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find three consecutive even integers such that the sum of the smallest integer and twice the second is 12 more than the third (only an algebraic solution will be accepted )



Answer :

jzakoian
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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