Solve the following system [ f(k+1)+g(k)=5 , f(k)-g(k+1)=3 ,f(0)=g(0)=1]?

Answer 1

Let f(x)=ax+b
g(x)=cx+d

Then
f(0)=b=1
g(0)=d=1

f(k+1)=a(k+1)+b
g(k+1)=c(k+1)+d

a(k+1)+b+ck+d=5
ak+b-(c(k+1)+d)=3

Answer 2

We can re-write this in the following way:
f(k+1) = 5 - g(k)
g(k+1) = f(k) - 3
f(0) = 1 = g(0)

We can get:
g(1) = f(0) - 3 = 1 - 3 = -2
f(1) = 5 - g(0) = 5 - 1 = 4

We can separate the f from the g: Set j = k-1:
f(j+2) = 5 - g(j+1) = 5 - (f(j) - 3) = 8 - f(j)
g(j+2) = f(j+1) - 3 = 5 - g(j) - 3 = 2 - g(j)

So we get:
f(j+2) = 8 - f(j)
g(j+2) = 2 - g(j)
f(0) = 1 = g(0)

For even values of j:
j = 0, 2, 4, 6, 8
f = 1, 7, 1, 7, 1
g = 1, 1, 1, 1, 1

For odd values of j:
j = 1, 3, 5, 7
f = 4, 4, 4, 4
g = -2, 4, -2, 4

So the full sequences are:
j: 0 1 2 3 4 5 6 7 8
f: 1 4 7 4 1 4 7 4 1
g: 1 -2 1 4 1 -2 1 4 1

So they each do a 4-step cycle:
f(j) does: 1, 4, 7, 4, 1, 4, 7, 4..
g(j) does: 1, -2, 1, 4, 1, -2, 1, 4...

Answer 3

The "solution" by itiogrenci is uniquely stupid. There is no justification for assuming f and g are linear functions.