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.
D) 47,000
0.60 (<em>x</em>+38,000)= 51,000
(0.60<em>x</em>)+22,800= 51,000
51,000-22,800= 28,200
28,200÷0.60= 47,000
0.06 (47,000+38,000)= 51,000
28,200+22,800= 51,000
Answer:
= 14X + 6
Step-by-step explanation:
Answer:
Exact form: -7/10
Decimal form: -0.7
Step-by-step explanation:
Isolate the variable by dividing each side by by factors that don't contain the variable.
