = 0
\]

2. Factor the quadratic equation:
\[
(2x + 1)(x + 2) = 0
\]
So, \( x = -\frac{1}{2} \) or \( x = -2 \).

3. Find \( y \) using \( y = x^2 + 4x + 5 \):
- If \( x = -\frac{1}{2} \):
\[
y = \left(-\frac{1}{2}\right)^2 + 4\left(-\frac{1}{2}\right) + 5
\]
\[
y = \frac{1}{4} - 2 + 5 = \frac{1}{4} + 3 = \frac{13}{4}
\]
- If \( x = -2 \):
\[
y = (-2)^2 + 4(-2) + 5
\]
\[
y = 4 - 8 + 5 = 1
\]

4. Solutions:
\[
(x, y) = \left(-\frac{1}{2}, \frac{13}{4}\right) \quad \text{and} \quad (x, y) = (-2, 1)
\]

---

#### System 6:
\[
\begin{cases}
y = 2x^2 - 8x + 7 \\
y = x^2 - 6x + 6
\end{cases}
\]

1. Set the two expressions for \( y \) equal to each other:
\[
2x^2 - 8x + 7 = x^2 - 6x + 6
\]
Rearrange into standard quadratic form:
\[
2x^2 - x^2 - 8x + 6x + 7 - 6 = 0
\]
\[
x^2 - 2x + 1 = 0
\]

2. Factor the quadratic equation:
\[
(x - 1)^2 = 0
\]
So, \( x = 1 \).

3. Find \( y \) using \( y = x^2 - 6x + 6 \):
\[
y = (1)^2 - 6(1) + 6
\]
\[
y = 1 - 6 + 6 = 1
\]

4. Solution:
\[
(x, y) = (1, 1)
\]

---

#### System 7:
\[
\begin{cases}
y = x^2 + 4 \\
y = 2x^2
\end{cases}
\]

1. Set the two expressions for \( y \) equal to each other:
\[
x^2 + 4 = 2x^2
\]
Rearrange into standard quadratic form:
\[
x^2 + 4 - 2x^2 = 0
\]
\[
-x^2 + 4 = 0
\]
\[
x^2 = 4
\]
\[
x = \pm 2
\]

2. Find \( y \) using \( y = 2x^2 \):
- If \( x = 2 \):
\[
y = 2(2)^2 = 2(4) = 8
\]
- If \( x = -2 \):
\[
y = 2(-2)^2 = 2(4) = 8
\]

3. Solutions:
\[
(x, y) = (2, 8) \quad \text{and} \quad (x, y) = (-2, 8)
\]

---

#### System 8:
\[
\begin{cases}
y = x^2 - 4 \\
y = 2x^2 - x
\end{cases}
\]

1. Set the two expressions for \( y \) equal to each other:
\[
x^2 - 4 = 2x^2 - x
\]
Rearrange into standard quadratic form:
\[
x^2 - 2x^2 + x - 4 = 0
\]
\[
-x^2 + x - 4 = 0
\]
Multiply through by \(-1\) to simplify:
\[
x^2 - x + 4 = 0
\]

2. Solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = -1 \), and \( c = 4 \):
\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(4)}}{2(1)}
\]
\[
x = \frac{1 \pm \sqrt{1 - 16}}{2}
\]
\[
x = \frac{1 \pm \sqrt{-15}}{2}
\]
Since the discriminant is negative (\( -15 \)), there are no real solutions.

---

Final Answers:

\[
\boxed{
\begin{aligned}
1. & \ (6\sqrt{3}, 3\sqrt{3}), (-6\sqrt{3}, -3\sqrt{3}) \\
2. & \ \text{No real solutions} \\
3. & \ (3, 0), (-2, -5) \\
4. & \ \text{No real solutions} \\
5. & \ \left(-\frac{1}{2}, \frac{13}{4}\right), (-2, 1) \\
6. & \ (1, 1) \\
7. & \ (2, 8), (-2, 8) \\
8. & \ \text{No real solutions}
\end{aligned}
}
\]