Question 84858: I have 7 bags of marbles. There are 200 marbles in total. Using the clues below, work out how many marbles are in each bag.
Bag 1 + Bag 2 = 57 marbles
Bag 2 + Bag 3 = 83 marbles
Bag 3 + Bag 4 = 71 marbles
Bag 4 + Bag 5 = 43 marbles
Bag 5 + Bag 6 = 66 marbles
Bag 6 + Bag 7 = 43 marbles
Thank you so much for your help. I tried to figure this out but came to no avail.
Found 2 solutions by jim_thompson5910, rapaljer:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Lets use these variables for each bag:
Bag 1 = a
Bag 2 = b
Bag 3 = c
Bag 4 = d
Bag 5 = e
Bag 6 = f
Bag 7 = g
Now set up the following system of equations
a + b = 57
b + c = 83
c + d = 71
d + e = 43
e + f = 66
f + g = 43
Also since we know how many marbles there are total, we can use this equation:
a+b+c+d+e+f+g=200
Now subtract (a+b=57), (c+d=71), and (f+g=43) from a+b+c+d+e+f+g=200 to eliminate everything but one bag (in this case bag 5 which is denoted "e")
a+b+c+d+e+f+g=200
-(a+b =57)
-( c+d =71)
-( f+g=43)
-------------------
e =29
So we know that bag 5 has 29 marbles
Now plug in e=29 to find f
Bag 6:
29 + f = 66
f = 37
So we know that bag 6 has 37 marbles
Now plug in f=37 to find g
Bag 7:
37 + g = 43
g = 6
So we know that bag 7 has 6 marbles
Now plug in e=29 to find d
Bag 4:
d + 29 = 43
d = 14
So we know that bag 4 has 14 marbles
Now plug in d=14 to find c
Bag 3:
c + 14 = 71
c = 57
So we know that bag 3 has 57 marbles
Now plug in c=57 to find b
Bag 2:
b + 57 = 83
b = 26
So we know that bag 2 has 26 marbles
Now plug in b=26 to find a
Bag 1:
a + 26 = 57
a = 31
So we know that bag 1 has 26 marbles
So here's a summary of all of the bags:
Bag 1 = 31 marbles
Bag 2 = 26 marbles
Bag 3 = 57 marbles
Bag 4 = 14 marbles
Bag 5 = 29 marbles
Bag 6 = 37 marbles
Bag 7 = 6 marbles
Check:
31+26+57+14+29+37+6=200
200=200
Answer by rapaljer(4671)  (Show Source):
You can put this solution on YOUR website! What you have here is 7 equations and 7 unknowns, unless someone sees a faster way to do this!!
Let the 7 variables be x1, x2, x3, x4, x5, x6, and x7 respectively.
Now, here are the equations:
x1+ x2+ x3+ x4+ x5+ x6+x7=200
x1+x2=57
x2+x3=83
x3+x4=71
x4+x5=43
x5+x6=66
x6+x7=43
If you have a graphing calculator, like a TI85, 86, 83+, or 84, you may have a program that will solve this system called [SIMLT] or perhaps [POLYSMLT].
If you don't have a calculator, then try getting everything in terms of x1. Do this by starting with
x2 =57-x1
Then
x3 = 83-x2
x3=83-(57-x1)
x3=26+x1
Next,
x4= 71-x3
x4=71-(26+x1)
x4=45-x1
x5=43-x4
x5=43-(45-x1)
x5=-2+x1
x6=66-x5
x6=66-(-2+x1)
x6=68-x1
x7=43-x6
x7=43-(68-x1)
x7=-25+x1
Now,
x1+ x2+ x3+ x4+ x5+ x6+x7=200
x1+(57-x1)+(26+x1)+(45-x1)+(-2+x1)+(68-x1)+(-25+x1)=200
x1+169=200
x1=31
x2=26
x3=57
x4=14
x5=29
x6=37
x7=6
That should do it!!!
R^2 at SCC
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