Question

If 2a+3b+5c+7d=34, 3(d+c)=b, 3b+c=a, c-1=d, what is a*b*c*d?

Answer 1

Answer:

a=3 b =5 c=16 d =24

Step-by-step explanation:

sorry I don't have step by step explanation

Answer 2

Answer:

0

Step-by-step explanation:

Let simply everything to c

Since 3(d+c) = 3d+3c=b (3d where d = c-1 so 3c-3+3c = b) b= 6c-3

And 3b+c = a Since a = 3b+c we will do 3(6c-3) which gives us 18c-9+c which is 19c-9

C-1=d

so, 2(19c-9)+3(6c-3)+5c+7(c-1) = 68c-34 = 34

68c = 68

68/68 = 1c/1 = 1

c=1 so let solve for a

you can do c-1=d so 1-1=d = 0 and everyone knows anything times 0 = 0