ne antifreeze solution is $$ 26 \% $$ alcohol and another is $$ 11 \% $$ alcohol. How much of each mixture should be added to make 54 L of a solution that is $$ 21 \% $$ alcohol? Part 1 of 2 L of $$ 26 \% $$ solution should be mixed. Part 2 of 2 L of $$ 11 \% $$ solution should be mixed.

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Let $$ x $$ be the volume (in L) of the $$ 26\% $$ solution and $$ y $$ be the volume (in L) of the $$ 11\% $$ solution. We have the system of equations:

1. Total volume: 
   $$
   x + y = 54
   $$

2. Total alcohol content (since $$ 21\% $$ of 54 L is alcohol):
   $$
   0.26x + 0.11y = 0.21 \times 54
   $$
   $$
   0.26x + 0.11y = 11.34
   $$

Substitute $$ y = 54 - x $$ from the first equation into the second:

$$
0.26x + 0.11(54 - x) = 11.34
$$

Distribute $$ 0.11 $$:

$$
0.26x + 5.94 - 0.11x = 11.34
$$

Combine like terms:

$$
0.15x + 5.94 = 11.34
$$

Subtract $$ 5.94 $$ from both sides:

$$
0.15x = 5.4
$$

Solve for $$ x $$:

$$
x = \frac{5.4}{0.15} = 36
$$

Now, find $$ y $$:

$$
y = 54 - 36 = 18
$$

Thus, the answers are:

- Part 1 of 2:
  $$
36 \text{ L of } 26\% \text{ solution}
  $$

- Part 2 of 2:
  $$
18 \text{ L of } 11\% \text{ solution}
  $$
36 liters of the 26% alcohol solution and 18 liters of the 11% alcohol solution should be mixed to obtain 54 liters of a 21% alcohol solution.

ne antifreeze solution is alcohol and another is alcohol. How much of each mixture should be added to make 54 L of a solution that is alcohol?
Part 1 of 2
L of solution should be mixed.
Part 2 of 2
L of solution should be mixed.

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ne antifreeze solution is alcohol and another is alcohol. How much of each mixture should be added to make 54 L of a solution that is alcohol? Part 1 of 2 L of solution should be mixed. Part 2 of 2 L of solution should be mixed.