Find two consecutive even integers such that the sum of the smaller and 5 times the larger is 202.

n; n+2 - two consecutive even integers
n + 5(n + 2) = 202
n + 5n + 10 = 202
6n + 10 = 202 |subtract 10 from both sides
6n = 192 |divide both sides by 6
n = 32
* Answer: 32 and 34. *

Answered by Anonymous | 2024-06-10

so the equation hmmm is this x+5(x+2)=202 now a little math x+5x+10=202 and hmm 6x=192 and x=32 and x+2 is 34 even number has spaces of 2 so i used x+2. Uh yeah whooooo! hows that babe ;)

Answered by STOPCHEATINGK12ER | 2024-06-10

The two consecutive even integers such that the sum of the smaller integer and 5 times the larger equals 202 are 32 and 34. This was found by setting up an equation and solving for the integers. Finally, we verified the solution is correct.
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Answered by Anonymous | 2024-09-30