EASY 10 POINTS???????

At the local sale yards some years ago, one farmer bought a group of six horses and
seven cows for $2 500. Another farmer bought a group of thirteen cows and one horse for
$2 310. If C is the cost in dollars of one cow and H is the cost in dollars of one horse:

Write two equations that relate the number of horses and cows to their costs??

Thanks for your help x

one farmer bought a group of six horses (6H) and seven cows (7C) for $2 500 =>
6H + 7C = 2500
Another farmer bought a group of thirteen cows (13C) and one horse (H) for $2 310.
H + 13C = 2310

There are your 2 equations. To solve by substitution:
H = 2310 - 13C

6(2310 - 13C) + 7C = 2500
13860 - 78C + 7C = 2500
13860-2500 = 71C
C = 11360/71 = 160
H = 2310 - 13*160 = 230

they set variables for you: c is cow; h is horse, so
6h+7c=2500 and h+13c=2310
then solve like system of equations
i prefer substitution
so you can set h=2310-13c by solving for h from h+13c=2310
then you plug that value of h into the other equation
so
6(2310-13c)+7c=2500
so
13860-78c+7c=2500
so
13860-71c=2500
-71c=(-11360)
c=160
then you plug this c value into either original equation
h+13(160)=2310
h+2080=2310
h=230
so
horses cost $230 each
and
cows cost $160 each

6H + 7C = 2500

H + 13C = 2310

Now if you need to solve just rearrange second equation like this: H = 2310 - 13C and substitute that into the first equation. 6(2310 - 13C) + 7C = 2500. solve for C

6c+7h equals $2500
13c+h equals $2310

Too confusing... try a math fourm?
Sorry!