Question 978177: we we're asked to form the inequality of this problem and solve.... can you pls help me??
This is the problem:
The ages in years of 3 brothers are consecutive multiples of 4. Three years ago, the sum of their ages was less than 39. Find their present ages.
thank you, in advance!!
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! consecutive ages are 4x, 4(x+1), 4(x+2)
4x-3 + 4x+4-3 + 4x+8-3 < 39
4x-3 + 4x+1 + 4x+5 < 39
12x + 3 < 39
12x < 36
x < 3
There are two solutions to this problem
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1) x = 2
ages are 8, 12, 16
5 + 9 + 13 < 39
27 < 39
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2) x = 1
ages are 4, 8, 12
1 + 5 + 9 < 39
15 < 39
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