Question

Hi Openstudy, I want to solve two coefficients:
2A + 2B = 6 and
2A + 2B = 4

i know i am usually suppose to subtract them but this is not linearly independent.

My friend suggested that I substitute A to equal 1, so my equation will become 1 = (6-2B)/2
which would lead B to equal 2.

My brother told me that this question is unsolvable, but i need to be able to solve this for Lando's equation. Can anyone please tell me which is true?

Answer 1

There is no solution. Think about it this way: how can (2A+2B) be equal to two different numbers? That would mean that 6 = 4.

Answer 2

my other friend said that same exact thing

Answer 3

hm what if i solve for A so it becomes A = (6-2B) / 2
then substitute it in the second problem?

Answer 4

Even if you use the substitution method, you won't get B = 2, you'll get 6 = 4. I think you made a mistake in your substitution.

A = (6-2B)/2 = 3 - B

put that into your other equation, and you'll still get 6 = 4.

2(3-B) + 2B = 4
6 - 2B + 2B = 4
6 = 4

Answer 5

ic..

Answer 6

thank you!

Answer 7

hey while we are on the topic, i need this answer to solve for the "2nd order homogenous linear recurrence relation with constant coefficients", you don't happen to know what I should do if this occurs do you?

Answer 8

hmm, sorry I can't help with that, too tired to look into it right now. This might help though?
http://people.uncw.edu/tompkinsj/133/recursion/homogeneous.htm

Answer 9

ill look into it, thanks a lot!