Question 834478: Find three consecutive integers such that the sum of the first and twice the second is 98 minus three times the third.
Find the smallest integer.
Answer by JulietG(1812)   (Show Source):
You can put this solution on YOUR website! Here's what we know:
B = A+1
C = A+2 (consecutive)
A + 2B = 98 - 3C
Substitute the known values
A + 2(A+1) = 98 - 3(A+2)
Distribute
A + 2A + 2 = 98 - 3A - 6
Add the As
3A + 2 = 98 - 3A - 6
Add 3A to each side
6A + 2 = 98 - 6
Subtract 2 from each side
6A = 98-6-2
Do the math
6A = 90
Divide each side by 6
A = 15
.
If A is 15, then B is 16, and C is 17
.
Let's see if it works.
(...the sum of the first and twice the second is 98 minus three times the third.)
15 + 32 = 98 - 51
47 = 47
Success!
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