SOLUTION: so this is the algebra problem of the week #12
the problem is as follows
A man has a bag of coins. He meets three people.The first one gets 1/2 the coins +2. The second get 1/2
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Question 969459: so this is the algebra problem of the week #12
the problem is as follows
A man has a bag of coins. He meets three people.The first one gets 1/2 the coins +2. The second get 1/2 the coins +2. The third gets 1/2 the coins +2. When the man returns home he has two coins in his bag. Write and solve an equation to determine how many he started with.
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
He began with 44 coins.
He started with x coins.
He gave away (x/2) +2, subtracted from x, leaves him with (x/2)-2
Second person gets half of that, which (x/4)-1+2=(x/4)+1
Now he is left with the difference, which is (x/4)-3
He gives away half of that (x/8)-3/2+2=(x/8)+(1/2)
He had (x/4)-3 and he gave away (x/8) +1/2. When you subtract, (x/4)-(x/8)=(x/8) And -3 -1/2=-(7/2)
He is left with (x/8)-(7/2)=2 because he has two coins left.
Multiply everything in the equation by 8 to clear fractions.
x-28=16 (7/2)*8=28
Add 28 to both sides.
x=44 coins.
Gives away half (22)+ 2 and has given away 24. He has 20 left.
He gives away half (10) +2 or 12, and he has 8 left.
He gives away half (4) plus 2 or 6, and he has 2 left.
Don't let the fractions of coins, or what appear to be fractional coins worry you. Just keep working it out and see what happens.
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