Question 633702

please help me solve this problem... find four consecutive even integers such that twice the third integer subtracted from five time the first equals 18 more than the sum of the second and the fourth odd integer


What you wrote is contradictory. You 1st stated that the integers are even, and then you stated that they are odd. They can't be both. It's either one or the other.


Let the smallest integer be S
Then the 2nd is: S + 2
The 3rd is: S + 4, and
The 4th is: S + 6


We therefore have: 5(S) - 2(S + 4) = (S + 2) + (S + 6) + 18

5S - 2S - 8 = S + 2 + S + 6 + 18


5S - 2S - 2S = 26 + 8


S, or smallest integer = {{{34}}}


Therefore, the integers (even) are: {{{highlight_green(34)}}}, {{{highlight_green(36)}}}, {{{highlight_green(38)}}}, and {{{highlight_green(40)}}}


You can do the check, I'm sure!!


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