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Question 436130: The sum of three consecutive integers is 21 larger than twice the smallest integer. Find the integers.
Answer by marilynh(7)   (Show Source):
You can put this solution on YOUR website! Let the first integer = x, since the three numbers are consecutive the second and third numbers are x + 1 and x + 2
Their sum is represented by x + (x + 1) + (x + 2)
We want to set that equal to "21 larger than twice the smallest integer"
"twice the smallest integer" is = 2x
therefore "21 larger than twice the smallest integer" = 2x + 21
Set the two sides equal to each other and we have:
x + (x + 1) + (x + 2) = 2x + 21
Now solve the equation for x:
3x + 3 = 2x + 21
1x = 18
So the first integer is 18. The next two are x + 1 = 19 and x + 2 = 20
Check your answer:
18 + 19 + 20 = 2(18) + 21
57 = 36 + 21
57 = 57
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