A math question I need help with

[ArizonaNessa]

I was playing a game and to open a chest I needed to answer this question. Can anyone figure it out and explain how you got the answer?

Question:

There is a farm with chickens and horses. There are 120 heads and 314 legs. How many chickens are on the farm?

 

[BarkerChickens]

Alright...120 heads and 314 legs
Horses have 1 head and 4 legs. Let's let Horses = X
Chickens have 1 head and 2 legs. Let's let Chickens = Y

So, 4X + 2Y = 314 legs
Similarly, 1X + 1Y = 120 heads

Let's set each one equal to Y (Chickens (because 314 is divisible by 2 easier than 4))

2Y = 314 - 4X (then, divide my two to get rid of the 2Y)
Y = 157 - 2X
and
Y = 120 - X

Now that they both equal Y, they must be equal to each other, right? So,

157 - 2X = 120 - X

Set the equation to equal X:

157 - 120 = -X + 2X, which equals

X = 37

So, if X equals 37, then plug it back in to see if it works.

4 (37) + 2Y = 314 legs
2Y = 314 legs - 148 horses
Y = 83

Similarly,

1 (37) + 1Y = 120 heads
Y = 120 heads - 37 horses
Y = 83

So, since Y is the same on both, the answer should be correct.

37 horses and 83 chickens

edited to add...if you need me to explain anything, just ask!

 

[ArizonaNessa]

Not in this lifetime or the next would I have ever figured that out!!!

 

[bheila]

My 11yr old son has those exact same problems every night for homework, I hate them because he always needs help from me

I love math but these are ridiculous problems

 

[BarkerChickens]

They are complicated, but if you think of it as 2 separate problems it is a little easier. There are legs and heads. Horses and chickens. Each horse and chicken has a head and legs.

So, # Chicken (2 legs) + # Horse (4 Legs) = XYZ total legs
and # Chicken (1 head) + # Horse (1 head) = XYZ total heads

They have horses and chickens in common.

If you make each one equal the same thing, then you can take that common one away.
For example,
Y = 1 + 5
Y = 2 + 4
therefore, 1 + 5 = 2 + 4

This lets you solve 2 separate equations with a common variable. Once you know how many Xs or Ys is there, then you can go back to your original equation and solve for the other variable. They should have the same answer.

It's a pain, but if you set back and make it two different problems, it is easier to handle.

Let add that I am not a math geek. In fact, I hated math growing up...I just happen to do well in the subject (well, up to trig anyway. Calculus and I don't get along to well).

Edited because my brain was having some shorting of wires!

 

[WhiteWingedDove]

I had a chicken with three legs once.


.. Just sayin'.

 

[Eggs4Sale]

The answer is BLUE!!!!

 

[BarkerChickens]

Quote:

In that case, it cannot be solved due to the unknown number of 3 legged chickens and 3 legged horses.

 

[Eggs4Sale]

Quote:

In that case, it cannot be solved due to the unknown number of 3 legged chickens and 3 legged horses.

Wouldn't that be Z?

 

[ArizonaNessa]

In school I was the geometry girl. I can do most math things that I can see but algebra is beyond me. My 7th grade teacher told me it is because I ask "Why?" too much. She said I would not just accept things as they are. Heck I don't know but I just don't get algebra but I can find you the volume, circumference, diameter, radius, and perimeter to just about anything. I need to be able to see it or it gets me so stressed and confused I eventually give up. I don't have a problem using formulas as long as you don't hold back information.

Edited to say: I also can't do those questions where you have two trains leaving on different days in different cities going different speeds with the same person on both trains and you want to know when they will get to Backwoods, Nowhere, USA!!!