Consecutive Integers: Find four cons. integers such that....

vonsmiley

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I'm back . . . no tears please. I've shed enough over this for all of us.
Can someone help me with this.

Find four consecutive integers such that, twice the sum of the two larger integers, exceeds three times the first by ninety-one.

. . .x = 2(3x + 4x) - 91
. . .x = 6x + 8x - 91
. . .x = 14x - 91

. . .-13x/13 = -91/13

. . .x = 7

Checking:

. . .2(3(7)+ 4(7)) - 91
. . .98 - 91 wrong

What am I doing wrong in the setup?

by the way, the answers are 81, 82, 83, and 84. Thank you!
 

stapel

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vonsmiley said:
x = 2(3x + 4x) -91
What does "x" stand for? How did you arrive that this equation?

For instance, if the exercise had said the following:

Find three numbers, where the second is three times as large as the first and the third is four times as large as the first, such that the difference of twice the sum of the second and third and 91 is equal to the first.
...then your equation would have fit:

. . . . .first number: x
. . . . .three times as large: 3x
. . . . .four times as large: 4x
. . . . .sum: 3x + 4x
. . . . .twice the sum: 2(3x + 4x)
. . . . .difference: 2(3x + 4x) - 91
. . . . .equals: 2(3x + 4x) - 91 = x

But this isn't what the exercise said, as near as I can tell. ("Consecutive integers", of course, are one apart from each other, not many times the size of each other, is part of my confusion.)

Please reply showing your step-by-step reasoning, as demonstrated above. Thank you.

Eliz.
 

vonsmiley

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I thought 4 Consecutive numbers if the first is x which is the same as 1x then the next would be 2x . . . 3x . . .
 

pka

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vonsmiley said:
I thought 4 Consecutive numbers if the first is x which is the same as 1x then the next would be 2x . . . 3x . . .
If the first is x then the next one is (x+1), the next one is (x+2), and the next one is (x+3).
 

vonsmiley

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x, x+1, x+2, x+3, x+4 <--I see my mistake here

x = 2[(x + 3) + (x+4)] -91
x= 4x + 14 -91
x = 4x -77
3x = 77
x = 25.666 wrong

I still don't understand what I am doing wrong over all.
 

pka

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vonsmiley said:
x, x+1, x+2, x+3, x+4 <--I see my mistake here
NO! Four consecutive integers are x, (x+1), (x+2), and (x+3)!
Count then 1, 2, 3, 4.
You wrote five numbers down.

2[(x+2)+(x+3)]=3x+91
 

vonsmiley

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:oops: but even with this change I'm still getting the wrong answer :shock:
 

pka

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SOLVE: 2[(x+2)+(x+3)]=3x+91.
 

vonsmiley

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Okay, but one last question the problem says exceeds 3x the firs By 91 how did you know it wasn't

2[(x+2)+(x+3)] - 91 <-- with 91 being the 3 times the first
the byis confusing. How did you know it meant plus instead of the 91 being the 3times
 

pka

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Look you must learn some basic facts.
a+b+c=0 is the same as a+b=-c is the same as a=-b-c.

So 2[(x+2)+(x+3)]=3x+91 is the same as 2[(x+2)+(x+3)]-91=3x is the same as 2[(x+2)+(x+3)]-3x=91.
 

vonsmiley

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thank you Eliz and Pka :D
 

stapel

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vonsmiley said:
...the problem says exceeds 3x the first by 91...
"This exceeds that by so much" means "(this) minus (that) equals (so much)".

Eliz.
 

vonsmiley

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Thank you Eliz :D
 
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