Question

Find conditions on k that will make the following system of equations have a unique solution. 3kx+2y = 2
12x+2ky = 4
and what is the unique solution?

Answer 1

Solve the two equations in terms of k first.

Answer 2

ok. should i put it into a matrix form? or solve for k independetly

Answer 3

Multiply the first equation by k and subtract the second equation from it to eliminate y.

Answer 4

hmmm i got 3k^2-12x = 2k-4.

Answer 5

or could it be k=4/11

Answer 6

let me try a different approach:
3kx+2y = 2 ----- (1)
12x+2ky = 4 ----- (2)
Multiply (1) by k
3k^2x + 2ky = 2k ----- (3)
Compare (3) and (2). For these two equations not to be a dependent
3k^2 must not equal 12
k^2 must not be 4
k must not be +/- 2

The next thing to try is to find the slope of the first equation and the slope of the second equation and they should not be parallel because parallel lines don't intersect and they won't have a unique solution. You may end up with the same constraint on k we got earlier but it is still worth doing it and finding out.

Answer 7

hmm. ok. i understand the first part. but I am not really understanding why should we obtain the slope

Answer 8

Both equations are straight lines. If we plot them and they intersect that means they have a unique solution. That means we have one point, an x,y pair, that is on both lines which will be the unique solution. But if we graph the two lines and find they are parallel (meaning their slopes are the same) the lines will never intersect and so there will not be a unique solution.

BTW, I did that and found out it gives the same constraint that k should not be +2 or -2.

Answer 9

ohhh i see now

Answer 10

So we found the condition on k. k must not be -2 or +2 for the set of equations to have a unique solution.

You still have to solve for x and y in terms of k.

Subtract (2) from (3) and that will eliminate y and you can solve for x.
Then put that x in and solve for y.

Answer 11

alright. so i got 3k^2-12x= 2ky-4. I am not quite sure on how to solve for x, because I am unsure of K

Answer 12

3k^2x + 2ky = 2k ----- (3)
12x + 2ky = 4 ----- (2)
subtract
(3k^2 - 12)x = 2k - 4
3(k^2 - 4)x = 2(k - 2)
3(k + 2)(k - 2)x = 2(k - 2)
3(k + 2)x = 2
x = 2 / (3(k+2))

You should be able to find y from (1) and simplify.

Answer 13

ok, so in that case can k be equal to anything in order to solve for y?

Answer 14

You will find y in terms of k much like I found x in terms of k.

Answer 15

that is brutal. ok, so in that case. y = (3k^2-2k)/-2k)

Answer 16

I am getting y = 2 / (k + 2)

Answer 17

can you show me the steps?

Answer 18

sorry that I have to continuously ask you questions.

Answer 19

3kx+2y = 2
2y = 2 - 3kx
x = 2 / (3(k+2))
2y = 2 - 3k * 2 / (3(k+2))
= { 2(3(k+2)) - 6k } / (3(k+2))
= { 6k + 12 - 6k } / (3(k+2))
= 12 / (3(k+2))
= 4/ (k + 2)
y = 2 / (k + 2)

Answer 20

So the unique solution is: x = 2 / (3(k+2)) and y = 2 / (k + 2)

Just to make sure there were no calculation mistake, pick a couple of values for k (anything other than -2 or + 2 which we determined earlier are not allowed) and solve the equation and see if it agrees with the answer we got for x and y.

Answer 21

If I pick k = 0, the original equations become:
2y = 2 ===> y = 1
12x = 4 ===> x = 1 / 3

And if I put k = 0 in the solution x = 2 / (3(k+2)) and y = 2 / (k + 2) we get
x = 2/6 = 1/3 and y = 2/2 = 1 which agrees with above.

Try k = 1 and see if it is still good.

Answer 22

ohhhh ok. thats awesome.thank you very much

Answer 23

you are welcome.