To make 20 kg of aluminum alloy with 70% aluminum, a metallurgist wants to use two metals with 55% and 80% aluminum content. How much of each metal should he use?

Soybean meal contains 16% protein and corn meal contains 9% protein. How many pounds of each should be mixed together to get a 350-lb mixture that is 12% protein?

Total weight equation:
[tex]x + y = 20[/tex]
Total aluminum content equation:
[tex]0.55x + 0.80y = 0.70 \times 20[/tex]
[tex]0.55x + 0.80y = 14[/tex]

To solve this system of equations, you can use substitution or elimination. Let's use substitution:

From the total weight equation, solve for [tex]y[/tex]:
[tex]y = 20 - x[/tex]

Substitute [tex]y[/tex] in the aluminum content equation:
[tex]0.55x + 0.80(20 - x) = 14[/tex]
[tex]0.55x + 16 - 0.80x = 14[/tex]
[tex]-0.25x = -2[/tex]
[tex]x = 8[/tex]

Now substitute [tex]x = 8[/tex] back into the equation for [tex]y[/tex]:
[tex]y = 20 - 8 = 12[/tex]

So, the metallurgist should use 8 kg of 55% aluminum metal and 12 kg of 80% aluminum metal.

Problem 2: Mixture of soybean meal and corn meal.

Let [tex]x[/tex] be the amount of soybean meal used (in pounds).
Let [tex]y[/tex] be the amount of corn meal used (in pounds).

We have the following scenarios:

Total weight equation:
[tex]x + y = 350[/tex]
Total protein content equation:
[tex]0.16x + 0.09y = 0.12 \times 350[/tex]
[tex]0.16x + 0.09y = 42[/tex]

Using substitution:

From the total weight equation, solve for [tex]y[/tex]:
[tex]y = 350 - x[/tex]

Substitute [tex]y[/tex] in the protein content equation:
[tex]0.16x + 0.09(350 - x) = 42[/tex]
[tex]0.16x + 31.5 - 0.09x = 42[/tex]
[tex]0.07x = 10.5[/tex]
[tex]x = 150[/tex]

Now substitute [tex]x = 150[/tex] back into the equation for [tex]y[/tex]:
[tex]y = 350 - 150 = 200[/tex]

So, the mixture should use 150 pounds of soybean meal and 200 pounds of corn meal.

These step-by-step solutions help ensure the correctness of the calculations and the understanding of the underlying mathematical principles.