Try to solve this math problem.

[j@cko]

There are cows, chickens and sheep total of 100.

x>0
y>0
z>0

the price of them are followed.

cow $ 10
sheep $ 3
chicken $ 0.5
-----------------
total $100

Can you write 3 equations to slove how many for each animal?

I can only get two so far.

 

[Woodie]

Simultaneous equations...what is this, homework???

the answer is:























Yes.

--Woodie

 

[Javelin]

x + y + z = 100
10x + 0.5y + 3z = 100

In general, there is no unique solution as you only have 2 equations and 3 variables. If you assume that x,y,z all have to be integers then it is conceivable that there might be a unique solution.

 

[goodoptics]

x + y + z = 100
10x + 3y + 0.5z = 100


# of chicken has to be even...

3y + 0.5z must be divisible by 10...and less than 90

tough problem....

 

[j@cko]

see, those are what I had. The third equation is needed to slove this question. Becasue if you only have two like goodoptics does and I. You still left with two variables.

 

[j@cko]

And no it's not HW. It's just a question teacher through at us for fun, since it's Friday...

 

[Woodie]

but you forget:

7x - 2.5z = -200

restated:

7x + 200 = 2.5z

therefore:

z= (7x + 200)/2.5


--Woodie

 

[j@cko]

but how are you going to solve Z and X? But what you have there is still a algebric expression with two variables...

 

[Capn]

it's solve, not slove.

 

[Woodie]

You don't (yet). Derive an equation for x & y, then one for y & z. Substitute eqn #3 (the one I gave you) into equation 1. Should read:

x + y + (7x + 200)/2.5 = 100

Now, restate this as y = (stuff)

now, derive another equation with only 2 variables...hint: y= fn(x)

 

[goodoptics]

x + y + z = 100
10x + 3y + 0.5z = 100


# of chicken has to be even...

3y + 0.5z must be divisible by 10...and less than 100

<< 3y + 0.5z must be divisible by 10...and less than 90 >>

thanks Woodie

tough problem....

there are nine eqns for the third eqn:

0x + 3y + 0.5z = 90
0x + 3y + 0.5z = 80
0x + 3y + 0.5z = 70
0x + 3y + 0.5z = 60
0x + 3y + 0.5z = 50 (this is the correct one!)
0x + 3y + 0.5z = 40
0x + 3y + 0.5z = 30
0x + 3y + 0.5z = 20
0x + 3y + 0.5z = 10


I know, my method sounds stupid

Solution: 5 cows, 1 sheep, 94 chickens

 

[Woodie]

<< 3y + 0.5z must be divisible by 10...and less than 90 >>



Why 90? Should be 100, no?

--Woodie

 

[j@cko]

Why do you have 0x??

 

[Javelin]

One solution is 0 cows, 20 sheep, 80 chickens.
Another is 5, 1, 94.

I think those are the only two possible solutions.

 

[j@cko]

um.. ya, I think woodie got it...

 

[goodoptics]

can the number of cows be zeros since the problem says there are cows, chickens and sheep?

 

[Woodie]

Goodoptics has a point:
Add:
x>0
y>0
z>0
That's more than enough to solve the problem...though it'll take a page or more to show it with a sequence of equations.

Have a nice weekend--I'M OUTTA HERE!!

--Woodie

 

[beat mania]

I think saying that


<<
cows, chickens and sheep >>



pretty much gave away that sheep = 1.

 

[Capn]

dude, plural of sheep is sheep.

It wouldn't say cows, chickens, and sheeps

If it was 1, it would have to say, cows, chickens, and a sheep.

 

[goodoptics]

<< but you forget:

7x - 2.5z = -200

restated:

7x + 200 = 2.5z

therefore:

z= (7x + 200)/2.5 >>

All three eqns must be linearly independent. Therefore, I don't think the third eqn you obtained will work.

 

[goodoptics]

<<I think saying that


<<
cows, chickens and sheep >>



pretty much gave away that sheep = 1.>>

Doh!!!

 

[MereMortal]

The following is not 3 eqns, but uses common sense:

x+y+z=100
20x+6y+z=200

We know that x<10, y<33, z<100. Subtract 1st from 2nd, get 19x+5y=100.
5*y gives number divisible by 5 or 10. To sum to 100, only multiple of 19 that is divisible by 5 or 10 is x={0,5}.

Therefore,

Case x=0 --> 5y=100, or y=20. Subsitute back into 1st eqn and get z=80.

Case x=5 --> 95+5y=100, or y=1. Substitute back and get z=94.

 

[Argo]

There are a lot of solutions, since you can always replace 6 chicks with one sheep (boy, did that sound wrong!).

 

[nd]

Here's the real answer. I'm not a math major, but I'm pretty sure I know what I'm talking about.

You can't get 3 equations, because 3 equations with 3 unknowns implies a single answer. This is a case where it's plausible for multiple solutions.

What you do have is 2 equations and 1 constraint.

The 2 obvious equations:

COWS*10 + SHEEP*3 + CHICKENS*0.5 = 100
COWS + SHEEP + CHICKENS = 100

And the constraint:

COWS, CHICKEN, and SHEEP are positive integers.

This constraint is simply accounting for the real-world scenario where you can't have partial animals, and you can't have negative animals. This is the only thing you can have for a third &quot;equation&quot;.

If you solve the two equations, you get 6 integer results that work, only 1 of which results in all three variables being in the range from 0 to 100 (COWS = 5, SHEEP=1, CHICKENS=94).

Some other results, if you care:

Sheep = 20, Chickens = 80
Sheep = 39, Chickens = 66
Sheep = 58, Chickens = 52
Sheep = 77, Chickens = 38
Sheep = 96, Chickens = 24

Note that all of these result in a negative value (or 0, which isn't possible according to the wording of the problem) for COWS, and hence won't work because it doesn't fit the constraint.

If your teacher tells you that there is a real 3rd equation, he/she is wrong

 

[mundania]

came to the same conclusion as nd