Answer:
b = c = 2, z = 5
Explanation:
You want the solution to the equations ...
- 2b + c +z = 11
- 3b + 4c + z = 19
- 3b + 6c + 5z = 43
Solution
A number of methods are available for solving this system of equations. We like the matrix method using a calculator, as shown in the attachment.
Since 'c' and 'z' appear with coefficients of 1, writing expressions for those that can be substituted into the other equations is pretty easy:
z = 11 -(2b +c) . . . . . . . using the first equation for an expression for z
3b +4c +(11 -2b -c) = 19 ⇒ b +3c = 8 . . . . . sub for z in 2nd equation
3b +6c +5(11 -2b -c) = 43 ⇒ -7b +c = -12 . . . . sub for z in 3rd equation
-7(8 -3c) +c = -12 ⇒ 22c = 44 ⇒ c = 2 . . . . sub (8-3c) for b in 3rd eqn
b = 8 -3(2) = 2
z = 11 -(2(2) +2) = 5
The solution is (b, c, z) = (2, 2, 5).