Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Experience the ease of finding reliable answers to your questions from a vast community of knowledgeable experts. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
Sure, let's analyze the given system of equations step-by-step to understand the greatest possible number of solutions.
The system of equations is:
[tex]\[ \begin{cases} x^2 + y^2 = 8 \\ 8x + 2y = 16 \end{cases} \][/tex]
Step 1: Analyze the Equations
1. The first equation [tex]\(x^2 + y^2 = 8\)[/tex] represents a circle centered at the origin [tex]\((0,0)\)[/tex] with a radius of [tex]\(\sqrt{8} = 2\sqrt{2}\)[/tex].
2. The second equation [tex]\(8x + 2y = 16\)[/tex] can be simplified by dividing everything by 2, resulting in [tex]\(4x + y = 8\)[/tex]. This is the equation of a line.
Step 2: Graphical Perspective
1. A circle can generally intersect a line in at most two points. However, this intersection can result in:
- Two distinct points (indicating two solutions).
- One point if the line is tangent to the circle (indicating a single solution).
- No points if the line does not intersect the circle (indicating no solutions).
Step 3: Solve the System Algebraically
To determine how many solutions there are, we need to solve the system of equations.
1. From the equation [tex]\(4x + y = 8\)[/tex], solve for [tex]\(y\)[/tex]:
[tex]\[ y = 8 - 4x \][/tex]
2. Substitute this expression for [tex]\(y\)[/tex] in the circle's equation [tex]\(x^2 + y^2 = 8\)[/tex]:
[tex]\[ x^2 + (8 - 4x)^2 = 8 \][/tex]
3. Expand and simplify:
[tex]\[ x^2 + (64 - 64x + 16x^2) = 8 \][/tex]
[tex]\[ 17x^2 - 64x + 64 = 8 \][/tex]
[tex]\[ 17x^2 - 64x + 56 = 0 \][/tex]
4. Solve this quadratic equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 17\)[/tex], [tex]\(b = -64\)[/tex], and [tex]\(c = 56\)[/tex]:
[tex]\[ x = \frac{64 \pm \sqrt{(-64)^2 - 4 \cdot 17 \cdot 56}}{2 \cdot 17} \][/tex]
[tex]\[ x = \frac{64 \pm \sqrt{4096 - 3808}}{34} \][/tex]
[tex]\[ x = \frac{64 \pm \sqrt{288}}{34} \][/tex]
[tex]\[ x = \frac{64 \pm 12\sqrt{2}}{34} \][/tex]
Simplifying further:
[tex]\[ x = \frac{32 \pm 6\sqrt{2}}{17} \][/tex]
5. Substitute these values of [tex]\(x\)[/tex] back into [tex]\(y = 8 - 4x\)[/tex] to find the corresponding [tex]\(y\)[/tex] values.
6. For [tex]\(x = \frac{32 - 6\sqrt{2}}{17}\)[/tex]:
[tex]\[ y = 8 - 4 \left(\frac{32 - 6\sqrt{2}}{17}\right) = 8 - \frac{128 - 24\sqrt{2}}{17} = \frac{136 - (128 - 24\sqrt{2})}{17} = \frac{8 + 24\sqrt{2}}{17} \][/tex]
7. For [tex]\(x = \frac{32 + 6\sqrt{2}}{17}\)[/tex]:
[tex]\[ y = 8 - 4 \left(\frac{32 + 6\sqrt{2}}{17}\right) = 8 - \frac{128 + 24\sqrt{2}}{17} = \frac{136 - (128 + 24\sqrt{2})}{17} = \frac{8 - 24\sqrt{2}}{17} \][/tex]
Thus, the solutions to the system are:
[tex]\[ \left( \frac{32 - 6\sqrt{2}}{17}, \frac{8 + 24\sqrt{2}}{17} \right) \quad \text{and} \quad \left( \frac{32 + 6\sqrt{2}}{17}, \frac{8 - 24\sqrt{2}}{17} \right) \][/tex]
Conclusion:
As the system has two distinct solutions, the greatest possible number of solutions for this system of equations is [tex]\(\boxed{2}\)[/tex].
The system of equations is:
[tex]\[ \begin{cases} x^2 + y^2 = 8 \\ 8x + 2y = 16 \end{cases} \][/tex]
Step 1: Analyze the Equations
1. The first equation [tex]\(x^2 + y^2 = 8\)[/tex] represents a circle centered at the origin [tex]\((0,0)\)[/tex] with a radius of [tex]\(\sqrt{8} = 2\sqrt{2}\)[/tex].
2. The second equation [tex]\(8x + 2y = 16\)[/tex] can be simplified by dividing everything by 2, resulting in [tex]\(4x + y = 8\)[/tex]. This is the equation of a line.
Step 2: Graphical Perspective
1. A circle can generally intersect a line in at most two points. However, this intersection can result in:
- Two distinct points (indicating two solutions).
- One point if the line is tangent to the circle (indicating a single solution).
- No points if the line does not intersect the circle (indicating no solutions).
Step 3: Solve the System Algebraically
To determine how many solutions there are, we need to solve the system of equations.
1. From the equation [tex]\(4x + y = 8\)[/tex], solve for [tex]\(y\)[/tex]:
[tex]\[ y = 8 - 4x \][/tex]
2. Substitute this expression for [tex]\(y\)[/tex] in the circle's equation [tex]\(x^2 + y^2 = 8\)[/tex]:
[tex]\[ x^2 + (8 - 4x)^2 = 8 \][/tex]
3. Expand and simplify:
[tex]\[ x^2 + (64 - 64x + 16x^2) = 8 \][/tex]
[tex]\[ 17x^2 - 64x + 64 = 8 \][/tex]
[tex]\[ 17x^2 - 64x + 56 = 0 \][/tex]
4. Solve this quadratic equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 17\)[/tex], [tex]\(b = -64\)[/tex], and [tex]\(c = 56\)[/tex]:
[tex]\[ x = \frac{64 \pm \sqrt{(-64)^2 - 4 \cdot 17 \cdot 56}}{2 \cdot 17} \][/tex]
[tex]\[ x = \frac{64 \pm \sqrt{4096 - 3808}}{34} \][/tex]
[tex]\[ x = \frac{64 \pm \sqrt{288}}{34} \][/tex]
[tex]\[ x = \frac{64 \pm 12\sqrt{2}}{34} \][/tex]
Simplifying further:
[tex]\[ x = \frac{32 \pm 6\sqrt{2}}{17} \][/tex]
5. Substitute these values of [tex]\(x\)[/tex] back into [tex]\(y = 8 - 4x\)[/tex] to find the corresponding [tex]\(y\)[/tex] values.
6. For [tex]\(x = \frac{32 - 6\sqrt{2}}{17}\)[/tex]:
[tex]\[ y = 8 - 4 \left(\frac{32 - 6\sqrt{2}}{17}\right) = 8 - \frac{128 - 24\sqrt{2}}{17} = \frac{136 - (128 - 24\sqrt{2})}{17} = \frac{8 + 24\sqrt{2}}{17} \][/tex]
7. For [tex]\(x = \frac{32 + 6\sqrt{2}}{17}\)[/tex]:
[tex]\[ y = 8 - 4 \left(\frac{32 + 6\sqrt{2}}{17}\right) = 8 - \frac{128 + 24\sqrt{2}}{17} = \frac{136 - (128 + 24\sqrt{2})}{17} = \frac{8 - 24\sqrt{2}}{17} \][/tex]
Thus, the solutions to the system are:
[tex]\[ \left( \frac{32 - 6\sqrt{2}}{17}, \frac{8 + 24\sqrt{2}}{17} \right) \quad \text{and} \quad \left( \frac{32 + 6\sqrt{2}}{17}, \frac{8 - 24\sqrt{2}}{17} \right) \][/tex]
Conclusion:
As the system has two distinct solutions, the greatest possible number of solutions for this system of equations is [tex]\(\boxed{2}\)[/tex].
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.