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Consider the system of simultaneous equations
x + 2y + z = 6
2x + y + 2z = 6
x + y + z = 5
This system has
x + 2y + z = 6
2x + y + 2z = 6
x + y + z = 5
This system has
- a)Unique solution
- b)Infinite number of solutions
- c)No solution
- d)Exactly two solution
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Consider the system of simultaneous equationsx + 2y + z = 62x + y + 2z...
(c ) 
∴ rank(A) = 2 ≠ 3 = rank(
) .
) .∴ The system is inconsistent and has no solution.
Most Upvoted Answer
Consider the system of simultaneous equationsx + 2y + z = 62x + y + 2z...
Given system of equations:
The given system of simultaneous equations is:
x + 2y + z = 6 ...(1)
2x + y + 2z = 6 ...(2)
x + y + z = 5 ...(3)
Explanation:
To determine the solution of the given system of equations, we can use the method of elimination or substitution.
Using Elimination Method:
- Let's start by eliminating the variable 'x' from equations (1) and (2).
- Multiply equation (1) by 2 and equation (2) by 1, to make the coefficients of 'x' equal in both equations.
- We get:
2x + 4y + 2z = 12 ...(4)
2x + y + 2z = 6 ...(5)
- Subtract equation (5) from equation (4) to eliminate 'x':
(2x + 4y + 2z) - (2x + y + 2z) = 12 - 6
3y = 6
- Simplifying the equation, we have:
3y = 6
y = 2
- Now substitute the value of 'y' back into equation (3):
x + y + z = 5
x + 2 + z = 5
x + z = 3 ...(6)
- Substituting the values of 'y' and 'z' into equation (6), we get:
x + 2 + z = 3
x + z = 1 ...(7)
- Now, we have two equations:
x + z = 1 ...(7)
x + z = 3 ...(6)
No Solution:
- Comparing equations (6) and (7), we can see that the coefficients of 'x' and 'z' are the same, but the constants on the right side are different.
- Since the coefficients are the same, the lines represented by these equations are parallel.
- Therefore, there is no solution that satisfies all three equations simultaneously.
- Hence, the correct answer is option 'C' - No solution.
The given system of simultaneous equations is:
x + 2y + z = 6 ...(1)
2x + y + 2z = 6 ...(2)
x + y + z = 5 ...(3)
Explanation:
To determine the solution of the given system of equations, we can use the method of elimination or substitution.
Using Elimination Method:
- Let's start by eliminating the variable 'x' from equations (1) and (2).
- Multiply equation (1) by 2 and equation (2) by 1, to make the coefficients of 'x' equal in both equations.
- We get:
2x + 4y + 2z = 12 ...(4)
2x + y + 2z = 6 ...(5)
- Subtract equation (5) from equation (4) to eliminate 'x':
(2x + 4y + 2z) - (2x + y + 2z) = 12 - 6
3y = 6
- Simplifying the equation, we have:
3y = 6
y = 2
- Now substitute the value of 'y' back into equation (3):
x + y + z = 5
x + 2 + z = 5
x + z = 3 ...(6)
- Substituting the values of 'y' and 'z' into equation (6), we get:
x + 2 + z = 3
x + z = 1 ...(7)
- Now, we have two equations:
x + z = 1 ...(7)
x + z = 3 ...(6)
No Solution:
- Comparing equations (6) and (7), we can see that the coefficients of 'x' and 'z' are the same, but the constants on the right side are different.
- Since the coefficients are the same, the lines represented by these equations are parallel.
- Therefore, there is no solution that satisfies all three equations simultaneously.
- Hence, the correct answer is option 'C' - No solution.
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Consider the system of simultaneous equationsx + 2y + z = 62x + y + 2z = 6x + y + z = 5This system hasa)Unique solutionb)Infinite number of solutionsc)No solutiond)Exactly two solutionCorrect answer is option 'C'. Can you explain this answer?
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