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For what value of ‘K’ will the system of equations: 3x + y = 1, (2K – 1) x + (K – 1) y = 2K + 1 have no solution
- a)3
- b)2
- c)1
- d)-2
Correct answer is option 'B'. Can you explain this answer?
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For what value of ‘K’ will the system of equations:3x + y ...
is a case of parallel lines which never meet. So there are no solutions obtainable for these equations. So equations are inconsistent3x + y = 1, (2K – 1) x + (K – 1) y = 2K + 1
b1=1,b2=k-1,c1=-1,c2=-2k-1
Most Upvoted Answer
For what value of ‘K’ will the system of equations:3x + y ...
Solution:
To find the value of K for which the given system of equations has no solution, we can use the determinant method.
1. Write the system of equations in matrix form:
$\begin{pmatrix}3 & -1 \\ 2K-1 & K-1 \\\end{pmatrix} \begin{pmatrix}x \\ y \\\end{pmatrix} = \begin{pmatrix}1 \\ 2K-1 \\\end{pmatrix}$
2. Find the determinant of the coefficient matrix:
$\begin{vmatrix}3 & -1 \\ 2K-1 & K-1 \\\end{vmatrix} = (3)(K-1) - (-1)(2K-1) = 3K - 4$
3. If the determinant is not equal to zero, then the system has a unique solution. If the determinant is equal to zero, then the system either has no solution or infinitely many solutions.
4. Set the determinant equal to zero and solve for K:
$3K - 4 = 0$
$K = \frac{4}{3}$
5. Since the determinant is not equal to zero for any value of K except K = 4/3, the system has no solution only for K = 4/3.
Therefore, the correct answer is option B) 2.
To find the value of K for which the given system of equations has no solution, we can use the determinant method.
1. Write the system of equations in matrix form:
$\begin{pmatrix}3 & -1 \\ 2K-1 & K-1 \\\end{pmatrix} \begin{pmatrix}x \\ y \\\end{pmatrix} = \begin{pmatrix}1 \\ 2K-1 \\\end{pmatrix}$
2. Find the determinant of the coefficient matrix:
$\begin{vmatrix}3 & -1 \\ 2K-1 & K-1 \\\end{vmatrix} = (3)(K-1) - (-1)(2K-1) = 3K - 4$
3. If the determinant is not equal to zero, then the system has a unique solution. If the determinant is equal to zero, then the system either has no solution or infinitely many solutions.
4. Set the determinant equal to zero and solve for K:
$3K - 4 = 0$
$K = \frac{4}{3}$
5. Since the determinant is not equal to zero for any value of K except K = 4/3, the system has no solution only for K = 4/3.
Therefore, the correct answer is option B) 2.
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For what value of ‘K’ will the system of equations:3x + y ...
Yes option b is correct.......first put the value of k as 2 in equation II .......i.e..........(2×2-1)x +(2-1)y = 2(2)+ 1.............. solving that we get.........3x+y=5.............. that can be written as...3x+y-5=0..........now the ratio of a1/ a2=3/3=1 ........b1/ b2= 1/1=1..... but C1/C2 = -1/-5= 1/5 ................ you can see that a1/a2=b1/b2 not equals C1/C2............. as in this type of equation pairs have no solution..........hence for value of k be 2 equations will have no solution......
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