Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Our platform connects you with professionals ready to provide precise answers to all your questions in various areas of expertise. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
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].
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We hope this was helpful. Please come back whenever you need more information or answers to your queries. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.
