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masha68 [24]
3 years ago
5

Two equations are given below:

Mathematics
1 answer:
goldenfox [79]3 years ago
4 0
Since the second equation gives a value for a, we can substitute it into the other equation to find a value for B.

Let's substitute b-2 into the first equation wherever there is an a.

a - 3b = 4
(b-2) - 3b = 4
b - 2 - 3b = 4
-2 - 2b = 4
-2b = 6
b = -3

Now let's find a by substituting -3 into either of the equations to find the value of a.

a = b - 2
a = -3 - 2
a = -5

So your solution set  is (-5, -3)
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A health clinic dietician is planning a meal consisting of three foods whose ingredients are summarized as follows: One Unit ofF
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                             Protein      Carbohydrates    Iron      calories

Food 1  (x₁)               10                       1                   4              80

Food 2 (x₂)               15                       2                  8              120

Food 3 (x₃)               20                       1                  11              100

Requirements         40                        6                 12

From the table we get

Objective Function z :

z  =  80*x₁   +  120*x₂   +   100*x₃     to minimize

Subjet to:

Constraint 1.  at least 40 U of protein

10*x₁  +  15*x₂ + 20*x₃  ≥  40

Constraint 2. at least  6 U of carbohydrates

1*x₁  +  2*x₂  + 1*x₃    ≥  6

Constraint 3.  at least  12 U of Iron

4*x₁   +  8*x₂  +  11*x₃  ≥  12

General constraints:

x₁    ≥  0     x₂   ≥  0    x₃   ≥  0    all integers

With the help of an on-line solver after 6 iterations the optimal solution is:

z (min) = 360      x₁  =  x₃ = 0   x₂ = 3

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3 years ago
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