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PSYCHO15rus [73]
3 years ago
8

find three consecutive even integers so that the twice the sum of the second and third is twelve less than six times the first

Mathematics
1 answer:
zysi [14]3 years ago
4 0

n, n + 2, n + 4 - three consecutive even integers

the twice the sum of the second and third: 2[(n + 2) + (n + 4)]

twelve less than six times the first: 6n - 12

The equation:

2[(n + 2) + (n + 4)] = 6n - 12

2(n + 2 + n + 4) = 6n - 12

2(2n + 6) = 6n - 12       <em>use distributive property</em>

(2)(2n) + (2)(6) = 6n - 12

4n + 12 = 6n - 12       <em>subtract 12 from both sides</em>

4n = 6n - 24       <em>subtract 6n from both sides</em>

-2n = -24        <em>divide both sides by (-2)</em>

n = 12

n + 2 = 12 + 2 = 14

n + 4 = 12 + 4 = 16

<h3>Answer: 12, 14, 16</h3>
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___________________

3x+5y=57

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minus the equations top to bottom=

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