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klasskru [66]
3 years ago
10

Find four consecutive even integers so that the sum of the first three is 2 more than twice the fourth

Mathematics
1 answer:
scoundrel [369]3 years ago
4 0

Answer:

The numbers are : -20,-18,-16,-14

Step-by-step explanation:

Let 1st even integer be =x

So, next consecutive even integer would be =x+2

3rd consecutive even integer would be =x+4

4th consecutive even integer would be =x+6

Sum of first three integers would be =x+x+2+x+4=3x+6

It says sum of the first three is 2 more than twice the fourth,

So we have

3x+6=2+4(x+6)

using distribution to simplify.

3x+6=2+(4\times x)+(4\times6)

3x+6=2+4x+24

3x+6=4x+26

Subtracting 4x from both sides

3x+6-4x=4x+26-4x

-x+6=26

Subtracting 6 from both sides

-x+6-6=26-6

-x=20

∴ x=-20

So first integer =-20

2nd integer  =-20+2=-18

3rd integer =-18+2=-16

4th integer  =-16+2=-14

So, the numbers are : -20,-18,-16,-14

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