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JulijaS [17]
4 years ago
5

Find three consecutive even integers whose sum is 66. show work pls

Mathematics
2 answers:
Mnenie [13.5K]4 years ago
6 0
<span>three consecutive even integers are x, x+2 and x+4

x + x+2 + x+4 = 66
3x + 6 = 66
3x = 66-6
3x = 60
x = 60/3
x = 20  </span>← first integer
<span>
x + 2 = 20 + 2 = 22 </span>← second  integer

x + 4 = 20 + 4 = 24 ← third integer
kari74 [83]4 years ago
5 0
For our three integers to be even x%2=0 ∀x∈S where S= {our three integers}. To guarantee this set our first integer as 2n (ie. 2n%n≡0). Then to have three consecutive even integers increment from the initial value by 2 thus setting them as 2n, 2n+2 and 2n+4.

Thus:
2n+2n+2+2n+4=66

Simplifying:
6n+6=66

Solving for 'n':
6n=60
n=10

The integers are thus 2n=20, 2n+2=22 and 2n+4=24.

To check: 20+22+24=66.
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Using a linear function, the graph for the amount remaining of the trip is given at the end of this answer.

------------------------------

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A similar problem is given at brainly.com/question/16643850

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