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09-16-2009   #1 (permalink)
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Default Solve for x and y: System of Equations

I have a problem I need help working out considering I am not much of an expert on such questions. Does anyone know how to such system of equations?

The Word Problem is:
A zookeeper needs to mix feed for the prairie dogs so that the feed has the right amount of protein. Feed A has 12% protein. Feed B has 5% protein. How many pounds of each does he need to mix to get 100 pounds of feed that is 8% protein.


I believe equation to solve is:
x+y=100
100(.8)=x(.12)+y(.05)

But I am not sure how to work it out with going overly-broad with working it out.
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09-16-2009   #2 (permalink)
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a + b = 100
.12a +.05b = .8? Did you mean 80% of protein?

and then solve and whatnot

"100(.8)=x(.12)+y(.05)"
actually what does this mean? are the numbers in the paren. in place for the 2nd equation?
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Last edited by Laughatyou21; 09-16-2009 at 05:51 PM.
 
09-16-2009   #3 (permalink)
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Quote:
Originally Posted by Laughatyou21
x + y = 100
.12x +.05y = .8? Did you mean 80% of protein?

and then solve and whatnot

"100(.8)=x(.12)+y(.05)"
actually what does this mean? are the numbers in the paren. in place for the 2nd equation?
The numbers in parenthesis are being multiplied to the variable.


Give me a minute. I'm gonna write and scan my solution. I hate typing it up.
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09-16-2009   #4 (permalink)
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I dun think you can multiply 100 by .8 if there's no variable to substitute the .8 in.

fml, well w/e, I was never good with system of equations.
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09-16-2009   #5 (permalink)
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x+y=100
100(.8)=x(.12)+y(.05)

Uh, from what I think...

the second equation should be
100(.08)=x(.12)+y(.05)

to solve for x and y, take your first equation
x+y=100, and rewrite it as y=100-x and then put this into your second equation so

100(.8)=x(.12)+(100-x)(.05)
8=.12x+5-.05x
solve for x

then when you have x, put it in for x+y=100 to figure out y
if it's not this, then I don't know /o/
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09-16-2009   #6 (permalink)
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Quote:
Originally Posted by Laughatyou21
I dun think you can multiply 100 by .8 if there's no variable to substitute the .8 in.

fml, well w/e, I was never good with system of equations.
Oh whoops, didn't see that part.
No wonder it looked so weird on paper.




Anyone else getting -60 lbs. for Feed A?
`-`
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Last edited by kuyaBaka; 09-16-2009 at 06:07 PM.
 
09-16-2009   #7 (permalink)
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Yeah, I got a negative too...I think the TS is missing something. T_T
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09-16-2009   #8 (permalink)
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Quote:
Originally Posted by Yume
x+y=100
100(.8)=x(.12)+y(.05)

Uh, from what I think...

the second equation should be
100(.08)=x(.12)+y(.05)

to solve for x and y, take your first equation
x+y=100, and rewrite it as y=100-x and then put this into your second equation so

100(.8)=x(.12)+(100-x)(.05)
8=.12x+5-.05x
solve for x

then when you have x, put it in for x+y=100 to figure out y
if it's not this, then I don't know /o/
Yume is right.
Letting x= amt. of feed A and y = amt of feed B,

x + y=100 (1)
8= 0.12x + 0.05y (2)

From (1): y= 100 - x. (3)

Sub (3) into (2):

8 = 0.12x + 0.05(100 - x)
3 = 0.07x

x = 42.86. (4)

Sub (4) into (3)

y= 100 - 42.86 = 57.14
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09-16-2009   #9 (permalink)
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Quote:
The Word Problem is:
A zookeeper needs to mix feed for the prairie dogs so that the feed has the right amount of protein. Feed A has 12% protein. Feed B has 5% protein. How many pounds of each does he need to mix to get 100 pounds of feed that is 8% protein.
Quote:
Originally Posted by Javier
Yume is right.
Letting x= amt. of feed A and y = amt of feed B,

x + y=100 (1)
8= 0.12x + 0.05y (2)

From (1): y= 100 - x. (3)

Sub (3) into (2):

8 = 0.12x + 0.05(100 - x)
3 = 0.07x

x = 42.86. (4)

Sub (4) into (3)

y= 100 - 42.86 = 57.14


Totally skipped over that little part of the question.
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09-16-2009   #10 (permalink)
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Quote:
Originally Posted by Javier
Yume is right.
Letting x= amt. of feed A and y = amt of feed B,

x + y=100 (1)
8= 0.12x + 0.05y (2)

From (1): y= 100 - x. (3)

Sub (3) into (2):

8 = 0.12x + 0.05(100 - x)
3 = 0.07x

x = 42.86. (4)

Sub (4) into (3)

y= 100 - 42.86 = 57.14
Quote:
Originally Posted by Yume
x+y=100
100(.8)=x(.12)+y(.05)

Uh, from what I think...

the second equation should be
100(.08)=x(.12)+y(.05)

to solve for x and y, take your first equation
x+y=100, and rewrite it as y=100-x and then put this into your second equation so

100(.8)=x(.12)+(100-x)(.05)
8=.12x+5-.05x
solve for x

then when you have x, put it in for x+y=100 to figure out y
if it's not this, then I don't know /o/
Thanks you two. Also.
Quote:
Originally Posted by Sanichi
x+y=100
100(.08)=x(.12)+y(.05)
I forgot to add the 0 when typing so I guess for those looking the word problem. . . .
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09-16-2009   #11 (permalink)
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Here's how I solved it. Complete with a scan and colored ink.

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09-16-2009   #12 (permalink)
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Quote:
Originally Posted by kuyaBaka
Here's how I solved it. Complete with a scan and colored ink.

I believe I understand the information now.
Thanks all of you!
kuyaBaka, Yume, and Javier~

The way I have been doing these problems. . . `-`
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09-16-2009   #13 (permalink)
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Ah, yes.. using the (1) +/- (2) approach to eliminate a variable also works but it's not that obvious sometimes lol. I tend to prefer substitution, can't go wrong with it :>
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