Can anyone help me with this algebra problem?

Question

The problem is...

"If a rooster is worth 5 coins, a hen 3 coins, and 3 chicks together 1 coin, how many roosters, hens, and chicks, totaling 100, can be bought for 100 coins?"

I have these equations so far:

5r+3h+c/3=100

r+h+c=100

but... now what?? Substitution?

I need to find 2 solutions, but what do i do next?

Answer 1

15r +9h +c =300

r+h+c=100

14r +8h =200

7r +4h =100

You have two equations, three unknowns.

Best to plug in values for r (they have to be even, otherwise you will get an odd number that you can't divide by 4)

r=2, h =86/4 no

r=4, h =76/4 =19 c=77 not divisible by 3

r=6. h =58/4

r=8, h =44/4=11 c= 81 one solution

r=10, h =30/4 no

r=12, h =16/4 =4 c=84 second solution

Answer 2

I can help

Answer 3

Try this site for help,it's pretty good.

http://www.mathway.com/problem.aspx?p=al...

Answer 4

Since you have three variables and only two equations there is no one answer.  I would multiply both sides of the first equation by 3, getting

(1a)  15r + 9h + c + 300

(2)      r + h + c = 100      and subtract equation (2) from equation(1a)

--------------------------------------...

         14r + 8h = 200

Now just pick them:

Solution #1

If r = 4 then

56 + 8h = 200

8h = 144

h = 18        (c becomes 100 - 4 - 18 = 78)

Solution #1  r = 4, h = 18, c = 78   <<<<<<<<<<

check    20 + 54 + 26 = 100  

--------------------------------------...

Solution #2

If r = 8 then

112 + 8h= 200

8h = 200 - 112 = 88

h = 11

So r = 8, h = 11 is a solution  (c becomes 100 - 8 - 11 = 81)

Solution #2 r = 8, h = 11, c = 81   <<<<<<<<<<

check  (5 x 8) + (3 x 11) + (81/3) = 40 + 33 + 27 = 100

--------------------------------------...

Solution #3 r = 12, h = 4, c = 84     (60 + 12 + 28 = 100)  

You can fill in the steps for Solution #3 for better understanding.

Don't try for another solution, I'm pretty sure that r has to be divisible by 4 and that there are no other solutions.