1. 10 Give The General Solution Of The Linear System X+y = 2z = 0 2x + 2y3z = 1 3x + 3y + Z = 7
Understanding and solving linear systems is a fundamental aspect of algebra that has wide-ranging applications in mathematics, engineering, physics, computer science, and many other disciplines. The particular system in question presents an intriguing challenge that involves multiple equations with variables interconnected through linear relationships. In this comprehensive guide, we will explore the process of determining the general solution for the given linear system, ensuring clarity in each step and offering insights into the methods used.
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Understanding the Given System of Equations
Before diving into solutions, it is crucial to interpret the system correctly. The system provided is:
- X + y = 2z = 0
- 2x + 2y3z = 1
- 3x + 3y + Z = 7
At first glance, the system appears to have some notation issues, especially in the first equation. It is essential to clarify the equations to proceed accurately.
Interpreting the Equations Correctly
The original system seems to be:
- X + y = 2z = 0
- 2x + 2y3z = 1
- 3x + 3y + Z = 7
However, the first equation likely contains a typo or formatting error. It appears to be attempting to express that:
- X + y = 2z
- and that 2z = 0
Similarly, in the second equation, '2y3z' might be a typo or missing an operation. Possibly, it is '2y + 3z'.
The corrected and properly formatted system likely is:
- X + y = 2z
- 2x + 2y + 3z = 1
- 3x + 3y + Z = 7
If this assumption aligns with the intended system, then the set consists of three equations with three variables: x, y, and z.
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Step-by-Step Solution Approach
To find the general solution, we'll follow a systematic approach:
- Rewrite the system in standard form.
- Use substitution or elimination methods to reduce the system.
- Express variables in terms of free parameters, representing the solution set.
Expressing the System in Standard Form
Given the interpreted system:
- x + y - 2z = 0
- 2x + 2y + 3z = 1
- 3x + 3y + z = 7
Now, the system is in a clear form suitable for solving.
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Applying the Elimination Method
The elimination method involves manipulating equations to eliminate variables step-by-step.
Step 1: Eliminate one variable between equations
- Subtract equation 1 from equation 2:
Simplify:
(2x - x) + (2y - y) + (3z + 2z) = 1
x + y + 5z = 1
Call this equation 4.
- Similarly, subtract equation 1 from equation 3:
(3x + 3y + z) - (x + y - 2z) = 7 - 0
Simplify:
(3x - x) + (3y - y) + (z + 2z) = 7
2x + 2y + 3z = 7
Call this equation 5.
Now, the system reduces to:
- x + y + 5z = 1
- 2x + 2y + 3z = 7
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Step 2: Eliminate variables between equations 4 and 5
Divide equation 5 by 2 for convenience:
(2x + 2y + 3z)/2 = 7/2
which simplifies to:
x + y + (3/2)z = 7/2
Recall equation 4:
x + y + 5z = 1
Subtract the scaled equation:
(x + y + 5z) - (x + y + (3/2)z) = 1 - 7/2
Simplify the left:
(5z - (3/2)z) = 1 - 7/2
Calculate:
(5z - 1.5z) = 1 - 3.5
which gives:
(3.5z) = -2.5
Solve for z:
z = -2.5 / 3.5 = -5/7
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Back-Substitution to Find x and y
Having found z, we can substitute back into earlier equations to find x and y.
Find x and y from equation 4:
x + y + 5z = 1
Substitute z = -5/7:
x + y + 5(-5/7) = 1
x + y - 25/7 = 1
Express 1 as 7/7:
x + y = 1 + 25/7 = 7/7 + 25/7 = 32/7
Now, express y in terms of x:
y = (32/7) - x
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Use the first original equation to relate x and y
Recall the first equation:
x + y - 2z = 0
Substitute y:
x + [(32/7) - x] - 2(-5/7) = 0
Simplify:
x + 32/7 - x + (10/7) = 0
x cancels out:
32/7 + 10/7 = 0
Sum:
(32 + 10)/7 = 42/7 = 6
But this equals zero, which is a contradiction unless 6 = 0, which is false.
This indicates an inconsistency in the system, implying that the system has no solution.
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Conclusion: System Consistency and Solution Set
Based on the above calculations, the system appears inconsistent because the derived condition leads to a contradiction. When solving linear systems, such contradictions indicate that the system has no solution — it is inconsistent.
Summary:
- The interpreted system was:
- x + y - 2z = 0
- 2x + 2y + 3z = 1
- 3x + 3y + z = 7
- Eliminations led to a value of z = -5/7.
- Substituting back into the equations revealed a contradiction.
Thus, the system has no solution and is inconsistent.
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Additional Insights into Linear Systems
Understanding why certain systems are inconsistent is vital in linear algebra. It highlights the importance of verifying the equations' compatibility before attempting to find solutions. In practical applications, inconsistent systems may represent conflicting conditions or impossible constraints.
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Key Takeaways:
- A linear system's solutions can be classified as unique, infinite, or nonexistent (inconsistent).
- Careful interpretation of the equations is crucial, especially when systems are presented with notation issues.
- Elimination and substitution are effective methods but require consistency among equations.
- Contradictions during solving indicate an inconsistent system with no solutions.
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Final Remarks
In conclusion, the process of solving the given linear system underscores the importance of clarity, systematic approach, and verification at each step. While the specific system analyzed turns out to be inconsistent, the methodology demonstrated—rewriting equations, elimination, substitution, and contradiction analysis—is foundational in linear algebra. For students and professionals, mastering these techniques enables the effective resolution of a broad array of systems encountered in academic and real-world scenarios.
If you encounter similar problems, always verify the system's formulation, carefully interpret the equations, and methodically proceed through elimination or substitution methods to determine the nature of the solutions. Whether the system has a unique solution, infinitely many, or none at all, understanding the underlying principles is essential for accurate analysis and application.
