It can't be simplified any further so it is:
2m + 2n + p
Two odd numbers added together will always be an even number. I'll show how: Let M and N be any two integers. 2M is even, and 2N is even. So (2M +1) is an odd number, and (2N + 1) is an odd number. (2M +1) + (2N +1) = 2M + 1 + 2N +1 = 2M + 2N + 2 = 2(M + N + 1), which is even.
2m(p - 6q)(p - 6q)
is eassy
An even number plus an odd number always results in an odd number. This is because an even number can be expressed as 2n (where n is an integer), and an odd number can be expressed as 2m + 1 (where m is also an integer). When you add them together, you get 2n + (2m + 1) = 2(n + m) + 1, which is in the form of an odd number.
5n+2n=7n If the variable is the same, n, then you can add the coefficients, 5 and 2. In General an+bn=(a+b)n FOR A CASE THAT DOES NOT WORK. 5n+2m=5n+2m You cannot combine unlike variables
N=l-m
0
Two odd numbers added together will always be an even number. I'll show how: Let M and N be any two integers. 2M is even, and 2N is even. So (2M +1) is an odd number, and (2N + 1) is an odd number. (2M +1) + (2N +1) = 2M + 1 + 2N +1 = 2M + 2N + 2 = 2(M + N + 1), which is even.
2m(p - 6q)(p - 6q)
Yes.
is eassy
5m+2n, n-3m, 7-3n (5m+2n)+(n-3m)+(7-3n)=2m+7
5n+2n=7n If the variable is the same, n, then you can add the coefficients, 5 and 2. In General an+bn=(a+b)n FOR A CASE THAT DOES NOT WORK. 5n+2m=5n+2m You cannot combine unlike variables
2n + 4m - 2n + m = 5m
Yes --------------------------------------------- Let n be an integer. Then 2n is an even number Let m be an integer. Then 2m is an even number and 2m + 1 is an odd number. Then: even + odd = (2n) + (2m + 1) = (2n + 2m) + 1 = 2(n + m) + 1 = 2k + 1 (where k = m + n) which is an odd number.
2n plus 4 = 2n + 4
This is a simple proof of what you ask. Any even number can be thought of as 2n for some integer n. So let's take two even numbers 2n and 2m and add them. 2n+2m=2(n+m) and since n and m are integers so is n+m, call it p, then the sum is 2p and that is even since it is a multiple of 2.