Question

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.

Answer 1

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'

:)

Answer 2

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