Math help please!?

Question

Can someone tell me how to solve the following two questions?

1. A rectangular city lot has a perimeter of 540 ft. The width of the lot is 150 ft less than its length. Find the length and width.

2. Find three consecutive odd integers such that the sum of the least and greatest is 78.

Thanks!

Answer 1

1. You have the perimeter. Expressed as an equation, you have 2w+2L=540 (w=width, L=length). You also know that the width is 150 less than the length. As an equation, this is expressed as w=L-150.

Substitute for w in the first equation and you have
2(L-150)+2L=540

Multiply the 2 through the first quantity to get
2L-300+2L=540

Combine like terms to get
4L-300=540

Add 300 to both sides and then divide by 4 tp solve for L. Once you have the answer for L, put it back into the first equation and you will be able to find the answer for w.

2. You know that even numbers are divisible by 2 and that any number multiplied by 2 is even. You also know that any even number plus one will give you an odd number. Let's say x is any number. you know that 2x will be even. If you add 1 to 2x you'll have an equation for any odd number: 2x+1.

So let's say the equation for the first odd integer is 2x+1, then the second equation for the second odd integer is 2x+3 and then for the third odd integer you have the equation 2x+5

You are give that the sum of the first integer and last integer equals 78.

As an equation you will have 2x+1+2x+5=78, which reduces down to 4x+6=78.

Subtract 6 from both sides to get 4x=72.

Divide each side by 4 to solve for x and then plug the answer into the three equations for the integers: 2x+1, 2x+3, 2x+5. Then you will have your integers.

Answer 2

Let's work the 1st problem with one variable.

Let w = width of the lot
Then w + 150 = length of the lot

The perimeter is the sum of the sides of the rectangle. So,
w + (w + 150) + w + (w + 150) = 540

4w + 300 = 540

4w = 240

w = 60 = width
w + 150 = 210 = length

Check: 60 + 210 + 60 + 210 = 540

Consecutive odd (even) integer problems can always be set up as follows:

Let n = 1st odd integer (least)
Then n + 2 = 2nd odd integer (odd integers are 2 units apart.)
and n = 4 = 3rd odd integer (greatest)

Now the equation: n + (n + 4) = 78

2n + 4 = 78

2n = 74

n = 37

n + 2 = 39

n + 4 = 41

Check: 37 + 41 = 78

Answer 3

1. 540 = 2L + 2w
540 = 2L + 2(L-150)
540 = 2L + 2L - 300
540 + 300 = 2L + 2L - 300+300
840 = 4L
840/4 = 4L/4
210= L, therefore

210-150 = w
60 = w

Length =210, width = 60

2. 78 = 37+41
Answer: 37, 39, 41

Answer 4

2nd

37 39 41

Answer 5

Sorry Can't help but look it up in the internet sites and get a tutor or something.