Answer:
Step-by-step explanation:
Let's call hens h and ducks d. The first algebraic equation says that 6 hens (6h) plus (+) 1 duck (1d) cost (=) 40.
The second algebraic equations says that 4 hens (4h) plus (+) 3 ducks (3d) cost (=) 36.
The system is
6h + 1d = 40
4h + 3d = 36
The best way to go about this is to solve it by substitution since we have a 1d in the first equation. We will solve that equation for d since that makes the most sense algebraically. Doing that,
1d = 40 - 6h.
Now that we know what d equals, we can sub it into the second equation where we see a d. In order,
4h + 3d = 36 becomes
4h + 3(40 - 6h) = 36 and then simplify. By substituting into the second equation we eliminated one of the variables. You can only have 1 unknown in a single equation, and now we do!
4h + 120 - 18h = 36 and
-14h = -84 so
h = 6.
That means that each hen costs $6. Since the cost of a duck is found in the bold print equation above, we will sub in a 6 for h to solve for d:
1d = 40 - 6(6) and
d = 40 - 36 so
d = 4.
That means that each duck costs $4.
Okay. For these types of problems, you must do order of operations (PEMDAS). Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction. Mind you that you do these steps from left to right, and multiplication and division is done from left to right. Same thing with addition and subtraction. With that being said, here are your answers if you do the expressions correctly.
1. 12
2. 106
3. 42
Answer:
y = -4
Step-by-step explanation:
See attached worksheet.