Find three consecutive integers...?

Question

Find three consecutive integers such that the sum of the second and third exceeds one-half of the first by 33.

Answer 1

The answer is 20, 21 and 22. How?

Let the numbers = x, x+1 and x+2.
x+1 + x+2 = x/2 + 33. (As per the given conditions)
= 2x + 3 = x/2 + 33
By multiplying both sides by 2, we get,
= 4x + 6 = x + 66
= 4x - x = 66 - 6
= 3x = 60
So x = 60/3 = 20. When the first number = 20, then
other numbers are 21 and 22. So,
the numbers are = 20, 21, 22

Answer 2

The three consecutives numbers are 20, 21, 22.

let x be the first integer
let x + 1 be the second integer
let x + 2 be the third integer

x + 1 + x + 2 = (1/2)x+ 33
2x + 3 = (1/2)x + 33 (combine the common variables)
2x = (1/2) x+ 30 (bring the three over to the other side)
3/2 x = 30 (subtract the (1/2)x on both sides)
3x = 60 (multiply both sides by two to get rid of the fraction)
x = 20 (then divide both sides by three)

Answer 3

Let x be the first consecutive integer. It follows that the second and third would be x + 1 and x + 2 respectively.

We're also given that the sum of the second and third (i.e. [x + 1] and [x + 2]) exceeds one half of the first (x) by 33.

[x + 1] + [x + 2] + 33 = (1/2)x

Simplifying,

2x + 3 + 33 = (1/2)x
2x + 36 = (1/2)x

Multiply both sides by 2 to get rid of that fraction, to get

4x + 72 = x

3x = -72
x = -24

Therefore, the three consecutive integers are -24, -23, -22

Answer 4

A. the three consecutive integers will be written as x, (x+a million), and (x+2). B. You requested the sum. So, it must be extra. x + (x+a million) + (x+2) = 36 C. x + (x+a million) + (x+2) = 36 x + x + a million + x + 2 = 36 3x + 3 = 36 3x = 36 - 3 3x = 33 x = 11 We already recognize that x is 11, so (x+a million) will be 12, and (x+2) will be thirteen. So the three integers are 11, 12, and thirteen. wish this helped :)

Answer 5

This is a simple matter of turning words into an equation.
Let the first number be n, and write everything else in terms of n

"the sum of the second and third" translates to
(n+1)+(n+2)
exceeds something by 33 means
= something plus 33
So n+1+n+2 = something plus 33
something = "one half of the first", or n/2
so n+2+n+2 = n/2+33
or
2n+3 = n/2 + 33
2n - n/2 = 30
5n/2 = 30
n = 30/5*2 = 12

Answer 6

Let the three numbers be x, x + 1 and x + 2

(x + 1) + (x + 2) = x/2 + 33
2x + 2 + 2x + 4 = x + 66
4x + 6 = x + 66
3x = 60
x = 20

20, 21, 22

Answer 7

Let x be the first of the three integers... Then
(x+1) + (x+2) = x/2 + 33.

Solve for x.

Answer 8

20 21 and 22

Answer 9

X + (X + 1) + (X + 2) <-- three consecutive integers

(X + 1) + (X + 2) = (x/2)+33 <-- setting it up

2X+3 = (x/2)+33 <-- solving left

2x-(x/2) = 33-3 <-- moving like terms around

1.5(X) = 30

X = 20

Answer: 20, 21, 22

Answer 10

(x+1) + (x+2) = (1/2) x + 33
2x - 30 = (1/2)x
4x-60=x
3x=60
x=20

20, 21, 22