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Find two consecutive even integers such that 5 times their sum is 10 less than their product. (Enter your answers as a comma-separated list.)

 Aug 27, 2020
 #1
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Let the lower number be n. Then

 

5*(n + n+ 2) = n*(n + 2) - 10      10n + 10 = n^2 +2n -10

 

Rearrange  

n^2 - 8n - 20 = 0

 

This factors as (n + 2)*(n - 10) = 0

 

I'll let you take it from here.

 Aug 27, 2020

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