Solve the system of equations
10b+5v+5z=55
3b+5v+5z=34
5b+6v+3z=42
I have gotten b=3 but I am stuck and Do not know how to proceed.
You're on the right track !!!!
10b+5v+5z=55 ⇒ 5v + 5z = 55 - 10b (1)
3b+5v+5z=34 ⇒ 5v + 5z = 34 - 3b (2)
5b+6v+3z=42 (3)
Note that (1) must = (2)....therefore
55 - 10b = 34 - 3b rearrange as
55-34 = 7b
21 = 7b
b = 3
Knowing this (1) becomes
10(3) + 5v + 5z = 55
30 + 5v + 5z = 55
5v + 5z = 25 [divide through by 5 ]
v + z = 5 → z = 5 - v (4)
Sub (4) into (3) and sub 3 for b and we have that
5(3) + 6v + 3(5 - v ) = 42
15 + 6v + 15 - 3v = 42
30 + 3v = 42
3v = 12
v = 4
So z = 5 - v = 5 - 4 = 1
So { b, v , z} = ( 3, 4 , 1}