ok this is what it looks like to me...I definitely could be wrong. one integer is three more than twice another integer if one integer is three more than another number (x+3=y) Then you just multiply "the other number" by 2. (x+3=2*y) Sum of two integers is 36....so add them (x+y=36) Solve that equation for one number so...(y=36-x) Then you would plug that equation into the first equation where there is a y: (x + 3 = 2*(36-x))
51
The sum of three negative integers is always negative.
The integers are 5 and 7.
9,11,13
It cannot be done. The basic rules of math. odd integer plus odd integer = even integer. odd integer plus even integer = odd integer. Always. odd integer plus odd integer plus odd integer = odd integer. Always.
That has no integer solution. Three times an integer is another integer; if you subtract to integers, you get an integer again, not a fraction.
The sum of three consecutive integers is -72
This would be impossible - since the mean of the three integers would have to be an integer, and if you divide -56 by 3, you do not get an integer.
Let the two consecutive even integers be (x) and (x + 2). According to the problem, the greater integer (x + 2) is equal to three times the smaller integer (x) minus four, which can be expressed as the equation (x + 2 = 3x - 4). Solving this gives (2 + 4 = 3x - x), leading to (6 = 2x) or (x = 3). However, since we're looking for even integers, we can set (x = 2n) for some integer (n), leading us to the conclusion that the integers are (2n) and (2n + 2). Thus, the specific integers will depend on the value of (n).
51
Suppose the smaller integer is x, then the larger one is x+1. x + 3*(x+1) = 43 That is x + 3x + 3 = 43 so that 4x = 40 and that implies that x = 10 and so the other integer is 11.
Let the smaller integer be x, then then larger integer is x + 2, and: 3x + (x + 2) = 58 → 4x = 56 → x = 14 → The two integers are 14 and 16.
The sum of three negative integers is always negative.
The sum of three consecutive integers can be expressed using ( l ) as the middle integer. The three integers can be represented as ( (l-1) ), ( l ), and ( (l+1) ). Therefore, the sum is ( (l-1) + l + (l+1) = 3l ).
They are 14, 16 and 18.
The numbers are 14, 16 and 18.
You can find this by dividing 2010 by three and adding two for the highest integer and subtracting 2 for the lowest integer. 2010/3=630, so the numbers are 628, 630 and 632