Find the value of k for which the system of linear equations
2x+5y=3
(k+1)x+2(k+2)y=2k
Will have infinite number of solutions.
Answers
Answered by
13
bonjour dear!
For there to be an infinite number of solutions, the
equation (k+1)x+2(k+2)y=2k must be a multiple of
the equation 2x+5y=3
Let the multiplier by "a".
Then the equation is 2ax+5ay=3a
Then the following set of equation must be true:
k+1=2a
2k=3a
Then k=3a/2
Substitute that into k+1=2a to get:
(3a/2)+1=2a
(1/2)a=1
a=2
If a=2 then 2k=3(2) and k=3
Try It:
Substituting for "k" you get:
(3+1)x+2(3+2)y=2(3)
4x+10y=6
This is a multiple of the equation 2x+5y=3
hope this helps'
keep smiling'
:)
Answered by
13
For infinite solution
2/k+1 = 5/2(k+2)=3/2k
Equating 1st one
4k+8=5k+5
=>k=3
Equating 2nd one
6k + 12=10k
=>k =3
So k =3 is the req. Ans
2/k+1 = 5/2(k+2)=3/2k
Equating 1st one
4k+8=5k+5
=>k=3
Equating 2nd one
6k + 12=10k
=>k =3
So k =3 is the req. Ans
Similar questions