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Radda [10]
4 years ago
10

How do i solve the sum of three consecutive integers is 48 less than twice the smallest of the three integers?

Mathematics
1 answer:
Dmitriy789 [7]4 years ago
4 0

So let's say the three consecutive integers are x-1,x,x+1.

2(x-1)=x-1+x+x+1+48

2x-2=3x+48

x=-50.

The three integers will be -49,-50,-51

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