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Find four consecutive integers such that the sum of the first three is 18 more than the fourth.

Sagot :

AlgebraicAardvarkidn

If x is the first integer, then

x+1 is the second,

x+2 is the third, and

x+3 is the fourth.

Now we need those to be such that adding the first three equals the fourth plus 18.

     x + x+1 + x+2 = x+3 + 18

Solving that equation, we'd do:

    3x + 3 = x + 21

          2x = 18

            x = 9

That means 9 is the first integer, so the others are 10, 11, and 12.

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