find three consecutive even integers such that the sum of the smallest integer and twice the second is 12 more than the third (o
nly an algebraic solution will be accepted )
1 answer:
Let us call them a, b and c
we know that
a+1=b
a+2=c
a+2b=12+c
we can deduce the following:
a+2(a+1)=12+(a+2)
3a+2=a+14
2a=12
a=6
so b would be 7 and c would be 8
Makes sense ?
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