SOLUTION: 1)find three consecutive integers such that the sum of the third equels 142 2)find four consecutive intagers such that 8 times the sum of the first and the third is 40 greater t

Algebra ->  Test -> SOLUTION: 1)find three consecutive integers such that the sum of the third equels 142 2)find four consecutive intagers such that 8 times the sum of the first and the third is 40 greater t      Log On


   



Question 411814: 1)find three consecutive integers such that the sum of the third equels 142
2)find four consecutive intagers such that 8 times the sum of the first and the third is 40 greater than 10 times the forth

Found 2 solutions by shree840, stanbon:
Answer by shree840(260) About Me  (Show Source):
You can put this solution on YOUR website!
1. divide by 3 so approx 47
46 47 48 142 does not fit it should b 141
2 8*(x+x+2)=10(x+3)+40
or 16x+16=10x+30+40
or 6x =70-16=54
or x=9
digits are 9 10 11 12

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
1)find three consecutive integers such that the sum of the third equals 142
Something is missing in your post.
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2)find four consecutive integers such that 8 times the sum of the first and the third is 40 greater than 10 times the forth
1st: x-2
2nd: x-1
3rd: x
4th: x+1
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Equation:
8(x-2+x) = 10(x+1)+40
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16x-16 = 10x+50
----
6x = 66
x = 11
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1st: x-2 = 9
2nd:10
3rd:11
4th:12
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Cheer