Question 50280
Find three consecutive even integers such that twice the sum of the first and third is twelve more than twice the second.
Three consecutive even intergers:
First integer=x
Second integer = x+2
Third integer = x+4
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Translation:
Twice the sum of the first and third: 2[(x)+(x+4)]
The "sum" means to add the first and third integer, than multiply the total by "2".
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"is" means "equals"
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Twelve more than twice the second: [(2)(x+2)]+12
Take the second integer (x+2), multiply it by "2" and add 12.

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So, putting it all together:
2[(x)+(x+4)]=[(2)(x+2)]+12
2(2x+4)=(2x+4)+12
4x+8=2x+16
4x-2x+8=2x-2x+16
2x+8-8=16-8
2x=8
2x/2=8/2
x=4
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Plug (x=4) into the integers and solve.

First integer=x=4
Second integer = x+2=4+2=6
Third integer = x+4=4+4=8
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2[(x)+(x+4)]=[(2)(x+2)]+12
2(4+8)=(2)(6)+12
2(12)=12+12
24=24 [checks out]