I couldn't figure out how to solve these two problems and have to show ALL my work for both, if someone could help me I would appreciate it thanks!

1. The perimeter of a rectangular field is 300 meters. This field is to be fenced along three sides only. The fencing along it's two lengths and one width is to be done at $50 per meter. If the total cost of the fencing is $12,000, what are the field's dimensions?

2. The admission fee at a fair is $2 for children and $4 for adults. On a certain day, 2,000 people enter the fair and $6,000 is collected. How many children and how many adults went to see the fair on that day?

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1. The rectangular field's perimeter can be represented by an equation:

2x + 2y = 300m (perimeter). Since the fence will not span one side, we can say the length of the fence could be 2x + y = f (fence).

Since we know each meter of fence costs $50 & the total cost is $12,000, we can find the length of the fence, in meters:

12,000/50 = 240

Now substitute 240m for the f in the fence equation & subtract the length of the fence from the entire perimeter:

2x + y = 240

(2x-2y)-(2x-y)=y

300-240=y

y=60

Thus the "missing" side is 60 meters long & the "y" side of the fence is 60 meters long.

Next substitute 60 for the y in the fence equation & solve for x:

2x+60=240

2x=240-60

x=180/2

x=90

The field is 60x90.

2. In this one we'll use x to represent a number of people & we'll designate them as children & adults by the dollar amounts:

2x+4x=$6000

Now solve for x:

6x=$6000

x=6000/6

x=1000

Check your answer:

1000+1000=2000

There were 1000 adults & 1000 children.

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