Find two consecutive even integers whose product is 1,520

Let x be the smallest number.
Let x + 2 be the other number (consecutive even integer)

Now to translate the rest;)
three times the smaller 3(x)
19 more -19
sum of the two integers - (x) +( x+2)

Pulling it together,
3x -19 = x + x +2
collect like terms
3x - 19 = 2x + 2
Put all the constants on one side by adding 19 to both sides.
3x - 19 +19= 2x + 2 + 19
3x = 2x +21
Subtract 2x from both sides to have all the x's on one side.
3x - 2x = 2x +21 - 2x
x = 21
The other number is 21 + 2, or 23
Check:three times the smaller = 3 x 21 = 63
sum of the two integers = 21 + 23 or 44
is 63 at least 19 more than 44

38.
The three numbers are 34, 36, and 38.

The two numbers are 32 and 34. The rest is your homework, you do it.

Let x be first (smaller) integer and y second (larger) integer.

1. They are consecutive (and even) : y=x+2
2. Twice the smaller exceeds the larger by 18: 2*x=y+18
3. Substitute first equation into second: 2x=x+2+18
4. 2x-x=2+18
5. x=20
6. y=x+2=22

These two numbers are 20 and 22.

We'll call x the smaller integer and y the larger. So we know x+3y=22. We can rearrange this to say:
x=22-3y. Now we guess, I'll guess y=6, now we solve and get x=22-3(6)=4. 4 and 6 are two even consecutive integers! So the answer is 4 and 6.

X + ( X + 2 ) + ( X + 4 ) = 105
3X + 6 = 105
3X = 105 - 6
3X = 99
X = 99/3
X = 33
So 33 + 35 + 37 = 105

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Let X be the lowest of the 3 integers
The sum of the 3 integers can be represented as X + (X+1) + (X+2).
Set this sum equal to 378 and solve for X.
X + (X+1) + (X+2) = 378
3X + 3 = 378
3X = 375
X = 125
Since X is the lowest of the 3 integers the other 2 will be X+1 and X+2 or 126 and 127.

Therefore, the answer is 125, 126 and 127

find two consecutive cube numbers or integers whose difference is a square number. i know it is 8 and 7 but hav to get the equation