5n + 1 = 10 + 2n5n - 2n + 1 - 1 = 10 - 1 + 2n - 2n3n = 93n/3 = 9/3n = 3
n = 1
2(10)+1
12+4n = 2n12+4n-4n = 2n-4n12 = -2n-1/2(12 = -2n)-6 = n
n=1 2/3
1 + 2 + 4 + ... + 2n = 2n+1 - 1
5n + 1 = 10 + 2n5n - 2n + 1 - 1 = 10 - 1 + 2n - 2n3n = 93n/3 = 9/3n = 3
the answer is 0
n = 1
6=(1-2n)+5 6-5=(1-2n)+5-5 1=1-2n 1-1=1-1-2n 0=-2n 0/-2=-2n/-2 0=n
Well, if 6 equals 1 plus five, the 2n you put in the middle must be nothing. Think. n=0 Sorry If it's not right
Solve for n. 6 = 1 - 2n + 5 6 = 6 - 2n 2n = 0 n = 0 Check it. 6 = 1 - 0 + 5 It checks.
2(10)+1
2(2n + 5) = 12 2n + 5 = 6 2n = 1 n = 1/2
12+4n = 2n12+4n-4n = 2n-4n12 = -2n-1/2(12 = -2n)-6 = n
n=1 2/3
This question can be expressed algebraically as: (1/n) + (1/(2n)) + 2 = 23, (1/n) + (1/(2n)) =21, ((1+2)/(2n)) = 21, (3/(2n)) = 21, or 2n = (3/21), 2n = (1/7), so n = (1/14). This, by the way, is an elementary algebraic proof that the solution to the above relation is (1/14). Anyway, to answer the question, reread the question: "[What integer is such that] the reciprocal of the integer...". notice, the reciprocal of (1/14) is 14, which is the integer in question! ^_^