Find three consecutive even integers such that 3 times the first is 26 less than twice the sum of the last two
1 answer:
Let's say our first integer is "a".
how to get the next consecutive EVEN integer? well, just add or subtract 2 from it, therefore, the second consecutive integer will be "a + 2".
and the next after that, will then be (a + 2) + 2, or "a + 4".
so those are are 3 integers, a a + 2 a+4
notice that, from any even or odd integer, if you hop twice either forwards or backwards, you'll land on another even or odd integer respectively.
2 + 2 is 4, or 8 + 2 is 10 some even ones
3 + 2 is 5, or 13 + 2 is 15, some odd ones
![\bf \stackrel{\textit{3 times the first}}{3a}~~=~~\stackrel{\textit{26 less than twice the sum of the others}}{2[~(a+2)+(a+4)~]~~~-26} \\\\\\ 3a=2[~2a+6~]-26\implies 3a=4a+12-26\implies 3a=4a-14 \\\\\\ 0=a-14\implies 14=a](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7B3%20times%20the%20first%7D%7D%7B3a%7D~~%3D~~%5Cstackrel%7B%5Ctextit%7B26%20less%20than%20twice%20the%20sum%20of%20the%20others%7D%7D%7B2%5B~%28a%2B2%29%2B%28a%2B4%29~%5D~~~-26%7D%0A%5C%5C%5C%5C%5C%5C%0A3a%3D2%5B~2a%2B6~%5D-26%5Cimplies%203a%3D4a%2B12-26%5Cimplies%203a%3D4a-14%0A%5C%5C%5C%5C%5C%5C%0A0%3Da-14%5Cimplies%2014%3Da)
what are the other two consecutive integers? well, a + 2 and a + 4.
how to get the next consecutive EVEN integer? well, just add or subtract 2 from it, therefore, the second consecutive integer will be "a + 2".
and the next after that, will then be (a + 2) + 2, or "a + 4".
so those are are 3 integers, a a + 2 a+4
notice that, from any even or odd integer, if you hop twice either forwards or backwards, you'll land on another even or odd integer respectively.
2 + 2 is 4, or 8 + 2 is 10 some even ones
3 + 2 is 5, or 13 + 2 is 15, some odd ones
what are the other two consecutive integers? well, a + 2 and a + 4.
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Answer: ≈ 196.53 m²
All work is shown in the picture.
