#### Luvvolleyball15

##### New member

1) Robert ran to the redoubt while wilbur walked to the parapet. both distances were the same, but Robert's speed was 6 miles per hour whil Wilbur's was 8 miles per hour. What was the time of each if Robert's time was 2 hours longer than Wilbur's? Write the distance equation.

2) Penelope and miranda found four consecutive odd integers such that 5 times the sum of the first two was 5 less than 19 times the fourth. What were the integers?

#### Denis

##### Senior Member
Luvvolleyball15 said:
1) Robert ran to the redoubt while wilbur walked to the parapet. both distances were the same, but Robert's speed was 6 miles per hour whil Wilbur's was 8 miles per hour. What was the time of each if Robert's time was 2 hours longer than Wilbur's? Write the distance equation.
["they" now introduce "redoubts" and "parapets" to teach speed...boy oh boy...]

LVB, formula is: speed = distance divided by time;
and that don't change, parapets or wood sheds!

t = time it took Wilbur; so Robert's time = t+2
d = distance

Wilbur: 8 = d / t : d = 8t [1]
Robert: 6 = d / (t + 2) : d = 6(t + 2) [2]

[1][2]: 8t = 6(t + 2)
8t = 6t + 12

Now finish it, ok?

#### tkhunny

##### Moderator
Staff member
For 1), just do what it says: "Write the distance equation."

Distance = Rate * Time

Do this more specifically for each traveller and see what falls out.

For 2), you must define or name your integers. It doesn't matter what you call them, just call the SOMETHING so that you can talk about them.

"Four consecutive odd integers"

I'm tempted to ignore that they are odd and see if I really care. If we have to be more specific, we can be. I'll still separate them by 2. Think about what Integers are. Why didn't it ask for Whole Numbers? We might get negative values, but will we get fractions like 1/4 or 12/7?

The first we'll call 'x', just because that it a popular choice. You can call it 'Bruce' if you like, but 'x' is a little simpler to write and say.

What's the NEXT integer 2 greater than the first one? x+2
The third one? x+4
The fourth one? x+6

Great, now they all have names. What does the problem tell us?

"5 times the sum of the first two"

5*(x + (x+2))

"5 less than 19 times the fourth"

19*(x+6) - 5

OK, now what does it want?

"What were the integers?"

5*(x + (x+2)) = 19*(x+6) - 5
5x + 5(x+2) = 19x + 114 - 5
5x + 5x + 10 = 19x + 109
10x + 10 = 19x + 109
10 = 9x + 109
-99 = 9x
-11 = x

x+2 = -9
x+4 = -7
x+6 = -5

It turns out the only solution does, indeed, have ODD numbers, so it really wasn't necessary to define that more carefully. It just worked out.

Now check.

"5 times the sum of the first two was 5 less than 19 times the fourth""

5*(-11 + (-9)) = 19*(-5) - 5 ??
5*(-20) = -95 - 5 ??
-100 = -100 !! Yes! I think we're done.

#### Gene

##### Senior Member
1) d = vt
8t = 6(t+2)

2) The numbers are
2x+1, 2x+3, 2x+5, 2x+7
5(2x+1 + 2x+3)= 19*(2x+7) - 5
Hint: The numbers are negative.

#### Denis

##### Senior Member
Wrong, Gene! Numbers are:
2x-3, 2x-1,2x+1,2x+3

#### Gene

##### Senior Member
Don't confuse me.
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Gene