Question 526891
Find four consecutive odd integers such that 10 more than 3 times the third is 40 less than the first?


Let the 1st integer be F


Then the others are: F + 2, F + 4, and F + 6 


We then have: 3(F + 4) + 10 = F - 40


3F + 12 + 10 = F - 40


2F + 22 = - 40


2F = - 62


F, or 1st integer = {{{(-62)/2}}}, or - 31


The 4 consecutive odd integers are: {{{highlight_green(-31_-29_-27_and_-25)}}}


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