The sum of two consecutive positive even integers is at most 15. What are the possible pairs of integers?
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I didn't learn this yet. Would that mean A+B <= 15? With pairs like 2,2; 4,2; 4,6????
2007-05-11 11:41:46 · 5 answers · asked by Toni 1 in Science & Mathematics ➔ Mathematics
Yes, you are right: 5,10; 4,11; 3,12....and so on.
2007-05-11 11:45:25 · answer #1 · answered by bruinfan 7 · 0⤊ 2⤋
The pair of numbers will be like 2 and 4 or 8 and 10. They are even and follow one another therefore we have 2 equations
a+b<=15
b=a+2
a + b <= 15
a + (a+2) <= 15 (substitute)
2a + 2 <=15
2a <=15 - 2
2a <= 13
a <= 13/2
a <= 6 1/2 and b = a + 2
so the possible pairs are:
6 and 8
4 and 6
2 and 4
2007-05-11 11:48:43 · answer #2 · answered by shanusav 2 · 0⤊ 0⤋
You don't actually need two variables. We know that the two numbers are "consecutive positive even integers". So if N is an even integer, then the next even integer is N+2. (It's NOT N+1, because that number would be odd if N is even.)
So we have N + (N+2) <= 15
2007-05-11 11:46:10 · answer #3 · answered by Anonymous · 0⤊ 0⤋
twice the sum of a number and 4 is less than 12
2015-01-12 02:41:53 · answer #4 · answered by tim 1 · 0⤊ 0⤋
x+(x+2)
2x+2 less than or equal to 15
2x less than or equal to 13
x less than or equal to 6.5
so 2,4 4,6 6,8
2007-05-11 11:45:38 · answer #5 · answered by Anonymous · 0⤊ 2⤋