ab+bc+cd=6 bc+cd=4 find ab+bc

Question

 ab+bc+cd=6

bc+cd=4

ab+cd=?

Answer 1

We have 4 unknowns in 2 equations, so there is no unique solution.

For real numbers there are infinite solutions.

If we limit a,b,c,d to positive integers, we can find a few that work:

Subtract the second equation from the first and we get

ab = 2

Then for positive integers, either a = 1, b = 2 or vice-versa.

Case 1: a=1, b=2
plug into bc+cd=4
2c+cd=4
c(2+d)=4
The only combinations of a product of two positive integers to get 4 are:
1x4, 4x1, 2x2
The only one that works with c(2+d) is 1x4
c=1, d=2
Therefore we get a solution:
(a,b,c,d)=(1,2,1,2) and ab + bc = 4

case 2: a=2, b=1
plug into bc+cd=4
c+cd=4
c(1+d)=4
The only combinations if c and d are positive integers are 1x4=4 and 2x2=4
c=1, d=3
c=2, d=1
Therefore we get two more solutions:
(a,b,c,d)=(2,1,1,3) and ab + bc = 3
(a,b,c,d)=(2,1,2,1) and ab + bc = 4

If you allow a,b,c,d to be negative integers and 0 you get more solutions.
If you allow a,b,c,d to be real numbers, you get infinite solutions.

Answer 2

4.

 

A = 2

B = 1

C = 2

D = 1