e.g. Find three consecutive numbers in which the sum is 482.
e.g.2. Find three consecutive numbers in which the sum of the first and the third is 949.
Please simplify it and give me an equation for each.
e.g.2. Find three consecutive numbers in which the sum of the first and the third is 949.
Please simplify it and give me an equation for each.
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a) 3n+3 = 482
n = 159.66 no integer solution
b) 2n+2 = 949
n = 473.5 no integer solution
n = 159.66 no integer solution
b) 2n+2 = 949
n = 473.5 no integer solution
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When talking about consecutive numbers, we usually mean integers, since consecutiveness means nothing for real or rational non-integer numbers
The two examples you provide do not have integer solution.
Any 3 consecutive numbers have a sum that is divisible by 3, but 482 is not
When you have 3 consecutive numbers, the first and third have the same parity (both odd or both even), so their sum must be even, but 949 is not.
The two examples you provide do not have integer solution.
Any 3 consecutive numbers have a sum that is divisible by 3, but 482 is not
When you have 3 consecutive numbers, the first and third have the same parity (both odd or both even), so their sum must be even, but 949 is not.
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1. Let x be the first number. The next number is (x + 1) and the next is (x + 2). So, (x) + (x + 1) + (x + 2) = 482.
the three consecutive numbers are x, x + 1, x + 2 (remember to substitute the values of x obtained!)
2. Let x be the first number. So, the third number is (x + 2) [follow reasoning from 1.]. then, x + (x + 2) = 949.
the three consecutive numbers are x, x + 1, x + 2 (remember to substitute the values of x obtained!)
the three consecutive numbers are x, x + 1, x + 2 (remember to substitute the values of x obtained!)
2. Let x be the first number. So, the third number is (x + 2) [follow reasoning from 1.]. then, x + (x + 2) = 949.
the three consecutive numbers are x, x + 1, x + 2 (remember to substitute the values of x obtained!)
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n = 1st number.....n + 1 = 2nd numb we.....and the 3rd number is n + 2
equation is... n + n + 1 = n = 2 = 482
ANOTHER WAY TO DO THIS PROBLEM:
because this number is divisible by 3 and we're looking for 3 consecutive numbers,,
divide 482 by 3 = 164. That is the middle number (n + 1)
so n = 163 ..... and the 3rd number (n + 2) = 165
proof
163 = n
164 = n + 1
165 = n + 2
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492
#2 n + (n + 2) = 949
2n + 2 = 949
...........2n = 947
............n = 947/2
.............n = 473.5 = 474
equation is... n + n + 1 = n = 2 = 482
ANOTHER WAY TO DO THIS PROBLEM:
because this number is divisible by 3 and we're looking for 3 consecutive numbers,,
divide 482 by 3 = 164. That is the middle number (n + 1)
so n = 163 ..... and the 3rd number (n + 2) = 165
proof
163 = n
164 = n + 1
165 = n + 2
-------
492
#2 n + (n + 2) = 949
2n + 2 = 949
...........2n = 947
............n = 947/2
.............n = 473.5 = 474
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1) x + (x + 1) + (x + 2) = 482
3x+ 3 = 482
3x = 479
I think that you have missing information on the first one, but if not
the solution is 159 2/3 , 160 2/3 and 161 2/3
are you sure that the problem dis not say evenor odd integers ???
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2)x + (x+2) = 949
2x + 2 = 949
2x = 947
x = 473.5 & 474.5 & 475.5
3x+ 3 = 482
3x = 479
I think that you have missing information on the first one, but if not
the solution is 159 2/3 , 160 2/3 and 161 2/3
are you sure that the problem dis not say evenor odd integers ???
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
2)x + (x+2) = 949
2x + 2 = 949
2x = 947
x = 473.5 & 474.5 & 475.5
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1. let x be the lowest number. Therefore, x+x+1+x+2=482
=3x+3=482
=3x=479
=x=whatever and the other numbers are x+1 and x+2
2. Same principle: let x be the lowest/first; x+x+2=949
=2x+2=949
=2x=947
Therefore x= 423.5 and the other numbers are x+1 and x+2
=3x+3=482
=3x=479
=x=whatever and the other numbers are x+1 and x+2
2. Same principle: let x be the lowest/first; x+x+2=949
=2x+2=949
=2x=947
Therefore x= 423.5 and the other numbers are x+1 and x+2