# Thread: find the value of k

1. ## find the value of k

find the value of k such that the system has infinite number of solutions

(k-1)x - y = 5
(k+1)x + (1-k)y = (3k+1)

plzz solve it wholly

thnks in advance

i am aquinted with the process of taking 3 cases and show the common value of k in all is the answer. so it will be helpful if u solve it with this process. i am having problem with factorising so plzz solve it wholly.

2. Originally Posted by saha.subham
find the value of k such that the system has infinite number of solutions

(k-1)x - y = 5
(k+1)x + (1-k)y = (3k+1)

plzz solve it wholly

thnks in advance

i am aquinted with the process of taking 3 cases and show the common value of k in all is the answer. so it will be helpful if u solve it with this process. i am having problem with factorising so plzz solve it wholly.
Someone will probably post a full solution after me, but I prefer to give a hint.

Consider the equations

x = y + 1
2x = 2y + 2

Note how this system has an infinite number of solutions.

So...

3. i no the process but just getting stuck in factorising so plzz anybody help me

4. Originally Posted by saha.subham
i no the process but just getting stuck in factorising so plzz anybody help me
Factorising???

Write

$\frac{k-1}{k+1}=\frac{-1}{1-k}=\frac{5}{3k+1}$

and solve for $k$.

5. Originally Posted by undefined
Factorising???

Write

$\frac{k-1}{k+1}=\frac{-1}{1-k}=\frac{5}{3k+1}$

and solve for $k$.
forget abt it plzz solve it fully

6. Have you even tried anything yourself? In particular, in the original system,
(k-1)x - y = 5
(k+1)x + (1-k)y = (3k+1)
if you multiply the first equation by 1- k and add to the second equation, what do you get?

7. i have solved all this type of sums accept 3 this is one of them plzz help

8. Originally Posted by saha.subham
i have solved all this type of sums accept 3 this is one of them plzz help
In what I wrote above, we have

$\frac{-1}{1-k}=\frac{5}{3k+1}$

Solve for $k$.. cross multiply.. easy stuff... plug the value you get back into the equations and see that it works out very nicely..

9. i understands it but i want to do all the sums by the same process
my process is
case1 -
for terms 1 and 2 i have to show the value of k

case 2

for terms 2 and 3 i have to show the value of k

case 3

for terms 1 and 3 i have to show the value of k

now the common value of k in all is the answer.

10. i have solved for case 2

for case 1 i have
k-1/k+1 = -1/1-k

for case 3

k-1/k+1 = 5/3k+1

plzz solve for k in both the case which i am not being able to do.

11. Hi saha.subham,

If $\frac{a}{b} = \frac{c}{d}$ then

ad = bc.

Similarly, can you simplify

$\frac{k-1}{k+1}$ = $\frac{-1}{1-k}$

12. Ah, it wasn't clear to me that you knew how to write the equations but not solve them. I did not know that the three cases you mentioned were the ones I wrote down. Please try to be a little more specific in the future, and show work; you would have gotten the guidance you needed a lot sooner that way.

I figured since solving systems of equations is more advanced than cross multiplying that you knew the procedure. Hopefully with sa-ri-ga-ma's post you will now have what you require to complete the problem.

Best regards.

Edit: Maybe by "factorising" you meant applying the distributive property? For example if you wanted to solve

$\frac{k-1}{k+1}=\frac{-1}{1-k}$

you would need to distribute as follows

$(k-1)(1-k)=-1(k+1)$

$k-k^2-1+k=-k-1$

Or further on, there is a place to factor out a $k$.

$3k-k^2=0$

$k(3-k)=0$

$k=0 \ \ \text{ OR }\ \ k=3$

Of course it's much easier to just solve

$\frac{-1}{1-k}=\frac{5}{3k+1}$

and plug in the value of $k$ you find to make sure it satisfies

$\frac{k-1}{k+1}=\frac{-1}{1-k}$

Anyway, this is a complete solution to one of the three cases you mentioned, so now you really should have all the tools you need to solve the other cases.