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Question 508678: the 2nd of the three numbers exceeds the smallest by 2. Whole the third number is twice the first. If the sum of the three number is 54, find the numbers.
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Call the three numbers X, Y, and Z with X being the first number, Y being the second, and Z being the third.
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The problem tells you that the second (that is Y) exceeds the first number by 2. That means that Y is equal to the first number (that is X) plus 2. So Y = X + 2.
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Substitute the equivalent X + 2 for Y and we can now say that the three numbers are:
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X, X + 2, and Z
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The problem also tells us that the third number (Z) is twice the first number (X). So we can say that Z equals 2 times X or Z = 2X. Substituting 2X for Z results in the three numbers being:
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X, X + 2, and 2X
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Finally, the problem tells us that if we add the three numbers, the total is 54. So adding the three numbers leads to the equation:
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X + X + 2 + 2X = 54
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Add the three terms containing the unknown X on the left side and the equation becomes:
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4X + 2 = 54
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Subtract 2 from both sides to eliminate the 2 from the left side and get:
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4X = 52
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Solve for X by dividing both sides by 4 and you get:
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X = 52/4 = 13
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So we know that the first number is 13. Then we know the second number is two greater, so it is 15. And finally the third number is twice the first number so it is two times 13 or 26.
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The three numbers are 13, 15, and 26. Check by adding to ensure that the sum of these three numbers is 54.
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Hope this helps you to understand the problem a little better.
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