Question 324893: 11:34:34 PM ] Guest : What are the three consecutive even integers such that ten time the first integer is equal to six times the sum of the second and third?
[ 11:38:12 PM ] Guest : Hi I have worked on this for hours. College, Susan Intermediate Algebra
[ 11:41:05 PM ] : This was my last attempt before asking for help 10x = 6(3x+6)
[ 11:42:48 PM ] : 10x = 12x+36 take 12x from both sides
[ 11:44:00 PM ] : you get -2x=36 divide both sides by -2
[ 11:44:46 PM ] : x = -18 is this one of the integers?
[ 11:47:26 PM ] : help please!!
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! let the integers be x, x+2, x+4
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10 times the first integer = 10x..............1
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sum of second and third
= x+2 +x+4
=2x+6
6 times the sum
6(2x+6)
=12x+36................2
equation 1 = equation 2
10x=12x +36
add -12x to both sides
10x-12x= 12x-12x+36
-2x=36
x= 36/-2
x=-18 . first integer
x+2 = -18+2=-16 second integer
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x+4 = -18+4
=-14 third integer
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Check 10x= 6(x+x+6)
10*-18=6(-18-18+6)
-180=-180
the integers are -18,-16,-14
you are right
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