Math Is Fun Forum

Monique

Member

Registered: 2007-02-17

Posts: 22

very confused help needed indeed??

can anyone helpme with this:

At the local sale yard some years ago, one farmer bought a group of 6 horses and 7 cows for $2 500. Another farmer bought a group of 13 cows and 1 horse for $2 310. If C is the cost in dollars of 1 cow and H is the cost in dollars of 1 horse:

a)  write 2 equations that relate the number of horses and cows to their costs?

b) solve the above equations simultaneoulsy to find the cost of a cow and the cost of a horse?

c)  how would you check that solution is correct?

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: very confused help needed indeed??

For a), you just need to convert the information into equations.

You're told that 6 horses and 7 cows cost $2500 altogether, so that means that 6H + 7C = 2500.
Similarly, the other sentence can be changed into H + 13C = 2310.

For b), you need to use those two equations to work out the values of H and C.
To solve simultaneous equations, you need to combine them in such a way that one of the unknowns is eliminated. One possible way to do this would be to multiply the second equation by 6: 6H + 78C = 13860.
Then subtract the other equation from this one: 71C = 11360.
This solves to give that C = 160.

Then you can plug this into either one of the original equations to find H.
6H + 7*160 = 2500 --> 6H = 1380 --> H = 230.

For c), you just need to check the values of H and C that you worked out and make sure that both of your original equations work out correctly.

6*230 + 7*160 = 2500, and 230 + 13*160 = 2310, which means that the values are correct.

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Why did the vector cross the road?
It wanted to be normal.

pi man

Member

Registered: 2006-07-06

Posts: 251

Re: very confused help needed indeed??

a)
6H + 7C = 2500
1H + 13C = 2310

b) rewriting the second equation to be H = 2310 - 13C.   Now substitute 2310-13c in for H in the 1st equation:
6(2310-13C) + 7C = 2500
13860 - 78C + 7C = 2500
71C = 11360
C = 160

Now substitue 160 in for C in either equation to find H:
6H + 7(160) = 2500
6H + 1120 = 2500
6H = 1380
H = 230

c)  Take those value for C and H and substitute that back into one of the equations and see if it's true:
1H + 13C = 2310
230 + 13(160) = 2310
230 +2080 =2310
2310 = 2310

That's it!