Q. 145.0( 6 Votes )
For what value of k does the system of equations
x + 2y = 3, 5x + ky + 7 = 0
have (i) a unique solution, (ii) no solution?
Also, show that there is no value of k for which the given system of equations has infinitely many solutions.
Answer :
(i) Given: x + 2y = 3 – eq 1
5x + ky + 7 = 0 – eq 2
Here,
a1 = 1, b1 = 2, c1 = - 3
a2 = 5, b2 = k, c2 = 7
Given systems of equations has a unique solution
∴ ≠
≠
k ≠10
∴ k ≠ 10
(ii) Given: x + 2y = 3 – eq 1
5x + ky + 7 = 0 – eq 2
Here,
a1 = 1, b1 = 2, c1 = - 3
a2 = 5, b2 = k, c2 = 7
Given that system of equations has no solution
∴ =
≠
Here,
=
Here,
k = 10
∴ k = 10
For the system of equations to have infinitely many solutions
=
=
=
=
which is wrong.
That is, for any value of k the give system of equations cannot have infinitely many solutions.
Rate this question :






















Find the value of k for which each of the following systems of equations has no solution:
kx + 3y = k - 3,12x + ky = k.
RS Aggarwal - MathematicsFind the value of k for which each of the following systems of equations has no solution:
kx + 3y = 3,12x + ky = 6.
RS Aggarwal - MathematicsFind the value of a for which the following system of equations has unique solution:
4x + py + 8 = 0, 2x + 2y + 2 = 0
KC Sinha - MathematicsA man purchased 47 stamps of 20 p and 25 p for Z 10. Find the number of each type of stamps.
RS Aggarwal - Mathematics