Math help?

Question

Find four consecutive integers such that the sum of the two greatest is 17 less than twice the sum of the two smallest.

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Hi everyone, I could reall appperciate some help here, I just can't figure out this one. Help will greatly be apperciated.

Answer 1

since the integers are consecutive you can write them as
a, a+1, a+2, a+3
which implies that
a+2+a+3=2(a+a+1)-17
2a+5=4a-15
20=2a
so a=10
and the integers are 10,11,12,13
and you can check
12+13=25
10+11=21 , 21 times 2 is 42 which is 17 more than 25

Answer 2

Let
x = first integer
x + 1 = 2nd integer
x + 2 = 3rd integer
x + 3 = 4th integer
(x + 2) + (x + 3) = "sum of the two greatest"
x + (x + 1) = "sum of the two smallest"
2[x + (x + 1)] = "twice the sum of the two smallest"
2[x + (x + 1)] - 17 = "17 less than twice the sum of the two smallest"


From the problem, we know that the "sum of the two greatest" is "17 less than twice the sum of the two smallest". so:

(x + 2) + (x + 3) = 2[x + (x + 1)] - 17



Now, we can simplify

2x + 5 = 2(2x + 1) - 17

2x + 5 = 4x + 2 - 17



Transpose the x's to the left, and the constants to the right

2x - 4x = 2 - 17 - 5

-2x = -20



Therefore,

x = 10

x + 1 = 11

x + 2 = 12

x + 3 = 13



Therefore, the 4 integers are 10, 11, 12 and 13.
^_^

Answer 3

let the
first # = x
second # = x+1
third # = x+2
fourth # = x+3
then set an equation
(x+2+x+3)-17=2(x+x+1)
solve the prenthes
x+2+x+3-17=2x+2x+2
combine like terms
2x-12=4x+2
get x by itself
add 12 to both sides which cancel out -12 on the left side and it becomes
2x=4x+14
now subtract 4x from both sides which cancel out 4x from the right side and it becomes
-2x=14
now divide both sides by -2 and you will get
x=-7
so
first # = -7
second # = x+1
-7+1= -6
third # = x+2
-7+2= -5
fourth # = x+3
-7+3= -4
you can even check it by pluging in the x
(x+2+x+3)-17=2(x+x+1)
(-7+2-7+3)-17=2{-7(+)-7+1}
(-9)-17= -14-14+2
-26= -26

Answer 4

let x = the first
let x+1 = the second
let x+2 = the third
let x+3 = the fourth

set it up as and equation:

x+2+x+3=2(x+x+1)-17
2x+5=4x-15
5=2x-15
20=2x
x=10
x+1=11
x+2=12
x+3=13

Answer 5

integers n, n+1, n+2, n+3
2(n+n+1)=n+2+n+3+17
4n+2=2n+22
2n=20
n=10

integers are 10, 11, 12, 13

check
2(10+11)-17=12+13
2*21-17=25
42-17=25
25=25

Answer 6

10, 11, 12, 13.
12+13=25
10+11=21x2=42
42-25=17

Answer 7

let the integers be x.+1,x+2 and x+3
given x+2+x+3=2(x+x+1)-17
=>2x+5=4x+2-17
adding -5 and simplifying
2x+5-5=4x-15-5
2x=4x-20
adding -4x
-2x=-20
dividing by -2
x=10
so the integers are 10,11,12,13

Answer 8

2(x)(x+1)=(x+2)(x+3) - 17

x, x+1, x+2, ...etc would be consecutive, right? So the two higher ones, minus 17, are equal to the two smaller ones times two.

Multiply and solve for x, that'll give you the lowest of the four consecutive #s.... make sense?