a = b + 7; 2a = b2 - 1. Substituting: 2b + 14 = b2 - 1 or b2 - 2b - 15 = 0
Factors are (x - 5)(x + 3)
b = 5, a = 12. Also works with b = -3 and a = 4
The smaller integer is 6, the larger integer is 32
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.
If the integer subtracted is smaller than or equal to the first integer, then the answer is positive. Otherwise, if the integer subtracted is larger, then the answer is negative.
Let x equal the smaller integer. Three times the larger is 3(x + 1) or 3x + 3. Added all together, 4x + 3 = 31. Solve for x. 4x = 28 x = 7 Check it. 7 + 24 = 31 It checks.
Let x = 1st integer, since consecutive even integer differs by 2 then the 2nd integer = x + 2. So we have, x + 4(x + 2) = 48 x + 4x + 8 = 48 5x = 40 x = 8 (1st integer) Thus, the integers are 8 and 10. Check.
The smaller integer is 6, the larger integer is 32
The larger integer is 30. The smaller is 28.
x > 16
No.A positive integer is always larger than a negative integer. In the case of two negative integers, the integer with the larger absolute value is actually smaller.
There can be no such integers: a smaller integer cannot be 5 times the larger number.
-1
no one wants to know the answer. its freaking math
You get another integer that will take the sign from the larger of the two integers that were combined.
The product of two consecutive positive integers can be found by multiplying the smaller integer by the larger integer. If the smaller integer is represented as ( n ), then the larger integer would be ( n + 1 ). Therefore, the product of two consecutive positive integers is ( n \times (n + 1) ).
x+3 and x+4 would be consecutive integers.
The sum will be a positive integer.It will be at least as large as double the smaller integer and at most as large as double the larger integer.
Divide the smaller into the larger. If the quotient is an integer, the smaller is a factor of the larger.