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Urgent help needed [Math Question]

Picnore

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Hey guys, I urgently need this solved.
The problem is:

"The Sum of three consecutive odd integers is -39. Find the numbers."

So far I did:

let i represent the integers

1i + 3i + 5i = -39

----------------------------------------

Now Im stuck. Did I do it correctly to far?
 
No, you are going about this the wrong way. The correct way would be to say
Code:
Let a = the first integer
Let b = the second integer
Let c = the third integer
a > b
b > c
|a| < |b|
|b| < |c|
a - b = 2
b - c = 2
a + b + c = -39

We can now solve this system of equations for a.
Code:
a + b + c = -39
a = -39 - b - c
Let us substitute this value of a into the first equation.
Code:
a - b = 2
(-39 - b -c) - b = 2
-39 - 2b - c = 2
c = -2b - 41
Let us now substitute this value of c into the second equation.
Code:
b - c = 2
b - (-2b - 41) = 2
b + 2b + 41 = 2
3b + 41 = 2
3b = -39
b = -13
Now we can substitute this value for b into the first two equations to find a and c.
Code:
a - b = 2
a - (-13) = 2
a + 13 = 2
a = -11
And
Code:
b - c = 2
-13 - c = 2
-c = 15
-1(-c) = -1(15)
c = -15

a = -11, b = -13, c = -15

Tada!
(Also, this really should have gone in the homework help section located here).
 
No, you are going about this the wrong way. The correct way would be to say
Code:
Let a = the first integer
Let b = the second integer
Let c = the third integer
a > b
b > c
|a| < |b|
|b| < |c|
a - b = 2
b - c = 2
a + b + c = -39

We can now solve this system of equations for a.
Code:
a + b + c = -39
a = -39 - b - c
Let us substitute this value of a into the first equation.
Code:
a - b = 2
(-39 - b -c) - b = 2
-39 - 2b - c = 2
c = -2b - 41
Let us now substitute this value of c into the second equation.
Code:
b - c = 2
b - (-2b - 41) = 2
b + 2b + 41 = 2
3b + 41 = 2
3b = -39
b = -13
Now we can substitute this value for b into the first two equations to find a and c.
Code:
a - b = 2
a - (-13) = 2
a + 13 = 2
a = -11
And
Code:
b - c = 2
-13 - c = 2
-c = 15
-1(-c) = -1(15)
c = -15

a = -11, b = -13, c = -15

Tada!
(Also, this really should have gone in the homework help section located here).

Wow. This is waay harder.
The method I learnt is school is 30x simpler.
I just cant find it in my textbook and I forgot it completely. Too much homework :(
 
What the christ, that's way too much work. The easiest way is letting x be the first integer.

x
x + 2
x + 4

So in the end you have

x + x + 2 + x + 4 = -39
3x + 6 = -39
3x = -45
x = -15

So the smallest is -15 then common sense tells you the other two are -13 and -11. You can also see this if you do x+2 and x+4.
 
What the christ, that's way too much work. The easiest way is letting x be the first integer.

x
x + 2
x + 4

So in the end you have

x + x + 2 + x + 4 = -39
3x + 6 = -39
3x = -45
x = -15

So the smallest is -15 then common sense tells you the other two are -13 and -11. You can also see this if you do x+2 and x+4.
Yeah, this is so much simpler than what Jimmy said.
 
What the christ, that's way too much work. The easiest way is letting x be the first integer.

x
x + 2
x + 4

So in the end you have

x + x + 2 + x + 4 = -39
3x + 6 = -39
3x = -45
x = -15

So the smallest is -15 then common sense tells you the other two are -13 and -11. You can also see this if you do x+2 and x+4.

This. Jimmy's way is too complicated.

However, if the last step isn't "common sense" to you: just plugin the -15 to x+2 and x+4 respectively, to get -13 and -11
 
wow Jimmy that was intense hahaha, Good job koot's you explained it well.
 
Or a simpler method... you did multiplication when it's really addition/subtraction of 2.

From: 1i + 3i + 5i = -39
To: (i+0)+(i-2)+(i-4) = -39
3i-6 = -39
3i = -33
i = -11
So: -11, -13, -15

I think the other guy did it the same way but... I felt like explaining what was wrong O_o

I like how Jimmy put it into like code pretty much ^^
 
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